How we can show $\mathbb{E}[T]=0$ and $\operatorname{Var}(T)=\frac{n}{n-2}$.

Question

I need help with this question.

Let $Z\sim N(0,1)$ and $Y\sim X^2_{(n)}$ be independent variables, and define$$T\stackrel{\rm def}{=}\frac{Z}{\sqrt{\frac{Y}{n}}}.$$ Prove that $\mathbb{E}[T]=0$and $\operatorname{Var}(T)=\frac{n}{n-2}$.

I am not sure how to do this.

Edit: I only know that $\mathbb{E}[Z]=0$, $\operatorname{Var}(Z)=1$, $\mathbb{E}[Y]=n$, and $\operatorname{Var}(Y)=2n$.

thanks

Answer 1

Hints.

• By independence of $Z$ and $Y$, we have $$ \mathbb{E}[f(Z)g(Y)] = \mathbb{E}[f(Z)]\mathbb{E}[g(Y)] $$ for any measurable functions $f,g$. Can you use that to show that $\mathbb{E}[T]=0$, knowing that $\mathbb{E}[Z]=0$?

• From there, $\operatorname{Var} T = \mathbb{E}[T^2] - \mathbb{E}[T]^2 = \mathbb{E}[T^2] = n\mathbb{E}\left[\frac{Z^2}{Y}\right]$. Using the same independence property, and the expectations $\mathbb{E}[Z^2]$ (it's $1$ since $Z\sim\mathcal{G}(0,1)$) and $\mathbb{E}[\frac{1}{Y}]$, you should be able to conclude.