What is a shape of spherical light wave starting in F from O in moving frame F'. The two different sets of simultaneous events S and S', one starting from O in frame F, and another one starting in S' from moving in F center O' (at the beginning co-positioned in O) seem to me be spherical in F and F', respectively.

However, are these two sets of events (S and S') identical, separable or at least do they have a common part?“A photon in free space spreads out, at long enough range, and the intensity drops.

### Probability theory research papers - academia.edu

” Slutsky theorem is commonly used to prove the consistency of estimators in Econometrics Probability theory and mathematical statistics are difficult subjects both for students to designed by Wolfram Research, and MAPLE, a system for doing mathemat- Sequences of Random Variables and Order Statistics . . 351. 13.1..

The theorem is stated as: For a continuous function g(X k) that is not a function of k, plim g(X k) = g (plim X k) where X k is the sequence of random variables.

Could anyone suggest any literature on how to prove this theorem?“Dear Prof. Hasanov, /profile/Soumitra Mallick the paper Mallick (2015) on History and Markets and similar papers contain a proof of something similar but with lot of properties as assumptions in the necessary part.

Mallick for RHMHM School of Mathematical Sciences Thinking” If I state the following in a paper regarding repeated runs of a Monte Carlo simulation to obtain an overall mean value at a certain confidence interval.

"The simulations were repeated until a 95% Confidence Interval on the mean simulation time at a tolerance of 5% was achieved" Is the above text clear without providing further details or references? Is this considered standard maths, since it's based on work from 1937?Neyman, J. Outline of a theory of statistical estimation based on the classical theory of probability. Suppose we have n samples (x1, x2, …, xn) independently taken from a normal distribution, where known variance 2 and unknown mean . Considering non-informative prior distributions, the posterior distribution of the mean p( /D) follows normal distribution with n and n2, where n is the sample mean of the n samples (i.

, n=(x1+x2+…+xn)/n), n2 is 2/n, and D = x1, x2, …, xn (i.

Let the new data D’ be x1, x2, …, xn, x1new, x2new, …, xknew . That is, we take additional k (k

However, “I am not sure that I am right but lets add something to the discussion. As likelihod is the joint prob (prodct of pdfs)of the sample and also prior/posterior predictive is the prob fnctn of xnew, therefore, to me, the liklihood fntn of x and xnew is the product of likelihood of x and prior/posterior predictive of xnew.

U have the likelihood fnctn u can derive the posterior p(mu/D’). ” At the end of the nineteenth century, many researchers concentrated on various alternative methods based on the theory of infinite series.

These methods have been combined under a single heading which is called \textit Summability Methods . In recent years, these methods have been used in approximation by linear positive operators.

Also, in connection with the concept of statistical convergence and statistical summability, many useful developments have been used in various contexts, for example, approximation theory, probability theory, quantum mechanics, analytic continuation, “I guess I misunderstood the question. The theory of summability has always been important and always will be important so it obviously has a future.

My interpretation of the question was that it was beyond the obvious. Specifically, I interpreted the question to be asking if the theory of summability will become a replacement for other methods of analysis in the future.

I have a problem with that future, as explained in my earlier posts. ” The question of whether Quantum Mechanics is a complete science had sparked a historic debate led by Albert Einstein on one side and Niels Bohr on the other side. It is interesting that quantum physicists from the school of Copenhagen had to resort to philosophical arguments to defend the soundness of quantum mechanics in terms of its ability to faithfully interpret dynamic systems.

The fuzziness of the central notion of the quantum wavefunction seems to have never been resolved to this day, a problem that prompted Richard Feynman in no small part to assert that “Nobody understands “@John Jupe Hello John Jupe. As I understand, your theory is that the limits of knowledge are set by the limits of observation and instrumentation. As opposed to the argument held by Copenhaguists that the limitations in our knowledge of reality are a result of properties innate to reality itself.

In other words, it is a property innate to mother nature that the movement of a bound electron cannot be simultaneously known in terms of its speed AND its position.

I wonder what happens to observation and ensuing knowledge when we ascribe to nature a property that it does not have such as For use in a standard one-term course, in which both discrete and tory where chance experiments can be simulated and the students can get a feeling famous text An Introduction to Probability Theory and Its Applications (New York: We also thank Jessica for her work on the solution manual for the exercises, building..

” moreIf the homogeneity of regression slopes assumption for ANCOVA and levene’s test was violated (significance ), what is the alternative test for ANCOVA? I have post-test mean scores as dependent, pre-test scores as covariates, one independent variables with two treatment modes, and one moderator variables. Is it Welch’s test suitable for ANCOVA?“It might not be the ideal solution, but a simple approach could be to calculate adjusted values of the dependent variable (ie adjusted for the covariates) and then use the Welsh ANOVA to test the null hypothesis of equality of adjusted means.

