5
$\begingroup$

Problem: Show that if $X$ and $Y$ are independent $N(0,1)$-distributed random variables, then $X/Y ∈ C(0,1)$.

Question: I don't know how to proceed below. I want to prove that the PDF of $X/Y$ is Cauchy. PS. I looked on wiki and they defined $C$ as $X/Y$

Attempt: enter image description here

The expression looked hairy and wolfram could not integrate it.

Note to self: problem 8.

$\endgroup$
4
$\begingroup$

Let $Z=X/Y$. Then $$ Ef(Z) = \int f(x/y)n(x)n(y) dxdy = \int f(z) n(yz)n(y)|y|dydz $$ With $n(x) = \frac1{\sqrt{2\pi}} \exp(-x^2/2)$ and $\sigma(z)^2 = 1/(1+z^2)$, $$ n(y)n(yz) = \frac1{2\pi} \exp((1+z^2)(-y^2/2)) = \frac 1{2\pi} \exp(-y^2/2\sigma(z)^2) \\ \int n(yz)n(y)|y|dy = 2\int_0^\infty \frac 1{2\pi} \exp(-y^2/2\sigma(z)^2)ydy\\ = \frac{\sigma(z)^2}{\pi} \int_0^\infty \exp(-y^2/2\sigma(z)^2)\frac{y}{\sigma(z)} \frac{dy}{\sigma(z)} = \frac{\sigma(z)^2}{\pi} \\ Ef(Z) = \int f(z) n(yz)n(y)|y|dydz = \int f(z) \frac{\sigma(z)^2}{\pi} dz = \int f(z) \frac{dz}{\pi(1+z^2)} $$

$\endgroup$
  • $\begingroup$ Did you really need the last line $Ef(Z) = ...$ ? Since you derived the correct PDF shouldn't the proof be done then? $\endgroup$ – jacob Sep 6 '14 at 9:36
  • $\begingroup$ In the last step, when you conclude $= \frac{\sigma(z)^2}{\pi}$ did you recognize it as an exponential PDF summing to 1? $\endgroup$ – jacob Sep 6 '14 at 9:48
  • 1
    $\begingroup$ the method with the computation of $Ef(Z)$ is general. $\endgroup$ – mookid Sep 6 '14 at 13:06
  • 1
    $\begingroup$ $\int_0^\infty \exp(-y^2/2\sigma(z)^2)\frac{y}{\sigma(z)} \frac{dy}{\sigma(z)} = \int_0^\infty \exp(-u^2/2)u du =\left[ -\exp(-u^2/2) \right]_0^\infty = 1$ $\endgroup$ – mookid Sep 6 '14 at 13:27
  • 1
    $\begingroup$ pdf may not exist. $\endgroup$ – mookid Sep 6 '14 at 19:18
2
$\begingroup$

An alternative proof taken from here - I find it helpful as most steps are explained intuitively. Please correct me if something is incorrect.

Proof

Start with the CDF which is $P(\frac{X}{Y} \leq t)$ which is an event, and we want to find the probability of this event.

$$ P(\frac{X}{Y} \leq t) = P(\frac{X}{|Y|} \leq t) = P(X \leq |Y|t) \text{ by symmetry of } N(0,1) $$

To get a probability, integrate the joint pdf over the region of interest. For $Y$ this is between $- \infty$ to $\infty$ but for $X$ this is from $- \infty$ to $t |y|$ !

$$P(X \leq |Y|t) = \int_{- \infty}^{\infty} \int_{- \infty}^{t |y|} f(x,y) \ dx \ dy $$

As $X$ and $Y$ are i.i.d. the joint PDF can be found by multiplying the pdf : $f_{X,Y}(x,y) = f_X(x) f_Y(y)$

\begin{aligned} P(X \leq |Y|t) &= \int_{- \infty}^{\infty} \int_{- \infty}^{t |y|} \frac{1}{\sqrt{2 \pi}} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2}(x^2 + y^2)}\ dx \ dy \\ & = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} y^2} \int_{- \infty}^{t |y|} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} x^2}\ dx \ dy \end{aligned}

Recognize that the inner integral cannot be computed analytically, however this is the normal CDF $\Phi(t |y|)$

\begin{aligned} P(X \leq |Y|t) & = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} y^2} \Phi(t |y|) \ dy \end{aligned}

This is an even function so we can remove the absolute value and multiply by 2.

\begin{aligned} P(X \leq |Y|t) & = \frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{- \frac{1}{2} y^2} \Phi(t y) \ dy \end{aligned}

The PDF is the derivative of the CDF, so we can take the derivative of the above equation. Under mild technical conditions we can swap the derivative and the integral. Note that $y$ is a dummy variable so here we need to take the derivative w.r.t. $t$

\begin{aligned} PDF = F'(t) & = \frac{d}{dt} \frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{- \frac{1}{2} y^2} \Phi(t y) \ dy \\ & = \frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{- \frac{1}{2} y^2} \ \frac{d}{dt} \Phi(t y) \ dy \end{aligned}

To compute the derivative use the chain rule: $\frac{d}{dt} \Phi(t y) = \frac{d \Phi(t y) }{d(ty)} \frac{d (t y) }{d(t)} = y \frac{d \Phi(t y) }{d(ty)} = y \phi(ty)$ where $\phi(ty)$ is the normal pdf of $(ty)$

\begin{aligned} F'(t) & = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} y \ e^{- \frac{1}{2} y^2} \ \phi(t y) \ dy \\ & = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} y \ e^{- \frac{1}{2} y^2} \frac{1}{\sqrt{2 \pi}} e^{- \frac{t^2y^2 }{2}} \ dy \\ & = \frac{1}{\pi} \int_{0}^{\infty} y e^{ - \frac{1}{2}(1 + t^2)y^2 } dy \end{aligned}

Which can be solved using substitution $u = \frac{1}{2}(1 + t^2)y^2$ so $du = y (1 + t^2) dy$

\begin{aligned} F'(t) & = \frac{1}{\pi} \int_{0}^{\infty} y e^{ - \frac{1}{2}(1 + t^2)y^2 } dy \\ & = \frac{1}{\pi(1 + t^2)} \int_{0}^{\infty} e^{-u} \ du \\ & = \frac{1}{\pi(1 + t^2)} \ \ \forall t \end{aligned}

Additional detail about the symmetry of $N(0,1)$ - Taken from here

The probability $P( X/Y\leqslant t)$ may be viewed as \begin{aligned} &P(\lbrace X/Y\leqslant t, Y>0 \rbrace\cup\lbrace X/Y\leqslant t, Y<0 \rbrace) \\ =\; &P(X/|Y|\leqslant t, Y>0) + P(-X/|Y|\leqslant t, Y<0)\\ =\; &P(X/|Y|\leqslant t, Y>0) + P(X/|Y|\leqslant t, Y<0)\\ =\; &P(X/|Y|\leqslant t) \end{aligned} by symmetry of the standard normal distribution, independence and the fact that $P(Y=0)=0$.

$\endgroup$
  • 1
    $\begingroup$ See this post for intermediate details $\endgroup$ – Harry49 May 30 '18 at 14:57
  • $\begingroup$ Great thanks ! I was also uncertain about the symmetry part. Will edit my post to add the intuition. $\endgroup$ – Xavier Bourret Sicotte May 30 '18 at 15:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.