91Ó°ÊÓ

X and Y are independent standard normal random variables.

$U=XV=\frac{x}{y}$

It is given that X and Y are two independent standard normal random variables.

That is, $\mathrm{μ}=0and\mathrm{σ}=1$

The probability density function of standard normal random variable X and Y as shown below:

$$\begin{gathered} f\left(x\right)=\frac{1}{\sqrt{2\mathrm{Ï€}}}\exp \left\{\frac{-x^2}{2}\right\} \\ f\left(y\right)=\frac{1}{\sqrt{2\mathrm{Ï€}}}\exp \left\{\frac{-y^2}{2}\right\} \end{gathered}$$

The joint density function of X and Y is shown below:

$f(x,y)=\frac{1}{2\mathrm{Ï€}}\exp \left\{\frac{-\left(x^2+y^2\right)}{2}\right\}$

It is given that

$$\begin{gathered} U=X,V=\frac{x}{y} \\ ⇒Y=\frac{x}{v}=\frac{U}{V} \end{gathered}$$

The Jacobean transformation is shown below:

$J=\left|\begin{array}{ll}\frac{∂x}{∂u}& \frac{∂x}{∂v}\\ \frac{∂y}{∂u}& \frac{∂y}{∂v}\end{array}\right|$

$$\begin{gathered} =\left|\begin{array}{cc}1& 0\\ \frac{1}{v}& \frac{-u}{v^2}\end{array}\right| \\ =\frac{-u}{v^{2^{}}}-0 \\ =\frac{-u}{v^2} \end{gathered}$$

The joint density function of u and v is,

$$\begin{gathered} f\left(u,v\right)=f_{xx}\left(x,y\right)\left|J\right| \\ =\frac{1}{2\mathrm{Ï€}}exp\left\{\frac{-\left(u^2+{\left(\frac{u}{v}\right)}^2\right)}{2}\right\}\frac{u}{v^2} \\ Findf\left(\frac{X}{Y}\right) \end{gathered}$$

$$\begin{gathered} f_v\left(V\right)={∫}_{-∞}^{∞}{f}_{UV}(u,v) \\ ={∫}_{-∞}^{∞}\frac{1}{2\mathrm{π}}\exp \left\{\frac{-\left(u^2+{\left(\frac{u}{v}\right)}^2\right)}{2}\right\}\frac{u}{v^2}du \\ =2{∫}_0^{∞}\frac{1}{2\mathrm{π}}\exp \left\{\frac{-\left(u^2+{\left(\frac{u}{v}\right)}^2\right)}{2}\right\}\frac{u}{v^2}du\left(\begin{array}{l}\text{since the integrated is an}\\ \text{even function of}u\end{array}\right) \end{gathered}$$

Using the change of variable ${u^2}_1=u^2\left(1+v^{-2}\right)$we have

$$\begin{gathered} f_v\left(v\right)={∫}_0^{∞}\frac{1}{\mathrm{π}\left(1+v^2\right)}{e}^{\frac{-{u^2}_1}{2}}\frac{u}{v^2}d{u}_1 \\ =\frac{1}{\mathrm{π}\left(1+v^2\right)}{∫}_0^{∞}{e}^{\frac{-{u^2}_1}{2}}d\left(\frac{{u^2}_1}{2}\right) \\ =\frac{1}{\mathrm{π}\left(1+v^2\right)}\left(1\right) \\ =\frac{1}{\mathrm{π}\left(1+v^2\right)} \end{gathered}$$