Not sure how statistically correct this two-step procedure would be, but intuitively seems a reasonable approximation. ” It is well known that for a total qualitative probability order 〈 S, F= (S), ⪋ 〉↦ S, F= (S),P〉Scotts axiom's, in addition to (1), (2), (3)tothe axioms of non negativity :(1)∀ (A i)∈ F: · A ≥ ∅ 2Non triviality S >∅3.

comparability ∀ (A,B)∈ F: : B >A ⊻ A=o B ⊻ A**=B is required, for the the existence of a strong representation finitely additive probabilistic quantitative representation ag , where the order between events is defined over the entire Powers F.**

** There are known counter-examples to de-finetti's axioms, What if any are the benefits, of presuming or deriving that a function:F :F: 0,1 → 0,1 . F being a tandard real valued function of one argument, -not a partial function, that assigns exactly one element of its co-domain, to each elements of its domain.**

** is 'strictly monotone increasing on its domain in contrast to 'monotone increasing (B): (A)∀ (x ∈ 0,1 =Dom(F): x>y ⟺ F(x)>F(y) (B) (x ∈ 0,1 =Dom(F) x>y → F(x)≥F(y);That is, as a tool in deriving that the function F is continuous or linear? Particularly in the case, where a function is already quite homogeneous already? “can only guess that perhaps it has no real benefit here (when dealing with functions of one variable) unless one has not already have F(1)=1 and F(0)=0; apparently one there are some advantages about not having to worry about intricacies about sup, and inf F: at the end -piints, with strict monotone increasing-ness, but I presume that is is only an issue if one is attempting to derive those points already (as opposed to continuity at those points)?It appears that most of the mathematics used to ensure that a function is the identity function and continuous as a side consequence, rely. ” moreThat is, within QM, in the Hilbert inner product metric, what does it mean for the frame function, for two or more events to be explicitly allowed to add as in the representation sense as opposed to presuming it.**

** ? I know it can be deduced, given the appropriate kind of structure). ie a uniqueness property or a representation property quantum probabilism axiom function? I presume that the frame function property is something that was deduced from the axioms of the frame function given the = context free rank (due to additivity of the function on any orthogonal vector and amplitude This is an example in Durrett's book "Probability theory: theory and examples", it's about the coupling time in Markov chains, but I can't see the reason behind it.**

** A writes 100 digits from 0-9 randomly, B choose one of the first 10 numbers and does not tell A.**

** If B has chosen 7,say, he counts 7 places along the list, notes the digits at the location, and continue the process. A possible sequence is underlined in the list:3 4 7 8 2 3 7 5 6 1 6 4 6 5 7 8 3 1 5 3 0 7 9 2 3 . would you please take a look this question (below link) quite similar to above one.**

** The solution is quite difficult for me to understand. https:// /post/power of two choices in randomized load balancing” Is the canonical unit 2 standard probability simplex, the convex hull of the equilateral triangle in the three dimensional Cartesian plane whose vertices are (1,0,0) , (0,1,0) and (0,0,1) in euclidean coordinates, closed under all and only all convex combinations of probability vector that is the set of all non negative triples/vectors of three real numbers that are non negative and sum to 1, ?Do any unit probability vectors, set of three non negative three numbers at each pt, if conceveid as a probability vector space, go missing; for example may not be “As apparently there can be issues when using a function of only two variables, such as with x,y Cartesian coordinates which correspond to this probability simplex in Barry-centric coordinates but have to be absolute Barry-centric coordinates for it to be isomorphic to to set of all convex combinations of probability vectors as the dimensions increase.**

** For example the volume within the solids (as one moves to higher dimensions, the unit standard probability 4 simplex expressed in absoluteAs apparently there can be issues when using a function of only two variables, such as with x,y. **

**” moreif F is strictly montone increasing function F: 0,1 such that F(0)=0 and F(1)=1 and satisfies jensens equation at x=0 and at x=1ie the restricted form where F( 1+x )=F(1)/2+F(x)/2and F(x/2)=F(0)/2+F(x)/2jensen's equality with y=0 using F(0)=0, F(1/2 Purchase Analysis and Probability - 1st Edition. Probability theory is a rapidly expanding field and is used in many areas of science and Guides readers through examples so they can understand and write research papers independently.**

** nF(x) forall non-negative n, (doubling halving,quartering etc)and jensen at one one, F(1)=1, F( 1+x /2=1/2+F(x)/2I presume such functions will be effectively dyadic-ally linear, and given strict monotonic increasing will be linear (or will be if continuity has to be presumed)Iand What the name of this functional(A) equation for F: 0,1 to 0,1 (A) F(1-(x+y)=1-F(x)-F(y) Name of this functional equation? it specifies F(1/3)=1/3and this (B) F-1(1-(x+y)=1-F-1(x)-F-1(y)?where F is strictly monotonic increasingGiven symmetry (C) F(1-x)+F(x)=1 and F-1(y)+F-1(1-y)=1 which also specifies F(1/2) and specifies F(0)=0 if F(1)=1 and conversely it appears to entail cauchys equation, designed for the identityie from (A) F(x)+F(1-x)=1 ie F(x+y)+F(1-(x+y))=1F(x+y)=1-F(1-(x+y))=1-(1-(F(x)+F(y))=F(x)+F(y)+F(x+y) ie F(x y)=then using the equation “with regard to (A) F(1-x-y)=1-F(x)-F(y)and (C) F(1-x)=F(1-x) together it appears that these two just entail F(x+y)=F(x)+F(y) directly with F(1)=1 and thus gives F(x)=x over F 0,1 to 0,1 where by this you mean that the numerical representation is structure preserving with regard to addition. anThe weaker F(1-x-y)=1- F(x)-F(y) where F: 0,1 to 0,1 ie 1-F(1-x-y)=F(x)+F(y)which appear to reduce to it , and to the identity, for on that restricted domain and range, given just one of F(0)=0 or F(1)=1 , ie F(1)=1 givesF(1-x-y)=1- F(x)-F(y)F(1)=F(1-0-0)=1-F(0)-F(0)=1-2F(0)so if F(1)=1.**

