# 2 suppose you have a random number generator which is

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2. Suppose you have a random number generator which is capable of generating random numbers distributed as X ∼ N (0 , 1). Describe how to generate random vectors Y N ( μ , Σ). Generate n -dimensional random vectors X by generating n realizations of the scalar random variable X . Then cov( X , X ) = I . Write Σ = CC T (the Cholesky factorization). Let Y = C X + μ . Then E [ Y ] = m and cov( Y , Y ) = C cov( X , X ) C T = CIC T = CC T = Σ . 3. Suppose X and Y are r.v.s. Show that E [( X - h ( Y )) 2 ] is minimized over all functions h be the function h ( y ) = E [ X | Y = y ] . Assume E [ X 2 ] < . (a) First approach: We will assume (only for convenience) that the r.v.s are continu- ous. We can write E [( X - h ( Y )) 2 ] = E [ E [( X - Y ( Y )) 2 | Y = y ]] = E Z ( X - h ( y )) 2 f X | Y ( X | Y = y ) dx = Z Z ( X - h ( y )) 2 f X | Y ( X | Y = y ) dxf Y ( y ) dy 2

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To minimize this, we can minimize the inner integral Z ( X - h ( y )) 2 f X | Y ( X | Y = y ) dx for each value of h ( y ). Since Y = y is a fixed value for the inner integral, h ( y ) is a fixed value for each value of y , and we can take the derivative with respect to h ( y ) and equate to zero to minimize: ∂h ( y ) Z ( X - h ( y )) 2 f X | Y ( X | Y = y ) dx = Z 2( X - h ( y )) f X | Y ( X | Y = y ) dx = 0 which leads to h ( y ) Z f X | Y ( X | Y = y ) dx = Z Xf X | Y ( X | Y = y ) dx The integral on the left is equal to 1 (since it integrates over the entire set of X ), and the integral on the right is equal to E [ X | Y = y ]. (b) Approach 2: This one is more appealing, because it is not expressed in terms of integrals (making it immediately applicable to all types of distributions), it does not require interchanging limiting operations (i.e., taking derivatives inside
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• Fall '08
• Stites,M
• Randomness, σx σy, |Y, fX |Y

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