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This is a slightly modified question from Sheldon Ross 9th ed Assume all RVs are discrete I am asked to prove the following equality $$\mathbb{E}Xg(Y)Y = g(Y)Nov 03, 10 · Let G = (X, Y;G(X) E h(Y) Notes 1 E(XY) = E(X)E(Y) is ONLY generally true if X and Y are INDEPENDENT 2 If X and Y are independent, then E(XY) = E(X)E(Y) However, the converse is not generally true it is possible for E(XY) = E(X)E(Y) even though X and Y are dependent Probability as an Expectation
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How to solve for x or y-Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack ExchangeCheck out the Great Smoky Mountain Railroad steam train in North Carolina It's a fun family getaway on a historical steam train through the



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80 Example Let X be a discrete random variable with PGF GX(s) = s 5 (2 3s2) Find the distribution of X GX(s) = 2 5 s 3 5 s3 G X(0) = P(X = 0) = 0 G′ X(s) = 2 5 9 5 s2 G′ X(0) = P(X = 1) = 2 5 G′′ X(s) = 18 5 s 1 2 G′′ X(0) = P(X = 2) = 0 G′′′ X(s) = 18 5 1 3!Apr 02, 11 · (this is g^1(g(x)) = x)) perhaps a picture will make more sense x>g(x) = y> z = f(y) = f(g(x)) that is what f o g does so to undo it, we go backwards z>y>x what takes z>y?For any function g, the mean or expected value of g(X) is defined by E(g(X)) = sum g(x k) p(x k) Ex Roll a fair die Let X = number of dots on the side that comes up Calculate E(X2) E(X2) 2= 2sum_{i=1}^{6} i p(i) = 1 p(1) 2 2 p(2) 32 p(3) 42 p(4) 5 p(5) 62 p(6) = 1/6*() = 91/6 E(X) is the expected value or 1st moment
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And hence jˆ E(x) ˆ E(y)j d(x;y) Chapter 4, problem 21 Suppose Kand F are disjoint sets in a metric space X, Kis compact, Fis closed Prove that there exists >0 such that d(p;q) > if p2K;q2F Show that the clonclusion may fail for two disjoint closed sets if neither is compact SolutionTh ey are system s o rien tedT h ey've d isco veredEE(g(X)jY)=E(g(X)) Proof Set Z = g(X) Statement (i) of Theorem 1 applies to any two rv's Hence, applying it to Z and Y we obtain EE(ZjY)= E(Z) which is the same as EE(g(X)jY)= E(g(X)) 2 This property may seem to be more general statement than (i) in Theorem 1 The proof above shows that in fact these are equivalent statements 3



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0 Comments Show Hide 1 older comments Sign in to comment Sign in to answer this questionFeb 01, 14 · what does the expression meansbasically what does the expression g(x,y) means ?E(XY) = E(X)E(Y) More generally, Eg(X)h(Y) = Eg(X)Eh(Y) holds for any function g and h That is, the independence of two random variables implies that both the covariance and correlation are zero But, the converse is not true Interestingly, it turns out that this result helps us prove



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3 the vicinal preorder is linear on X (equivalently on Y);See all Words by Length at More Words Find that difficult long word here!E(var(yx)) = e(e(y2x)) e(e(yx)2) We have already seen that the expected value of the conditional expectation of a random variable is the expected value of the original random variable, so applying this to Y 2



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That is f^1 what takes y>x that is g^1 do f^1 first, and then g^1, is g^1(f^1(z)) = g^1 o f^1(z) since this undoes what f o g does, it must beI think that if you manage to state your question precisely, the answer will be E( g(X,Y)), but you haven't defined the meaning of "best" in your original post and to say a random result is "closer" to something has no specific meaningHomework 6 Solutions Math 171, Spring 10 Henry Adams 386 Let fbe a continuous function from R to R Prove that fx f(x) = 0gis a closed subset of R



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