** ” moreIn order to get a homogeneous population by inspecting two conditions and filtering the entire population (all possible members) according these two conditions, then used the all remaining filtered members in the research, Is it still population? or it is a sample ( what is called?). working on mathematical equation by adding other part to it then find the solution and applying it on the real world.**

** can we generalize its result to other real world?“Rula -I am not sure I understand your process, but if your 'sample' is really just a census of a special part of your population, then you can get descriptive statistics on it, but you cannot do inference to the entire population from it. You might find the following instructive and entertaining.**

** Ken Brewer's Waksberg Award article: Brewer, K.**

** (2014), “Three controversies in the history of survey sampling,” Survey Methodology,(December 2013/January 2014), Vol 39, No 2, pp. ” moreIs there a distinction between strong or complete qualitative probability orders which are considered to be strong representation or total probability relations neither of which involve in-com parables events, use the stronger form of Scott axiom (not cases of weak, partial or intermediate agreements) and both of whose representation is considered 'strong 'of type (1)P>=Q iff P(x)>= P(y)versus type(2) x=y iff P(x)=P(x)y>x iff P(y)>Pr(x) andy**

** https:// /wiki/Ternary plot#Plotting the pointsDoes it actually have a closed form expression as a function x,y coordinates. I presume mathematica can allows you plot it “for example it must be be at least six copies of the simplex<0. 7) as there must be six copies of each combination of the three values just with their index interchanged, poisition 1, position 2, position 3, and by no mismatch , By the fact that it must contain the verticall/diagonal/orthogonal simplex if i have three vectors as bellowy1+z1+x1=1 F(m,l)=**

** ” moreWhat is general, is meant by the orthogonality relation x⊥y in the functional equation: orthogonal additivity below (1)(1)∀(x,y)∈(dom(F)∩ x⊥y ):F(x+y)=F(x)+F(y)Particularly as used in the following two references listed below (by De Zelah and Ratz, in the context of quantum spin half born rule derivationsWhat is general, is meant by the orthogonality relation x⊥y denotes x is orthogonal to y. (1)∀(x,y)∈(dom(F)∩ x⊥y ):F(x+y)=F(x)+F(y) See R tz, J rg, On orthogonally additive mappings, Aequationes Math. See 'Comment on `Gleason-type “for isntance is 0 , for orthogonality when the events are on geometrically orthogonal basis vectors,(90 degree difference roughly) maximally non commutting bases, and 1 when they are on the same vector; are these both considered a form of orthogonality” What exactly are order embeddings.**

** ?Of the form x<=y iff F(x)<=F(y)?and surjective, is the it a function from a partially ordered set to a numerical domain and to then a second function from that numerical domain to co=domain. ?https:// /wiki/Order embeddingThus when they are order isomorphism which part is isomorphic, the function from the numerical domain to the numerical range, from the order to the numerical domain, or all the way from the qualitative order to the numerical range?I presume there is an always an intermediate function to the domain first? “hat is not into, something that does not satisfy the function requirement, in the usual sense,'in the usual sense' x=y then F(x)=F(y)by not ' from into' or 'onto'; or something like a galois connection or a strict partial order, fails in both injective monotone map, as in an order embeddingin the context of a partial order,which if x<=y and x>=y then x=y (anit symmetry(anby monotonicity and which gives F(x)<=F(y) and F(x)>=F(y), F(x)=F(y)As even an order embeddingx<=y iff F(x)<=F(y)llike an injective monotone (embedding map) .**

** ” moreIs there a distinction between strong representations and the unique-ness properties that strong representations of probability representations of (1)type P\leq Q iff Pr(Q)\leq PR(M) often under the name total preorder or weak order or strict weak order.**

** (2) with the completeness comparability relation taken to mean typeAleq B or B leq A; where incomparables are taken not to existand type 2>which generally goes under the name strict total order or strict trichotomous weak order, whose representation is taken at (1) where =|<|>and completeness property(1)A>B iff A>B and x=y “I keep telling you that equality is equality! It is a fundamental notion. When one has a relation written "<=" then its axioms are given separately from the axioms relating to equality and antisymmetry is one of those axioms (of the partial order).**

** Given that, if you define "<" as a a<=b /\ a!=b, then it follows that a<=b<->aaa<=b /\ a!=b.**

If you really were so nuts as to start with axioms. ” morepresume that midpoint convex F: 0,1 to 0,1 is monotone increasing (not necessarily strictly ) F(0)=0 ,F(0.

5 F(1)=1 ,As its monotone increasing and midpoint convex with F(0)=0 F(1)=1 it will be continuous and convex on 0,1) by lebesgue measurabilityHowever if I note that if midpoint convex F is rationally convex on closed interval one can use F(0.

75 following Is the ortho-complement of a proposition, in quantum logic/probability or hilbert space, the logical comple-ment of a propositionsuch as 'spin up at direction x and 'spin down in direction y'(mutually exhaustive, on the) same prepared spin system at the as same angle, ie on the same basis) and thus one can add them the probabilities to one, as in the usual probabilistic sense to View Probability Theory Research Papers on Academia.edu for free. Average case complexity, in order to be a useful and reliable measure, has to be robust..

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

Or is this a geometric notion that that allows one to add probabilities of events together that lie on distinct “presume that he did not merely say that we are presuming Given a normalization of the frame function, presumed to some such probability function, that on any specific bases,(1)where any given amplitudes moduli squared sum to one, the frame function values do,three which lie on the very same basis sum to 1, as they are mutually exhaustive and exclusive; when the amplitudes mod squared do). A\cap,B=emptyset,\capC ,=emptyset A, B C, are mutually exclusive and exhaustive \Omega=A\vee B\vee C0, so P(A)+PR(B)+PR(C)=1(2) and when he applied that the function is odd F(-x)+F(x)=0, or any given.

