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I'm having a rather difficult time with these types of problems.

Say we are are considering two random variables $X_1$, $X_2$ ~ $N(0,1)$.

We want to compute the distribution of $U=\frac{X_1}{X_1 + X_2}$.

My first attempts at this sought to use the the joint density of $U$, $V=X_2$ and then compute the marginal density. This brings us to a seemingly unsolvable integral. Is there some way to compute this without performing a wildly difficult gaussian integral?

Algebraic
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1 Answers1

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Hints:

  • $U=\frac{1}{1 + Y}$ where $Y=\frac{X_2}{X_1}$
  • Consider the distribution of $\frac1U$
  • If $X_1,X_2$ i.i.d. $N(0,1)$ then $\frac{X_2}{X_1} \sim \operatorname{Cauchy}(0,1)$
  • If $Z \sim \operatorname{Cauchy}(x_0,\gamma)$ then $\frac1Z \sim \operatorname{Cauchy}\left(\frac{x_0}{x_0^2+\gamma^2},\frac{\gamma}{x_0^2+\gamma^2}\right)$
Henry
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  • Ah, this really makes sense. Would distribution of 1+Y just be a Cauchy(1,1)? I'm trying to make an analogy from what I would expect of the normal distribution, but the Cauchy distribution doesn't actually have a mean. – Algebraic Oct 25 '20 at 02:03
  • No, wait... That would just be exactly the same distribution, wouldn't it? Since the support set is all real numbers, the addition of any one number doesn't affect the distribution at all. – Algebraic Oct 25 '20 at 02:19
  • Okay, now I'm like 100% certain that it's X+1~Cauchy(-1,1) given that X~Cauchy(0,1). All of this finally makes sense now. Thank you so much for your answer! – Algebraic Oct 25 '20 at 02:37
  • @Algebraic: $Y=\frac{X_2}{X_1}\sim \operatorname{Cauchy}(0,1)$ so $1+Y \sim \operatorname{Cauchy}(1,1)$. You want $\frac{1}{1+Y}$ – Henry Oct 25 '20 at 09:47
  • Are you sure about that? I did the transformation and found the median to be int_{-infinity}^{-1}. Of course, I understood to introduce the transformations one at a time, then use the inverse property of the Cauchy distribution. – Algebraic Oct 28 '20 at 01:19
  • @Algebraic The median of $Y=\frac{X_2}{X_1}$ is $0$, the median of $1+Y=\frac{X_1+X_2}{X_1}$ is $1$, and the median of $\frac1{1+Y}=\frac{X_1}{X_1+X_2}$ is $\frac12$, and simulation will confirm this – Henry Oct 28 '20 at 02:55