question NO:9 Suppose that a random variable X has a standard normal distribution, and the conditional distribution of a Poisson random variable Y, given the value of X = x, has expectation g(x) = x2 + 1. Determine E[Y] and Var[Y]. Ans: E[Y] = E[E(Y|X)] = E[X2 + 1] = V[X] + {E[X]}2 + 1 = 1 + 0 + 1 = 2 please clarify how come var[x]=1 and E[X]=0 2ND PART Var[Y] = Var[E(Y|X)] + E[Var(Y|X)] = Var[X2 + 1] + E[X2 + 1] = Var[X2 ] + E[X2 ] + 1 but Z = X2 is 2 χ1 so has variance 2 and expectation 1 Thus Var[Y] = 2 + 1 + 1 = 4 PLEASE CLARIFY HOW VAR[X^2]=2 AMD [E(X^2)]=1
Hi there, Since X is a std normal variable N(0,1), it has mean 0 and variance 1. So E(X)=0 and Var(X)=1 E(x^2)=Var(x) + {E(x)}^2, by formula =1 + 0^2 =1
