This is what I'm supposed to prove...

Suppose that $Y$ and $Z$ are independent random variables with $Y \sim z^2(v)$, and $Z \sim N(0,1)$, and $T = \frac{Z}{\sqrt{\frac{Y}{v}}}$. Let $X = T^2$. Prove that $X \sim f(1,v)$.

How would I start to prove this?

I know I should be using the theorem that states: If $U$ and $V$ are independent variable having square distribution with $V_1$ and $V_2$ degrees of freedom then $$F= \frac{\frac{U}{V_1}} {\frac{V}{V_2}}$$ is a random variable having an $F$ distribution for $f > 0$ and $g(f) = 0$ elsewhere.


1 Answer 1


$X=T^2=Z^2/(Y/v)$. $Z^2$ is the square of a standard normal random variable, so it is $\chi^2$ with $1$ degree of freedom. $Y$ is (I assume) $\chi^2$ with $v$ degrees of freedom. Now, apply the theorem you mention above.


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