” moreThere are traditional ways of getting the cut-off value that maximize the sensitivity and specificity of a given test from the ROC curve. But I think those ways posses some arbitrariness in doing the job.

How about this;Can we use Bayes formula to calculate the value of the variable which equates the probability of the disease to the probability of the control, assuming that both classes are normally distributed? This should be the ideal cut-off, without considering any factors related to the cost,,,etc that can affect the conventional method. So from a set of observations, we can Is the following function F: 0,1 to 0,1 F strictly monotonic increasing F(1)=1,(i presume this unnecessary as its specified by the first two(1)ie x+y=1 if and only iof F(x)+F(y)=1 F(x)+F(1-x)=1; F-1(x)+F-1(1-x)=1(2)F homomorphic(2) x+y+z=1 if and only F(x)+F(y)+F(m)= 1(3) x+y+z+m=1; if and only ifF(x)+F(y)+F(z)+F(M)=1Give F(x)=x and F continuous (it appears to entail cauchys equation with F(1)=1 over ethe unit triangle due tot he common term).

F(x+y)+F(z)=1, F(x)+F(y)+F(z)=1, etc, F(x+y)=1-F(z)=F(x)+F(y). ILet F1,F2 be two increasing functions on a,b and F1(x)=F2(x) on a dense subset of a,b and F2 is continuous.

?In my case F: 0,1 to 0,1 and F agrees with the identity function for all rationals, and F(1)=1=G(1)=1 and G(0)=F(0)=0; Does one have to anything to show continuity of F at 1 and zero, given that its now identical to the identity on (0,1) and just continuous (0,1) one, where F is strictly monotonic increasing and has the same end points with the same values.

Wont its limiting values just be G's? as one never gets outside of (0,1) on “Between x and y arbitrary there is a rational. By the Archimedean axiom, there is an N such that N(y-x)>1.

ThereforeNy - Nx >1Therefore there is an integer M such thatNx< M< Nyand hencex< M/N< ywhich was to be proved. Cardinality, in such issues, is a bit of a red herring, as shown above,” With regard to Jensens equality (before continuity is applied) where F(0)=0 and (and F(1)=1 if need be, which is not required );Jensen's equation beingF(x+y)/2=F(x)/2 +F(y)/2Is this literally equivalent to cauchy's functional equation F(x+y)=F(x)+F(y) over all rationals in the non quite--continuous case?,That is continuity and linearity immediate for a jensen's function that satisfies this(1)F: 0,1 to 0,1 (2)F(0)=0 F(1)=1(3)\forall(x,y)\in dom(F):F(x/2+y/2)=F(x)/2 +F(y)/2That is, one does not have to assume anything, further such mono-tonicity, bounded on a set of https:// /questions/2265396/can-this-strictly-increasing-convex-function-f-meet-its-linear-line-segment-in three places.

See my postIs this correct that this result or intuition would follow even if the convex function neither of the two differentiability constraints. But only F: 0,1 to 0,1 1F is Convex and strictly monotonically increasing2.

I presume that bijectivity will follow via surjectivity resulting the continuity of convexity, given the min and max points, F(0)=0 F(1)=1 and F strictly increasing, INjectivity being “I hope its not true. It would render my symmetry equations (almost) obsolete, and I DO NOT want to get rid of them, or have them rendered redundant .

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>

**a a<=b /\ a!=b.**

If you really were so nuts as to start with axioms. ” morepresume that midpoint convex F: 0,1 to 0,1 is monotone increasing (not necessarily strictly ) F(0)=0 ,F(0.

5 F(1)=1 ,As its monotone increasing and midpoint convex with F(0)=0 F(1)=1 it will be continuous and convex on 0,1) by lebesgue measurabilityHowever if I note that if midpoint convex F is rationally convex on closed interval one can use F(0.

75 following Is the ortho-complement of a proposition, in quantum logic/probability or hilbert space, the logical comple-ment of a propositionsuch as 'spin up at direction x and 'spin down in direction y'(mutually exhaustive, on the) same prepared spin system at the as same angle, ie on the same basis) and thus one can add them the probabilities to one, as in the usual probabilistic sense to View Probability Theory Research Papers on Academia.edu for free. Average case complexity, in order to be a useful and reliable measure, has to be robust..

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

Or is this a geometric notion that that allows one to add probabilities of events together that lie on distinct “presume that he did not merely say that we are presuming Given a normalization of the frame function, presumed to some such probability function, that on any specific bases,(1)where any given amplitudes moduli squared sum to one, the frame function values do,three which lie on the very same basis sum to 1, as they are mutually exhaustive and exclusive; when the amplitudes mod squared do). A\cap,B=emptyset,\capC ,=emptyset A, B C, are mutually exclusive and exhaustive \Omega=A\vee B\vee C0, so P(A)+PR(B)+PR(C)=1(2) and when he applied that the function is odd F(-x)+F(x)=0, or any given.

” moreThere are traditional ways of getting the cut-off value that maximize the sensitivity and specificity of a given test from the ROC curve. But I think those ways posses some arbitrariness in doing the job.

