How to find E(Y), Var(Y) when Y=(X2)^2/(X1)^2 , X1, X2 independent

Question

Let X1 , X2 be independent and X1~N(0,1), X2~N(0,1) (cf * N($\mu ,\sigma^2$))

Let Y=$\frac{X_2^2}{X_1^2}$ , then find E(Y) and Var(Y)

Help me

Y = F(1,1)

Y is F-distribution Right?

Answer 1

If $X$ is standard normal distributed, $X \sim N(0,1)$, then $X^2$ is chi squared distributed with one degree of freedom, $X^2 \sim \chi^2(1)$ and $\frac{1}{X^2}$ is inverse chi squared distributed with one degree of freedom, $\frac{1}{X^2} \sim$ Inv $\chi^2(1)$.

If $X_1$ and $X_2$ are independent, then $\frac{1}{X_1^2}$ and $X_2^2$ are independent, so $E\left(\frac{X_2^2}{X_1^2}\right) = E\left(\frac{1}{X_1^2}\right) \cdot E\left(X_2^2\right)$.

We know that if $Z_1 \sim \chi^2(k)$, $k=1,2,\dots$, then $$EZ_1 = k, \quad \text{for } k=1,2,\dots$$ and $$E\frac{1}{Z_1} = \frac{1}{k-2}, \quad \text{for } k>2.$$

As pointed out in the comments, the expectation and variance does not exist. I hope this helps you see why - you do have not enough degrees of freedom!

In general it holds, that if $X_i \sim N(0,1)$, for $i = 1,2,\dots,n$ are i.i.d., then $\sum_{i=1}^n X_i^2 \sim \chi^2(n)$.