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

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.

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 $$t \geq 0$$.

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.