sept 2006, q.9

[achiless]

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) = x^2 + 1.
Determine E[Y] and Var[Y]

I don't understand hov they come to E[Var(Y|X)] =E[X^2 + 1] in the result:
Var[Y] = Var[E(Y|X)] + E[Var(Y|X)] = Var[X^2 + 1] + E[X^2 + 1]
= Var[X^2] + E[X^2] + 1

Could you please help me with that?

 

[Michael]

Y|X is Poisson with expectation x^2 + 1

From the tables, the poisson distribution has E[Y|X] = u
u = x^2 + 1
So Y|X ~ Poisson(x^2 + 1)

From the tables, var(Y|X) = u
var(Y|X) = x^2 + 1

So E[var(Y|X)] = E[x^2 + 1]

 

[achiless]

Thanks Michael, that was helpful.