Showing that the ratio of two standard independent normals is a Cauchy using Characteristic Functions

Question

Question

Let $X$ and $Y$ be independent standard normals. Use characteristic functions to find the distribution of $X/Y$.

My attempt

We will attempt to show that $Ee^{itX/Y}=e^{-|t|}$ (the c.f. of a Cauchy random variable) from which the claim will follow. To this end, note that $$ Ee^{itX/Y}=\int\frac{1}{\sqrt{2\pi}}e^{-y^2/2}\int e^{itx/y}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\, dx\, dy=\int \frac{1}{\sqrt{2\pi}}e^{-y^2/2}\exp\left(-\frac{t^2}{2y^2}\right)\, dy $$ where we used the fact that $Ee^{it X}=\exp(-0.5t^2)$. We can write it as $$ Ee^{itX/Y}=\int \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}\left[\frac{t^2}{y^2}+y^2\right]\right)\, dy. $$ But at this point I don't know how to evaluate the integral. I tried completing the square in the exponent but I couldn't make progress from there.

Any help is appreciated.

Answer 1

Set $\phi(t):= \mathbb{E}\exp(itX/Y)$. Since $\phi(0)=1$ and $\phi$ is even, it suffices to show that $\phi(t)=e^{-t}$ for all $t>0$.

Fix $t>0$. Following your calculations we have

$$\phi(t) = \sqrt{\frac{2}{\pi}} \int_{(0,\infty)} \exp \left(- \frac{1}{2} \left[ \frac{t^2}{y^2}+y^2 \right] \right) \, dy \tag{1}$$

Performing a change of variables, $z:=t/y$, we find

$$\phi(t) = t \sqrt{\frac{2}{\pi} } \int_{(0,\infty)}\frac{1}{z^2} \exp \left(- \frac{1}{2} \left[ z^2 + \frac{t^2}{z^2} \right] \right) \, dz \stackrel{(1)}{=} - \phi'(t).$$

This shows that $\phi$ solves the ODE $$\phi' = - \phi \quad \text{on $(0,\infty)$}.$$

Since $\phi(0)=1$ we conclude that $\phi(t)=e^{-t}$ for $t \geq 0$.

Answer 2

Switch to polar coordinates. It is slightly tidier to evaluate $E(e^{iY/X})$, which is expressible by symmetry as $$ E(e^{iY/X})=2\int_{x=0}^\infty\int_{y=-\infty}^\infty e^{ity/x}\frac1{2\pi}e^{-(x^2+y^2)/2}\,dy\,dx. $$ In polar coordinates this equals $$ 2\int_{\theta=-\pi/2}^{\pi/2}\int_{r=0}^\infty e^{it\tan \theta}\frac1{2\pi}e^{-r^2/2}r\,dr\,d\theta=\int_{\theta=-\pi/2}^{\pi/2}\frac1{\pi}e^{it\tan\theta}d\theta. $$ Apply the substitution $u=\tan\theta$, $du=\sec^2\theta\, d\theta=(1+u^2)d\theta$ to obtain $$ \int_{u=-\infty}^\infty\frac1\pi\frac{e^{itu}}{1+u^2}du. $$ At this point you can quit, because you've just written down the characteristic function of a Cauchy density.