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Then E(y g(X)) 2 is minimized when g(X) = EYjX Lecture 26 Examples I Toss 100 coins What's the conditional expectation of the number of heads given the number of heads among the rst fty tosses?CGI C ɃT C g ^ c ɖ𗧂 e v O z z Ă ܂ B t Ŏg CGI,PHP 𑽐 Ă ܂ ̂ň x BCGI/PHP ECGI/PHP J X ^ } C Y T g S ł BProof lnexy = xy = lnex lney = ln(ex ·ey) Since lnx is onetoone, then exy = ex ·ey 1 = e0 = ex(−x) = ex ·e−x ⇒ e−x = 1 ex ex−y = ex(−y) = ex ·e−y = ex 1 ey ex ey • For r = m ∈ N, emx = e z }m { x···x = z }m { ex ···ex = (ex)m • For r = 1 n, n ∈ N and n 6= 0, ex = e n n x = e 1 nx n ⇒ e n x = (ex) 1 • For r rational, let r = m n, m, n ∈ N

H h s wk h x q g lv s x wh g lq j r i wk h r ii u r d g d g y h q wx u h lq y lwh v \ r x wr f olp e lq wr wk h g u ly h u v v h d w d q g h s h u lh q f h h h s oln h q h y h u e h ir u h lwk r y h u f x v wr p l d e oh h h s y h k lf oh v lq f ox g lq j ix wx u lv wlf fAnswer (1 of 2) Conditional expectation is difficult to work with in the most general case Here is a link to the proof in the general case, but it may not be that informative if you are not familiar with measure theory Law of total expectation I will give you a "proof" in the special caseEX2jY = y = 1 25 (y 1)2 4 25 (y 1) Thus EX2jY = 1 25 (Y 1)2 4 25 (Y 1) = 1 25 (Y2 2Y 3) Once again, EX2jY is a function of Y Intuition EXjY is the function of Y that bests approximates X This is a vague statement since we have not said what \best" means We consider two extreme cases First suppose that X is itself a function of

Substituting 2T for T in X(w), we have Y(w) = 2(2T) sin(w2T) The zero crossings are at w,2T = nr, or w = n S96 (a) x(t) = Substituting t X(w)et dw = 0 in the preceding equation, we get x(0) = 1 27r X( ) de (b) X(w) = f x(t)e jwt dt Substituting o = 0 in the preceding equation, we get X(0) =x(t2 2 > \ V X V Z Y X W V U T S R ^ 1 ` # 2 % 2 # / _ 2 > 3 3 8 3 a n j ~ j f } e { z f f f y n g f x w e v u l t s f f f q r q k o p o n j k m l k j i h g f e d c b ~ y f t f n f e c f f f t } c t t b f b s f f f g i} { ꒆ w k 6 ̃G X e T B g ^ q O X y XAlfabia i A t @ r A j B t F C V A u C _ A A } A J C v N e B b N ̃ j B

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J Katich, X Qian, Y X Zhao, K Allada, K Aniol, J RM Annand, T Averett, F Benmokhtar, W Bertozzi, P C Bradshaw, P Bosted, A Camsonne, M Canan, G DA curve in the plane is said to be parameterized if the set of coordinates on the curve, (x,y), are represented as functions of a variable tNamely, x = f(t), y = g(t) t D where D is a set of real numbers The variable t is called a parameter and the relations between x, y and t are called parametric equationsThe set D is called the domain of f and g and it is the set of values t takesIe, f Rn!Ris convex if g( ) = f(x 0 v) is convex 8x 0 2dom(f) and 8v2Rn We just proved this happens i g00( 2) = vTrf(x 0 v)v 0;

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11 PH ¬g¢G lsjO¬L "hl¢VH Hg'¢F Uhg¢m Hg¬rm lK k±VM Uhlm Ugn "hl¢VH Hg'¢F jajlG "hl¢VH Hg'¢F Ugn Hgl¢«HJ Hgjhg¢m PH "hl¢VH Hg'¢F lK j§hk¢kh Ugn aVHz" gl«¢¬ lK Hglug'lhJ Hgjw¢g¢m P'G "hl¢VH PH T Ugn HgVrL Hgjsgsgd 'VrL lkjµ ½V¥n HgV¥'c îgn lgwR Hglkjµ ggjuV ¢ Hg'¢F lK PH' VH¥v Hgjug¢lhJ UfV H™kjVkJ3 The Probability Transform Let Xa continuous random variable whose distribution function F X is strictly increasing on the possible values of X Then F X has an inverse function Let U= F X(X), then for u;1, PfU ug= PfF X(X) ug= PfU F 1 X (u)g= F X(F 1 X (u)) = u In other words, U is a uniform random variable on 0;1́A ̗l ȊF l ̂ t ̗l ɁA ܂ ́u I v ۂ̋삯 ݎ Ƃ āA F l ł C y ɕs Y Ɋւ 鑊 k ł A T r X 󂯂 ̐ 肽 A Ă X ̂ ɗ Ǝv 0 N Ԃ̉ Ј Ƀs I h ł A 犔 Г A G X e g E T r X ݗ ܂ B @ ܂łɒ~ ς s Y Ɋւ m A m E n E F l ̂ ߂Ƀt p ăT r X 񋟂 ĎQ ܂ B

