Assume Y=1​+X+u​, where X​, Y​, and ​u=v+X are random​ variables, v is independent of X​; E(v​)=0, Var(v​)=1​, E(X​)=1, and Var(X​)=2. a) E(u|X​) b) E(Y|X)

Assume that the random variables X1 and X2 are bivariate normally distributed with mean µ and covariance matrix Σ. Show that if cov(X1, X2) = 0 then X1 and X2 are independent.

Suppose that X is a random variable for which E(X) = μ and Var(X) = σ2. Show that E[X(X − 1)] = μ(μ − 1) + σ2.

1) Assume a relationship Y = g2(91(X)) between the two random variables X and Y where gi and g2 are two functions. To compute the mean of Y, we can choose any of the following equivalent expressions: E(Y)= | fr (u)ydy = | fn(x)(u)g2(u)du = / fx(#)g2(9n(x)dr (1) -00 where fy (y), f91(x)(u) and fx(x) are the probability density functions (p.d.f.s) of Y, 91(X) and X, respectively. Now consider Y = X? where X is N(0, o2). a) Determine the p.d.f. fy (y) of Y. Hint: for y > 0, fy (y) Prob(Y < y) = Prob(-< X < VI) = % x (x)dx. b) Then evaluate explicitly E(Y)= S fy(y)ydy.