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), f1(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.

If X is a random variable with pdf f(x) = 2x − 2 where x = (1, 2), find the variance of Y = 2X - 3.

Suppose X and Y are independent. X has a mean of 1 and variance of 1, Y has a mean of 0, and variance of 2. Let S=X+Y, calculate E(S) and Var(S). Let Z=2Y^2+1/2 X+1 calculate E(Z). Hint: for any random variable X, we have Var(X)=E(X-E(X))^2=E(X^2 )-(E(X))^2, you may want to find EY^2  with this. Calculate cov(S,X). Hint: similarly, we have cov(Z,X)=E(ZX)-E(Z)E(X), Calculate cov(Z,X). Are Z and X independent? Are Z and Y independent? Why? What about mean independence?

If Var(X1) = 2, Var(X2) = 4, Var(X3) = 3, Cov(X1, X2) = 1, Cov(X1, X3) = -2, and X2 and X3 are independent, find the mean and variance of Y = X1 – 2X2+3X3.