Show that if $X \sim N(0,1)$ and $Y \sim N(0,1)$ then $\frac{X}{Y} \sim \mathcal{C}(0,1)$

Question

Let $X,Y$ be independent random variables. Show, without using the change of variable theorem, that if $X \sim N(0,1)$ and $Y \sim N(0,1)$ then $\frac{X}{Y} \sim \mathcal{C}(0,1)$

So this problem is supposed to be solved using conditional expectation. So Im always getting messed in the same step.

We want to see that $\frac{X}{Y} \sim \mathcal{C}(0,1)$.

Now

$$P(\frac{X}{Y} \leq z)=E(1\{\frac{X}{Y}\leq z\})=E(E(1\{\frac{X}{Y}\leq z\}|Y))$$

Now

$$E(1\{\frac{X}{Y}\leq z\}|Y)=\int 1\{\frac{X}{Y}\leq z\}f_{\frac{X}{Y}|Y=y}(z)dy$$

And therefore, to obtain $f_{\frac{X}{Y}|Y=y}(z)$

Y compute

$$P(\frac{X}{Y} \leq z|Y=y)=P(\frac{X}{y} \leq z|Y=y)=P(X \leq yz|Y=y)=P(X \leq yz)$$

Since $Y,X$ are independent. Now since $X\sim N(0,1)$ I obtain

$$f_{\frac{X}{Y}|Y=y}(z)=(P(\frac{X}{Y} \leq z|Y=y))'=(P(X \leq yz))'=\frac{y}{\sqrt{2\pi}}e^{-\frac{(yz)^2}{2}}$$ So far, no problem (unless thats what I think, but I don't know how to compute

$$\int 1\{\frac{X}{Y}\leq z\}f_{\frac{X}{Y}|Y=y}(z)dy=\int 1\{\frac{X}{Y}\leq z\}\frac{y}{\sqrt{2\pi}}e^{-\frac{(yz)^2}{2}}dy$$

How should I proceed? Is my working out ok so far? Thanks!

Answer 1

Denote $\frac{1}{\sqrt{2\pi}}$ by $c$. If $X$ and $Y$ are independent, then \begin{align} & F_{X/Y}(z) \\ = & P\left[\frac{X}{Y} \leq z\right] \\ = & \int_{-\infty}^\infty P\left[\frac{X}{y} \leq z\right]f_Y(y)dy \\ = & \int_{-\infty}^0 P[X \geq y z]ce^{-y^2/2}dy + \int_0^\infty P[X \leq y z]ce^{-y^2/2}dy \\ = & \int_{-\infty}^0 \left[\int_{yz}^{\infty} ce^{-x^2/2}dx\right]ce^{-y^2/2}dy +\int_{0}^\infty \left[\int_{-\infty}^{yz} ce^{-x^2/2}dx\right]ce^{-y^2/2}dy \\ \end{align} If we differentiate the right hand side of the expression under the integration with respect to $z$, we have \begin{align} & F'_{X/Y}(z) \\ = & -\int_{-\infty}^0 \left[ce^{-(yz)^2/2}y\right]ce^{-y^2/2}dy + \int_0^\infty \left[ce^{-(yz)^2/2}y\right]ce^{-y^2/2}dy \\ = & -\frac{1}{2\pi}\int_{-\infty}^0 ye^{-\frac{1}{2}(1 + z^2)y^2}dy + \frac{1}{2\pi}\int_0^\infty ye^{-\frac{1}{2}(1 + z^2)y^2}dy \\ = & \frac{1}{\pi(1 + z^2)}, \end{align} which is exactly the density of Cauchy distribution. So the proof is done if the differentiation to the integration is legitimate, which follows from that $ye^{-\frac{1}{2}(1 + z^2)y^2}$ is dominated by $|y|e^{-\frac{1}{2}y^2}$ that is integrable over the real line (for a rigorous theorem related to this, see Theorem $16.8$ in Probability and Measure by Patrick Billingsley).