Q. 6.35 - A First Course in Probability

Question

If X and Y are independent standard normal random variables, determine the joint density function of 

$$\begin{gathered} U=XV \\ =\frac{X}{Y} \end{gathered}$$

Then use your result to show that $X/Y$ has a Cauchy distribution.

Step-by-Step Solution

Answer

The joint density function of u and v$=\frac{1}{\pi \left(1+v^2\right)}$

Step 1: Given information

X and Y are independent standard normal random variables.

$U=X V =\frac{x}{y}$

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

That is, $\mu =0 and \sigma =1$

Step 2: Explanation

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

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

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

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

It is given that

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

Step 3: Explanation

The Jacobean transformation is shown below:

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

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

Step 4: Explanation

The joint density function of u and v is,

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

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

Step 5: Explanation

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

$$\begin{gathered} f_v\left(v\right)={\int }_0^{\infty }\frac{1}{\pi \left(1+v^2\right)}{e}^{\frac{-{u^2}_1}{2}}\frac{u}{v^2}d{u}_1 \\ =\frac{1}{\pi \left(1+v^2\right)}{\int }_0^{\infty }{e}^{\frac{-{u^2}_1}{2}}d\left(\frac{{u^2}_1}{2}\right) \\ =\frac{1}{\pi \left(1+v^2\right)}\left(1\right) \\ =\frac{1}{\pi \left(1+v^2\right)} \end{gathered}$$