Question

Let the independent random variables X₁ and X₂ be N(0,1) and $\chi$ ²(r) respectively. Let Y₁=X₁/(X₂)^(1/2)/r and Y₂=X₂

i) Find the joint pdf of Y₁ and Y₂

ii) Determine the marginal pdf of Y₁ and show that Y₁ has a t distribution

Answer 1

$X_1\sim N(0,1)$ and its probability density function is given as,

$g(x_1)={1\over\sqrt{2\pi}}e^{-{x_1^2\over2}},\space 0\lt x_1\lt\infin$

and,

$X_2\sim \chi^2(r)$ and its probability density function is given as,

$h(x_2)={1\over\varGamma({r\over2})2^{r\over2}}x_2^{{r\over2}-1}e^{-{x_2\over2}},\space 0\lt x_2\lt\infin$

Since the random variables $X_1$ and $X_2$ are independent, their joint probability density function is given as,

$f(x_1,x_2)=g(x_1)\times h(x_2)$. So,

$f(x_1,x_2)={1\over\sqrt{2\pi}}e^{-{x_1^2\over2}}{1\over\varGamma({r\over2})2^{r\over2}}x_2^{{r\over2}-1}e^{-{x_2\over2}},\space -\infin\lt x_1\lt \infin,\space 0\lt x_2\lt\infin\\0,\space elsewhere$

Define,

$Y_1={X_1\over\sqrt{(X_2)\over r}}$ and $Y_2=X_2$. They are one to one mapping of the set, $S=\{(x_1,x_2):-\infin\lt x_1\lt\infin, 0\lt x_2\lt\infin\}$ onto the set,

$R=\{(y_1,y_2),-\infin\lt y_1\lt\infin, 0\lt y_2\lt\infin\}$

So,

$X_2=Y_2$ and $X_1=Y_1\sqrt{Y_2\over r}={Y_1(Y_2)^{1\over2}\over \sqrt{r}}$

The Jacobian of transformation is,

$J=\begin{vmatrix} {d_{x_1}\over d_{y_1}} & {d_{x_1}\over d_{y_2}} \\ {d_{x_2}\over d_{y_1}} & {d_{x_2}\over d_{y_2}} \end{vmatrix}=\begin{vmatrix} {\sqrt{y_2\over r}} & {y_1\over2\sqrt{r\times y_2}} \\ 0 & 1 \end{vmatrix}=\sqrt{{y_2\over r}}$

Then, the joint pdf of $Y_1$ and $Y_2$ is given by,

$g(y_1,y_2)=g(y_1\sqrt{y_2\over r},y_2)|J|\\= {1\over\sqrt{2\pi}\varGamma({r\over2})2^{r\over2}}y_2^{{r\over2}-1}e^{-{{1\over2}y_1^2}{y_2\over r}}e^{-{y_2\over2}}\sqrt{{y_2\over r}}\\= {1\over\sqrt{2\pi}\varGamma({r\over2})2^{r\over2}}y_2^{{r\over2}-1}e^{-{y_2\over2}(1+{y_1^2\over r})}{y_2^{1\over2}\over\sqrt{r}},\space -\infin\lt y_1\lt\infin,\space 0\lt y_2\lt\infin$

The marginal pdf of $Y_1$ is given by,

$g(y_1)=\displaystyle\int^{\infin}_0 g(y_1,y_2)dy_2=\displaystyle\int^{\infin}_0{1\over\sqrt{2\pi}\varGamma({r\over2})2^{r\over2}}y_2^{{{1\over2}(r+1)}-1}e^{-{y_2\over2}(1+{y_1^2\over r})}{y_2^{1\over2}\over\sqrt{r}}dy_2$

We make the substitution,

$w={y_2(1+{y_1^2\over r})\over2}$ and ${dw\over dy_2}={1+{y_1^2\over r}\over 2}\implies dw={1+{y_1^2\over r}\over 2}dy_2$

We can now write the integral as,

$\displaystyle\int_0^{\infin}{1\over\sqrt{2\pi r}\varGamma({r\over2})2^{r\over2}}({2w\over1+{y_1^2\over r}})^{{1\over2}(r+1)-1}e^{-w}({2\over1+{y_1^2\over r}})dw$

Next, we introduce $\varGamma({r+1\over2})$

$={1\over\sqrt{2\pi r}\varGamma({r\over2})2^{r\over2}}({2\over1+{y_1^2\over r}})^{{1\over2}(r+1)}\varGamma({r+1\over2})\displaystyle\int_0^{\infin}{w^{{1\over2}(r+1)-1}e^{-w}\over\varGamma({r+1\over2})}dw$

The integral part is equal to 1.0 because it is a pdf of the Gamma distribution with parameters $\alpha={r+1\over2}$ and $\beta=1$

So we have,

${1\over\sqrt{2\pi}\varGamma({r\over2})2^{r\over2}}({2\over1+{y_1^2\over r}})^{{1\over2}(r+1)}\varGamma({r+1\over2})$ and we can write as,

$g(y_1)=({\varGamma({r+1\over2})\over\sqrt{\pi r}\varGamma({r\over2})})({1\over(1+{y_1^2\over r})^{r+1\over 2}}),\space -\infin\lt y_1\lt \infin\\0,\space elsewhere$

Thus, if $X_1\sim N(0,1)$ and $X_2\sim \chi^2(r)$ and $X_1$ and $X_2$ are independent then, the random variable $Y_1={X_1\over\sqrt{X_2\over r}}$ has a t distribution with $r$ degrees of freedom.

Therefore,

$Y_1\sim t-distribution(r).$