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EX2jY = 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 of91 If X and Y are ind ep end en t, then for all b oun ded, con tin uou s f ,g R d!1 If X and Y are independent rv's then E(XjY)=E(X) Proof As we know, X and Y are independent if and only if fX;Y(x;y) = fX(x)fY(y) or, equivalently, fXjY(xjy)= fX(x) But then E(XjY =y)=åx xfXjY(xjy)=åx xfX(x)=E(X) 2 2 EE(g(X)jY)=E(g(X)) Proof Set Z = g(X) Statement (i) of Theorem 1 applies to any two rv's Hence, applying
∂x,whereh(x,y) = f(u(x,y),v(x,y)) and f(u,v) = u2 v2 u2 −v2, u(x,y) = e −x y, v(x,y) = exy Solution First we will work out what the chain rule looks like in this context The mapping h (x,y) → h(x,y) is defined to be the composition of two mappings, h = f g g hx,yi → µFor any two rvs X and Y E(X Y) = E(X)E(Y) (4) If X and Y are independent, then Var(X Y) = Var(X)Var(Y) (5) The above properties generalize in the obvious fashion to to any finite number of rvs In general (independent or not) Var(X Y) = Var(X)V(Y)2Cov(X,Y), where Cov(X,Y) def= E(XY)−E(X)E(Y), is called the covariance between XE−λ = λ X∞ k=0 λk k!
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G(u,v)=∫∫g(x,y)e−i2π(xuyv)dxdy g(x,y)=∫∫G(u,v)ei2π(xuyv)dudv e−π(x 2y2)⎯FT⎯→ e−π(u 2v2) (x)δy ⎯FT→1 x)rect(y⎯FT→ sinc(u v circ(r)⎯FT⎯→ J 1(2πρ) ρ The function rect(x)rect(y) is shown on the left Its transform is the function sinc(u)sinc(v) shown onA g ro v e F o lk s to e Co l r id g e A s he b ro S il k H o e W t E s o s a t w W n s N h x H a a ncU(x,y) v(x,y) , and f hu,vi → f(u,v)



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(a) If there is a walk between two vertices x and y in some graph G, then there is also a path between x and y in G SolutionWe proceed by induction on the length of the walk (Recall that the length of a walk x = v1,v2,,vk = y from x to y is the number of its edges k −1) The base case is k = 2, ie, when x = v1,v2 = y This walk doesCharacter Residential Areas S Y O S T ER D R W CARUSO PL W 1ST STProof 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



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Change of Variables for Double Integrals We have already seen that, under the change of variables T(u, v) = (x, y) where x = g(u, v) and y = h(u, v), a small region ΔA in the xyplane is related to the area formed by the product ΔuΔv in the uvplane by the approximation ΔA ≈ J(u, v)Δu, ΔvLa profesora Leda Navarro Picado nos explica las normas ortográficas en la escritura de palabras con las letras b, v, c, z, s, g, j y h para asegurar una cThen, U = g(X) and V = h(Y) are also independent for any function g and h We will come back to various properties of functions of random variables at the end of this chapter 2 2 Moments and Conditional Expectation Using expectation, we can define the moments and other special functions of a random variable



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= λ X∞ k=1 λ λk−1 (k −1)!Example 5 X and Y are jointly continuous with joint pdf f(x,y) = (e−(xy) if 0 ≤ x, 0 ≤ y 0, otherwise Let Z = X/Y Find the pdf of Z The first thing we do is draw a picture of the support set (which in this case is the firstThe reason behind this is that the definition of the mgf of X Y is the expectation of et(XY ), which is equal to the product etX etY In case of indepedence, the expectation of that product is the product of the expectations • While for independent rv's, covariance and correlation are always 0, the converse is not true One



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