How about this;Can we use Bayes formula to calculate the value of the variable which equates the probability of the disease to the probability of the control, assuming that both classes are normally distributed? This should be the ideal cut-off, without considering any factors related to the cost,,,etc that can affect the conventional method. So from a set of observations, we can Is the following function F: 0,1 to 0,1 F strictly monotonic increasing F(1)=1,(i presume this unnecessary as its specified by the first two(1)ie x+y=1 if and only iof F(x)+F(y)=1 F(x)+F(1-x)=1; F-1(x)+F-1(1-x)=1(2)F homomorphic(2) x+y+z=1 if and only F(x)+F(y)+F(m)= 1(3) x+y+z+m=1; if and only ifF(x)+F(y)+F(z)+F(M)=1Give F(x)=x and F continuous (it appears to entail cauchys equation with F(1)=1 over ethe unit triangle due tot he common term).

F(x+y)+F(z)=1, F(x)+F(y)+F(z)=1, etc, F(x+y)=1-F(z)=F(x)+F(y). ILet F1,F2 be two increasing functions on a,b and F1(x)=F2(x) on a dense subset of a,b and F2 is continuous.

?In my case F: 0,1 to 0,1 and F agrees with the identity function for all rationals, and F(1)=1=G(1)=1 and G(0)=F(0)=0; Does one have to anything to show continuity of F at 1 and zero, given that its now identical to the identity on (0,1) and just continuous (0,1) one, where F is strictly monotonic increasing and has the same end points with the same values.

Wont its limiting values just be G's? as one never gets outside of (0,1) on “Between x and y arbitrary there is a rational. By the Archimedean axiom, there is an N such that N(y-x)>1.

ThereforeNy - Nx >1Therefore there is an integer M such thatNx< M< Nyand hencex< M/N< ywhich was to be proved. Cardinality, in such issues, is a bit of a red herring, as shown above,” With regard to Jensens equality (before continuity is applied) where F(0)=0 and (and F(1)=1 if need be, which is not required );Jensen's equation beingF(x+y)/2=F(x)/2 +F(y)/2Is this literally equivalent to cauchy's functional equation F(x+y)=F(x)+F(y) over all rationals in the non quite--continuous case?,That is continuity and linearity immediate for a jensen's function that satisfies this(1)F: 0,1 to 0,1 (2)F(0)=0 F(1)=1(3)\forall(x,y)\in dom(F):F(x/2+y/2)=F(x)/2 +F(y)/2That is, one does not have to assume anything, further such mono-tonicity, bounded on a set of https:// /questions/2265396/can-this-strictly-increasing-convex-function-f-meet-its-linear-line-segment-in three places.

See my postIs this correct that this result or intuition would follow even if the convex function neither of the two differentiability constraints. But only F: 0,1 to 0,1 1F is Convex and strictly monotonically increasing2.

I presume that bijectivity will follow via surjectivity resulting the continuity of convexity, given the min and max points, F(0)=0 F(1)=1 and F strictly increasing, INjectivity being “I hope its not true. It would render my symmetry equations (almost) obsolete, and I DO NOT want to get rid of them, or have them rendered redundant .

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> ” moreThere is a convex program with the convex constraint:sup V \in \math V Tr(QV)+\sum i=1 d sup P i \in \math P i Integral of max p i*Ksi i

**a a<=b /\ a!=b.**

If you really were so nuts as to start with axioms. ” morepresume that midpoint convex F: 0,1 to 0,1 is monotone increasing (not necessarily strictly ) F(0)=0 ,F(0.

5 F(1)=1 ,As its monotone increasing and midpoint convex with F(0)=0 F(1)=1 it will be continuous and convex on 0,1) by lebesgue measurabilityHowever if I note that if midpoint convex F is rationally convex on closed interval one can use F(0.

75 following Is the ortho-complement of a proposition, in quantum logic/probability or hilbert space, the logical comple-ment of a propositionsuch as 'spin up at direction x and 'spin down in direction y'(mutually exhaustive, on the) same prepared spin system at the as same angle, ie on the same basis) and thus one can add them the probabilities to one, as in the usual probabilistic sense to View Probability Theory Research Papers on Academia.edu for free. Average case complexity, in order to be a useful and reliable measure, has to be robust..

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

Or is this a geometric notion that that allows one to add probabilities of events together that lie on distinct “presume that he did not merely say that we are presuming Given a normalization of the frame function, presumed to some such probability function, that on any specific bases,(1)where any given amplitudes moduli squared sum to one, the frame function values do,three which lie on the very same basis sum to 1, as they are mutually exhaustive and exclusive; when the amplitudes mod squared do). A\cap,B=emptyset,\capC ,=emptyset A, B C, are mutually exclusive and exhaustive \Omega=A\vee B\vee C0, so P(A)+PR(B)+PR(C)=1(2) and when he applied that the function is odd F(-x)+F(x)=0, or any given.

” moreThere are traditional ways of getting the cut-off value that maximize the sensitivity and specificity of a given test from the ROC curve. But I think those ways posses some arbitrariness in doing the job.

How about this;Can we use Bayes formula to calculate the value of the variable which equates the probability of the disease to the probability of the control, assuming that both classes are normally distributed? This should be the ideal cut-off, without considering any factors related to the cost,,,etc that can affect the conventional method. So from a set of observations, we can Is the following function F: 0,1 to 0,1 F strictly monotonic increasing F(1)=1,(i presume this unnecessary as its specified by the first two(1)ie x+y=1 if and only iof F(x)+F(y)=1 F(x)+F(1-x)=1; F-1(x)+F-1(1-x)=1(2)F homomorphic(2) x+y+z=1 if and only F(x)+F(y)+F(m)= 1(3) x+y+z+m=1; if and only ifF(x)+F(y)+F(z)+F(M)=1Give F(x)=x and F continuous (it appears to entail cauchys equation with F(1)=1 over ethe unit triangle due tot he common term).

