Help Proving That $T_n \sim t_n$ with Z and Y Independent

[evinda]

Hello! (Wave)

Could you help me at the following exercise? (Thinking)

Let $Z \sim N(0,1) $ , $Y \sim \chi_n^2$ and $Z,Y$ independent .

Show that the random variable $T_n := \frac{Z}{ \sqrt{\frac{Y}{n}}} \sim t_n$.

 

[Valkarie]

Hint: Use the fact that $\frac{Z}{ \sqrt{Y/n}}$ has the same distribution as $Z \sqrt{n/Y}$.We need to prove that $T_n \sim t_n$. Let's start by noting that $T_n = \frac{Z}{ \sqrt{\frac{Y}{n}}}$, and that $Z$ and $Y$ are independent. We can then use the fact that $\frac{Z}{ \sqrt{Y/n}}$ has the same distribution as $Z \sqrt{n/Y}$.Now, we need to show that $Z \sqrt{n/Y}$ follows a $t_n$ distribution. To do this, we will calculate the probability density function (PDF) of $Z \sqrt{n/Y}$ and show that it is equal to the PDF of a $t_n$ distribution, which is given by\begin{align*}f_{t_n}(z) = \frac{\Gamma(\frac{n+1}{2})}{\sqrt{n\pi}\Gamma(\frac{n}{2})} (1+\frac{z^2}{n})^{-\frac{n+1}{2}} \end{align*}The PDF of $Z \sqrt{n/Y}$ is given by\begin{align*}f_{Z \sqrt{n/Y}}(z) &= \int_{0}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}\frac{1}{\sqrt{2\pi y}}e^{-\frac{y}{2}}\frac{1}{\sqrt{n/y}}e^{-\frac{n}{2y}}dy \\&= \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}\int_{0}^{\infty} \frac{1}{\sqrt{2\pi y}}e^{-\frac{y+nz^2}{2}} dy \\&= \frac{1}{\sqrt