Jensen inequality probability
WebMay 16, 2024 · Relative entropy is a well-known asymmetric and unbounded divergence measure [], whereas the Jensen-Shannon divergence [19,20] (a.k.a. the capacitory discrimination []) is a bounded symmetrization of relative entropy, which does not require the pair of probability measures to have matching supports.It has the pleasing property that … WebJun 21, 2024 · with the inequality becoming strict if convexity is strict, unless \(X\) is a constant with probability 1. The inequality is reversed if \(g\) is concave. Last edited: …
Jensen inequality probability
Did you know?
WebMIT OpenCourseWare is a web based publication of virtually all MIT course content. OCW is open and available to the world and is a permanent MIT activity WebOct 25, 2024 · Wikipedia lists Jensen's inequality as ϕ ( E [ X]) ≤ E [ ϕ ( X)] for a convex function ϕ. For ϕ ( x) = x ln x that expands to E [ X] ln E [ X] ≤ E [ X ln X]. That seems opposite in sign from what you and the original question have. Are you sure a typo in the question didn't lead to a sign error in the answer? – olooney Oct 25, 2024 at 16:09 2
Web6 Probability & Statistics with Applications to Computing 6.3 Theorem 6.3.2: Jensen’s Inequality Let Xbe any random variable, and g: Rn!R be a convex function. Then, g(E[X]) E[g(X)] Proof of Jensen’s Inequality. We will only prove it in the case Xis a discrete random variable (not a random vector), and with nite range (not countably in nite). WebTutorial 8: Jensen inequality 7 1. Show that for all x ∈ K,thereareopensetsV x,W x in Ω, such that y ∈ V x,x∈ W x and V x ∩W x = ∅. 2. Show that there exists a finite subset {x1,...,x n} …
The classical form of Jensen's inequality involves several numbers and weights. The inequality can be stated quite generally using either the language of measure theory or (equivalently) probability. In the probabilistic setting, the inequality can be further generalized to its full strength. Finite form For a real convex … See more In mathematics, Jensen's inequality, named after the Danish mathematician Johan Jensen, relates the value of a convex function of an integral to the integral of the convex function. It was proved by Jensen in 1906, … See more Form involving a probability density function Suppose Ω is a measurable subset of the real line and f(x) is a non-negative function such that $${\displaystyle \int _{-\infty }^{\infty }f(x)\,dx=1.}$$ See more • Jensen's Operator Inequality of Hansen and Pedersen. • "Jensen inequality", Encyclopedia of Mathematics, EMS Press, 2001 [1994] See more Jensen's inequality can be proved in several ways, and three different proofs corresponding to the different statements above will be offered. Before embarking on these mathematical derivations, however, it is worth analyzing an intuitive graphical argument … See more • Karamata's inequality for a more general inequality • Popoviciu's inequality • Law of averages See more WebJensen's Inequality (with probability one) Asked 9 years, 5 months ago Modified 5 years, 10 months ago Viewed 7k times 10 In the following theorem, I have a problem about the …
WebA consequence is the arithmetic geometric mean inequality: Proposition 7. For positive x 1;:::;x n, x 1+x 2+:::+xn n n p x 1 x 2:::x n. Proof Let Y be a random variable taking the value logx i with probability 1=n. Then the left hand side is E 2Y and the right hand side is 2E[Y ]. The inequality follows from the convexity of exponentiation.
WebJensen's inequality is an inequality involving convexity of a function. We first make the following definitions: A function is convex on an interval I I if the segment between any … hda2100hww hotpoint dishwasherWebJensen's inequality states that for a convex function f, the expectation of that function is greater than or equal to the function of the expectation. In our case, this means that Df(PQ) = E[f(p/q)] ≥ f(E[p/q]) Since the expectation of p/q is equal to 1 for any probability distribution P and Q, we have Df(PQ) ≥ f(1) = 0 Equality holds if ... golden city terryville ctWebDec 24, 2024 · STA 711 Week 5 R L Wolpert Theorem 1 (Jensen’s Inequality) Let ϕ be a convex function on R and let X ∈ L1 be integrable. Then ϕ E[X]≤ E ϕ(X) One proof with a nice geometric feel relies on finding a tangent line to the graph of ϕ at the point µ = E[X].To start, note by convexity that for any a < b < c, ϕ(b) lies below the value at x = b of the linear … golden city tenerifeWeb2.1.2 The Inequality. Jensen’s Inequality (JI) states that, for a convex function \(g\) ... Since any probability is bounded between 0 and 1, and variance must be greater than or equal to … golden city timber \u0026 hardware pty ltdWebOur first bound is perhaps the most basic of all probability inequalities, and it is known as Markov’s inequality. Given its basic-ness, it is perhaps unsurprising that its proof is essentially only one line. Proposition 1 (Markov’s inequality). LetZ ≥ 0 beanon-negativerandom variable. Thenforallt ≥ 0, P(Z ≥ t) ≤ E[Z] t. hd a1Web(1) the Jensen inequality: Suppose ψ(·) is a convexfunction and Xand ψ(X) havefinite expectation. Then ψ(E(X)) ≤ E(ψ(X)). Proof. Convexity implies for every a, there exists a … golden city threehttp://www.probability.net/jensen.pdf hda3440 hotpoint dishwasher