F(x+y)+F(z)=1, F(x)+F(y)+F(z)=1, etc, F(x+y)=1-F(z)=F(x)+F(y). ILet F1,F2 be two increasing functions on a,b and F1(x)=F2(x) on a dense subset of a,b and F2 is continuous.

?In my case F: 0,1 to 0,1 and F agrees with the identity function for all rationals, and F(1)=1=G(1)=1 and G(0)=F(0)=0; Does one have to anything to show continuity of F at 1 and zero, given that its now identical to the identity on (0,1) and just continuous (0,1) one, where F is strictly monotonic increasing and has the same end points with the same values.

Wont its limiting values just be G's? as one never gets outside of (0,1) on “Between x and y arbitrary there is a rational. By the Archimedean axiom, there is an N such that N(y-x)>1.

ThereforeNy - Nx >1Therefore there is an integer M such thatNx< M< Nyand hencex< M/N< ywhich was to be proved. Cardinality, in such issues, is a bit of a red herring, as shown above,” With regard to Jensens equality (before continuity is applied) where F(0)=0 and (and F(1)=1 if need be, which is not required );Jensen's equation beingF(x+y)/2=F(x)/2 +F(y)/2Is this literally equivalent to cauchy's functional equation F(x+y)=F(x)+F(y) over all rationals in the non quite--continuous case?,That is continuity and linearity immediate for a jensen's function that satisfies this(1)F: 0,1 to 0,1 (2)F(0)=0 F(1)=1(3)\forall(x,y)\in dom(F):F(x/2+y/2)=F(x)/2 +F(y)/2That is, one does not have to assume anything, further such mono-tonicity, bounded on a set of https:// /questions/2265396/can-this-strictly-increasing-convex-function-f-meet-its-linear-line-segment-in three places.

See my postIs this correct that this result or intuition would follow even if the convex function neither of the two differentiability constraints. But only F: 0,1 to 0,1 1F is Convex and strictly monotonically increasing2.

I presume that bijectivity will follow via surjectivity resulting the continuity of convexity, given the min and max points, F(0)=0 F(1)=1 and F strictly increasing, INjectivity being “I hope its not true. It would render my symmetry equations (almost) obsolete, and I DO NOT want to get rid of them, or have them rendered redundant .

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> 2 + 2q i' *Ksi i + r i' dP i <= epsilon*r,where Q, p i, p i', q i, q i', r i, r i', r are variables.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> \math V is a convex closed subset of a positive semidefinite cone\math P i is a family of possible probability distributions of a random variable Ksi iP i is a probability distribution of a random variable Ksi iThey deprive of the information on the methods of the solution of the problem: "This approximation “Thank you Pavel!I met this program while reading the info on convex programming and I was interested in general methods of building the effective solutions. I see from your remark that there are these methods.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> The problem that we have no specialists on convex programming in our institute, this forum is the way to get info from outside. The problems can be really hard just to notice, without advise it can be difficult to get through. ” Is there a general name for these 'pair' of symmetic probability functional equations (1) and (2); see https:// /posts/2239756/edit1.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> (-1)(1)=F(1);See https:// /wiki/Minkowski%27s question mark function, the second equation that is highlighted in the third attachment below. Its that property as bi-conditional or in other words, it applies to the inverse function as well.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> What is the official (if at all ) name of this functional equation.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> and for See attached document page 14;the equation expressed by sublinear functions when convex. Likwise am i correct taht 2F(x)=f(2x) is not usually a property of midpoint convexity or full convexity, even if F(0)=0, F(1)=1, F(1/2) unless other properties such as symmetry or sub-linearity are imposed (which I have noticed)Am i correct in presuming that by convex that mean just mid point convex. At the same time sub-linear functions appear to be asserting something very much closer to a homogeneity condition that cauchy equations do not have are midpoint convex functions F(x/2+y/2)<=F(x)/2+F(y/2);rationally super-additive with F(0)=0;?ie do they satisfy F(x+y)>=F(x)+F(y); for pairs of points or rational numbers (or something weaker)(and which are injective, with F(1)=1 if need be, ).