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 You can't actually estimate Z since Z is a random variable and not a parameter you need to be careful about using estimation in this context Z is a random variable, so if you wanted to get the population mean of Z then you calculate E Z = E g (X,Y) Remember that estimation concerns estimating something that is essentially fixed, like mu2 v K e _ ^ m j Z a e b q Z l g Z m q g h h h k g h \ i j h _ d l u, b f _ x s b g Z m q g h h h h k g h \ Z g b y « G Z m q g h h h k g h \ i j _ ^ k l ZE g(X) ≥ g( EX ) Source Wolfram MathWorld a) Proof Let g(X) be −√$, for all positive real numbers g(x) is a convex function as seen by the graph to the right Since g(X) satisfies the requirements for Jensen's inequality, we can use it to directly prove the inequality

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Sity function and the distribution function of X, respectively Note that F x (x) =P(X ≤x) and fx(x) =F(x) When X =ψ(Y), we want to obtain the probability density function of YLet f y(y) and F y(y) be the probability density function and the distribution function of Y, respectively Inthecaseofψ(X) >0,thedistributionfunctionofY, Fy(y), is rewritten as followsI n t e g r i t y S e r v i c e E x c e l l e n c e Headquarters US Air Force USAF Maintenance Metrics Looking Forward with Aircraft Availability (AA) Lt Col Jeff Meserve, USAF Chief, Congressionals, Studies & Analysis Branch Directorate of Maintenance DCS/Logistics, Installations and Mission Support315 f g h ` _ k l \ h l j m ^ h \ _ i h t e Z j k d Z b k l h j b y g Z g _ f k d b b t e Z j k d b _ a b d K i h j _ ^ ^ h k l h \ _ j g b ^ Z g g b _ i h q

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F(y) = f(x) f0(x)(y x) 1 2 f00(z)(y x)2;X iu y So, let's de ne g= u x iu y This is our candidate for f0 2Show that g(z) is analytic Write g = ˚ i , where ˚ = u x and = u y Checking the CauchyRiemann equations we have ˚ x ˚ y x y = u xx u xy u yx u yy Since uis harmonic we know u xx = u yy, so ˚ x = y It is clear that ˚ y = x Thus gsatis es the CauchyRiemann(a) T R2!R2, with T x y = x y y Solution This IS a linear transformation Let's check the properties (1) T(~x ~y) = T(~x) T(~y) Let ~x and ~y be vectors in R2 Then, we can write them as ~x = x 1 x 2 ;~y = y 1 y 2 By de nition, we have that T(~x ~y) = T x 1 y 1 x 2 y 2 = x 1 y 1 x 2 y 2 x 2 y 2 and T(~x) T(~y) = T x 1

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Ways to Visualize functions f R !R (eg f(x) = x2) (1) SetTheoretic Picture (2) Graph of f (Thinking y= f(x)) The graph of f R !R is the subset of R2 given by Graph(f) = f(x;y) 2R2 jy= f(x)g (3) Level sets of fE(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܂ u y w A T V e B X z ̓R ~ j P V o ̏ ł v Ƃ l Ă A w A P A E J b g E p } E J ̎U Ɋւ 邲 k 瑼 Ȃ Ԙb 肷 钆 ŁA Ⴆ ΁A l s E s } E ` k j ^ E E ȂǁA n ̂ ȏ A d ŕ Ă ɃA h o C X i w A T Ƃ A l X ȐE ƁE E ̕ X ɂȂ 邽 ߁j ̂ q l Ǝ҂ Љ ȂǁA E 𗬂̂ł w A T ڎw Ă ܂ B

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If X and Y are independent, then E(es(XY )) = E(esXesY) = E(esX)E(esY), and we conclude that the mgf of an independent sum is the product of the individual mgf's Sometimes to stress the particular rv X, we write M5 (Logan, 24 # 1) Solve the problem ut =kuxx, x >0, t >0, ux(0,t)=0, t >0, u(x,0)=φ(x), x >0, with an insulated boundary condition by extending φ to all of the real axis as an even function The solution is u(x,t)= Z ∞ 0 G(x −y,t)G(x y,t)φ(y)dy First note that the solution to the IVP ut = kuxx, −∞ < x < ∞, t > 0, u(x,0) = f(x), −∞X Y Z \ ^ _ ` a b c d e qqq f g h a i j d \ ^ k b e _ l m Y n o p q r s t u v w x y z {} ~ w {q u r y t v ~ z s x s u {r z y t g v q ~ x w g j i d \ ^ k b e

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" #" " #" E(g(X,Y))=g(x,y)f XY (x,y) It is important to note that if the function g(x,y) is only dependent on either x or y the formula above reverts to the 1dimensional case Ex Suppose X and Y have a joint pdf f XY(x,y) Calculate E(X This is a slightly modified question from Sheldon Ross 9th ed Assume all RVs are discrete I am asked to prove the following equality $$\mathbb{E}Xg(Y)Y = g(Y) \mathbb{E}XY$$ Here is my attemptThere is more variation in the height of the minuscules, as some of them have parts higher or lower than the typical sizeNormally, b, d, f, h, k, l, t are the letters with ascenders, and g, j, p, q, y are the ones with descenders In addition, with oldstyle numerals still used by some traditional or classical fonts, 6 and 8 make up the ascender set, and 3, 4, 5, 7 and 9 the descender set

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For some z2x;y)f(y) f(x) f0(x)(y x) Now to establish (ii) ,(iii) in general dimension, we recall that convexity is equivalent to convexity along all lines;Figure 485 The family of antiderivatives of 2x consists of all functions of the form x2 C, where C is any real number For some functions, evaluating indefinite integrals follows directly from properties of derivatives For example, for n ≠ −1, ∫xndx = xn 1

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