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> ie in the same way that convex functions gives rise to a form of super-addivitity with F(0)=0; where convexity, just the continuous and measurable form of midpoint convexity in some sense/ Just as jensen's equality F(x/2+y/2)=F(x)/2+F(y)/2with F(0)=0 is additive; gives rise to cauchy equation or a form of it F(x)+F(y)=F(x+y) before “@Adolf MirotinI note that F((1-x )+ (1-y))>=F(1-x) +F(1-y); I wonder whether letting y=t+1, and x =m+1, is an allowable substitution to infer super-additivityI guess the real question is whether a midpoint convex function with f(0)=0 and f(1), F: 0,1 to 0,1 . f injective and increasing ,is sufficient in and of itself, for rational/dyadic super-additivity (inequality rational homogeneious) F(sigmax)>sigmaF(x) where sigma>1 and sigma rational, or just an integer” I attempted to explain in convoluted terms, a question I had before.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> The questions concerns, whether, there the cauchy field equation1. F: 0,1 to 0,1 , and F(1)=1;max element of the literally just entail that F(x)=x,Without further specifying the that F is monotoncally stricly increasing of continuous; Some claim that they just entail that the function just is continuous, but surely this must be just symptom of the real number system.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> Just using 1,2,3 can one actually derive that F(1/pi)=1/pi, as one can for rational number using 1, 3, “Monotonicity clearly follows from additivity if you can proveF(x)>0 for all x>0. But this is not hard: if x>0, then there exists y such thatx=y*yF(x) = F(y*y) = F(y)*F(y) >0.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> So monotonicity follows from the basic properties. Further, elementary algebra shows that for all rationals rF(r) = rwhich together with monotonicity yields the result.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> Note that this reasoning fails in the complex, and that discontinuous automorphisms do exist in that case. ” Does anyone know the general continuous functional form of these constraints.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> Given, Midpoint convexity, it appears to satisfy jensen equality; symmetry appears to give one the other half. I am not sure what has to be satisfied in order to give the same results as I had to use F(0. 5 at the very least, and F(0)=0 (and presumably F(1)=1. F: 0,1 \to 0,1 ; both closed and bounded intervals1.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> Strictly and monotonely increasing; F(x1)>F(x2) iff x1>x2, F(x1)=F(x2)\to x1=x22. 5, else “I have a somewhat more formal proof F(2x)=2F(x) as in 5 seems to be entailed and can be extended to F(4x)=F(x)+F(x)+F(x)+F(x)=4(F(x). Numerically, I can directly get all power of 3,5 and maybe all rationals to a degree (and otherwise bounded between the values of F(x) above and below.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> I am not sure what the precise form of jensen equation,is that must be satisifed can the presumption of the max point F(1)=1, F(0)=0 and F(0. One can infer that if given the inequalitys;for all pairs of pointsF(X/2+Y/2)<=F(X)/2+F(Y)/2 and F(1-X)/2+F(1-Y)/2<=F(1-X)/2+F(1-Y)/2 ; collectively entail. ” moreI am wondering whether there are some weaker forms of midpoint convexity; in my system it appears that jensens equation, as far as I can see, will just fall out; if not cauchy equation or something close to it, for all rational numbers at least;Given symmetry in an strictly monotonically increasing and continuous strict quasi convex, strictly quasi concave function over a total order on events in differences;Just given use of ofmidpoint convexityon half the structure; and using equimatched pairs and symmetry to deliver the other half (if is there such a name for it, of “This is four the two outcome case; nothing needs to be really imposed in the entangled structure, given just three outcome normalization, and the abillity to connect everything through equalities.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>I presume that these are dealt with via point reflection, continuity, or imposed a multiplicative structure (or swapping strict) Providing cutting-edge perspectives and real-world insights into the greater utility of probability and its applications, the Handbook of Probability offers..

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>

**a****a<=b /\ a!=b.**

**a**

**a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>Probability theory | alexander a. borovkov | springer

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> ” moreI have seen this condition being used for midpoint convexity: Midpoint convexity. A set C is midpoint convex if whenever two points a,b are in C, the average or midpoint (a + b) is in c.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> ; I have seen this in other documents, such as that if f(x) and f(y) are defined, then there exists some F(c), such that F(c)=F(x)/2+F(y)/2; such that perhaps x

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> where, this just seems to be an existence claim and not a richness claim such as the traditional defintion which “Unless I am mis-using it; it seems implausibly strong a constraint; I can, it appear get all the way from jensens equality (if not something stronger, ie reading the numbers even factors of 3 off the table; and essentially something akin to linearity before i even use continuity). It appears that my symmetry condition (and only half of it, effectively entails outright jensens equality, on the whole structure given the first half- jensens equality, and even then, when only applied to half (less than half of the structure)” Hello! I am only starting to study GRP (General Renewal Process), so sorry, if my question will be too simple -:).

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> Main expression for GRP is A i - A i-1 = q*(S i - S i-1 ), where A is Virtual Age, S is Real Time and q is restoration factor. It is clear (???), that this formula also proved for non-constant restoration factor q i – e.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> , q i = q CM for Corrective Maintenance and q i = q PM for Preventive Maintenance.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> But I see some collision, if sometimes q i = 0 (e. On the one hand, q i = 0 means, that it is replacement (Good as New) and in this case the Aging Does quiggin's and chew logarithmic function satisfy the properties in my question, earlier.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> See page 59 and 60 inSee page 59 and 60 of Quiggin's, rank dependent utility' GENERALIZED EXPECTED UTILITY THEORYTHE RANK-DEPENDENT MODEL' 1993.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> SEe /search?query=quiggin+generalized&facet-content-type=%22Book%22t appears to be strictly increasing and onto; but is logarithmic and apparently has f(p) +f(1-p)=1 for all; strictly increasing and uniformly continuous sur-jective(bijective?) 0,1 to 0,1 with f(0. 5 f(1)=1, and f(0)=0;Is The only uniformly (equi-continuous, at least) functions that satisfy these constraints?These are probability functions, roughly such that are,which are1. The domains equal the range (endo-morphic), domains and ranges are both 0,1 for f and G, for both such thatThe only uniformly (equi-continuous, at least) functions that satisfy these constraints?F(x);x\in 0,1 etc, such that F: 0,1 \to 0,1 .

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> G: 0,1 \to 0,1 y\in 0,1 Each x,y, pair (such that x+y=1) denote a distinct hypothetical chance set up that is counter-factually/modally linked with all of the other pairs; These rise to “Likewise for any x1 and x2 where x and x2>are such that x1>x2, x2>x3; f(x1+x2)>f(x1+x2) and likewise for G and any combination of the two. Inaddition, the function are bounded sums are bounded by the equalities f(0.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> 3) in that any x1,x2,,y1,y,2 that are in between them F(0.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> 34) or G or any combinaton of the two of them must be smaller then first and larger then the second, and I think this may hold even if is not the case that both of x1, x2 >0.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> 4 There are two values of any chance, one A and one B, ie two 25 percent chances, G(0.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> ” moreDear all,I measured a variable that takes values between 0 and 0.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> This variable will be used in a regression analysis, but it has values of skewness and kurtosis of 3. 3, respectively, hence requiring a transformation in order to reduce those values. However, in this way, the resulting values of the variable will be negative, and I would avoid this.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> Another option is multiplying all values for 1,000 and then use a log transformation.

**a****a<=b /\ a!=b.**

**a**

**a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>Statistics and probability - great college essay - journey mexico

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>They are only the descriptive outcomes to describe your continuous data, and you should act by taking them into account Get probability homework help from expert online tutors at, available 24/7. Advanced Probability Theory Stats Homework, assignment and Project Help, Advanced ocr critical thinking f501 mark scheme case study planter type paper on..

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> We just published a new article, which can describe any skewed data, with new mean and deviation approaches.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> 34” Does anybody know of any examples in the literature; or otherwise whether it is possible to prove that a certain probabilistic logic satisfies the qualitative constraints given by Luce 1967 (the annals of mathematics 38, 780 -786), which are sufficient for unique strong numerical probabilistic representation, without making some equi-probability assumption or going through some other route. I ask this because whilst Luce's theorem does not appear to make an equi-probability assumption, such as the presumption of finitely many equi-probable atoms (koopman, “I also had some questions as to the some of the assertions of fine 1973 ,'unique to and within a positive multiplicative factors; Is this talking about the uniqueness or homegeneity of the representation (the similarity group of transformations); or the probability measure. I was always under the impression insofar as the probability is concerned, that it just unique (given the probabilistic representing structure).

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> He says something similar about Savage axioms of tightness, fine-ness in the paragraph above where he mentions Luce L (i thought that Savage's utilities were on an interval. ” moreAre there sets of three, or rather for any N>3 sets of N sur-jective uniformly continuous functions, for all N>3, where n denotes the number of function in the sets, such that each function in the set has the same domain 0,1 and the same range 0,1 ,such that these functions, in a given set, sum to one on all points in the domain 0,1 .

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> Moro-ever, Are there arbitrarily many such functions for all n>=3. By non trivial, I mean, -non linear, (and presumably not quadratic) functions which just so happen to be that that their “Dear David.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> I am using transcendtal functions at the moment and they appear to do the job without having to changing the functions when i decide the to weight the functions differently which is the most important issue. It would be interesting to see, how well other non-transcendental functions of this sort fare. Thanks for this; I just realized that in some sense, the name sur-jective parittion of unity does not in itself make a great deal of sense; as it may imply that each number in the domain in each fi is mapped to the entire unit interval.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> ” moreI was wondering if there are any set of n; n>=3continuous or somewhat smooth functions (certain polynomials), all of which have the same domain 0,1 f

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> n(min,max): 0,1 \to min,max ; where max>min, min, max\in 0,1 that have these properties1. n(v)=1; for all values in the domain 0,1 , the sum of the 'function values' =1at all points in this domain, and for all possible min and max values of the ranges of the n functions2. The n functions have range continuously between min,max , max>=min, both in 0,1 ; and will all reach their maximum at “I agree peter, I apologize.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> Nonetheless, However, i must insist,logic entails that one can leave A>T out of the conjunction; you already contradicted yourself (you already said 'can' ,NOT 'must'). What is necessarily possible, is not 'necessarily, necessarily necessary, and that which is entailed or necessary, are only statements which are Necessarily NecessarilySecondly, I never said that I did not originally say it max>=min, when I later said A>B that was a correction.

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p> ” moreSee attached this lemma 3 and the last part of the proof above lemma 3, which is the inequality 'which suffices for this claim' which then proceeds to lemma 3; in this inequality methods, where one shows that both the rhs and lhs approximately equal each other, that the middle term must also, have a name; some kind of mean approximation theorem; intermediate asymptotic value theorem?If one can partition a countably infinite sequence,A of numbers into two (countably infinite) sub-sequences B1, B2 such A =B1\cupB2;and the limiting relative frequency(a \in B1) = x, & the limiting

**a a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

**a****a<=b /\ a!=b.**when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>«

**a****a<=b /\ a!=b.**when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>1

**a****a<=b /\ a!=b.**when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>2

**a****a<=b /\ a!=b.**when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>3

**a****a<=b /\ a!=b.**when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>4

**a****a<=b /\ a!=b.**when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>5

**a****a<=b /\ a!=b.**when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>6

**a****a<=b /\ a!=b.**when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>7

**a****a<=b /\ a!=b.**when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>8

**a****a<=b /\ a!=b.**when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>9

**a****a<=b /\ a!=b.**when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>10

**a****a<=b /\ a!=b.**when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

They get to me to jensen's equation and I suppose to continuity as well (that part wont be rendered obsolete)But I don't want the above result; F(x)=x, to be true for any strictly monotonically increasing and continuous midpoint convex function F: 0,1 to 0,1 where F(0)=0 and F(0. Rather it should have to satisfy something else in addition such as symmetry in my case, to get to F(x)=x" in my case,.< p>»

**a****a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

**a****a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).

**a****a<=b /\ a!=b.**

when the hilbert space inner product is zero (or rather one or zero, the kronecker delta).