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X, Y are independent standard normal random variables, what is the distribution of $ \frac{X}{X+Y} $

Could anyone help me with this? Thanks.


I have worked the problem by multivariable transformation:

Let $Z=\frac{X}{X+Y} , W=X$

Consider transformation $(X,Y)\longrightarrow(Z,W)$

Then $X(Z,W)=W , Y(Z,W)=\frac{W(1-Z)}{Z}$ defines the inverse transformation.

The Jacobian is $J(Z,W)=\frac{w}{z^{2}} $

So $f_{Z,W}(z,w)=f_{X,Y}(w,\frac{w(1-z)}{z})\cdot\mid\frac{w}{z^{2}}\mid$

As X and Y are independent. Then the marginal pdf of Z is $f_{Z}(z)=\intop_{0}^{\infty}\frac{w}{z^{2}}\cdot f_{X}(w)\cdot f_{Y}(\frac{w(1-z)}{z})dw+\intop_{-\infty}^{0}-\frac{w}{z^{2}}\cdot f_{X}(w)\cdot f_{Y}(\frac{w(1-z)}{z})dw$ After calculation we get $f_{Z}(z)=\frac{1}{\pi\cdot\frac{1}{2}\cdot(1+(\frac{z-\frac{1}{2}}{\frac{1}{2}})^{2})}$

Hence $Z\sim \mathrm{Cauchy}(\frac{1}{2},\frac{1}{2}).$

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    The $f_Z(z)$ you have found is correct. If you simplify your expression a little, you will see that it is same as as the density given in Didier Piau's answer.2011-12-09

1 Answers 1

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Since $X$ and $Y$ are independent standard gaussian random variables, the distribution of $Z=\frac{X}{X+Y}$ has density $f_Z$, where for every $z$ in $\mathbb R$, $ \color{red}{f_Z(z)=\frac1\pi\,\frac1{z^2+(1-z)^2}}. $ The direct way to prove this (as is now done by the OP) is to rely on the change of variables method expanded here.

One can deduce from the expression of $f_Z$ that $Z=\frac12(1+T)$, where $T$ is standard Cauchy, that is, the distribution of $T$ has density $f_T$, where for every $t$ in $\mathbb R$, $ \color{purple}{f_T(t)=\frac1\pi\,\frac1{1+t^2}}. $ But the formulas for $f_Z$ and $f_T$ are also direct consequences of two facts:

  1. The ratio of two independent standard gaussian random variables is a standard Cauchy random variable.

  2. If $X$ and $Y$ are independent standard gaussian random variables, then the random variables $\frac1{\sqrt2} (X+Y)$ and $\frac1{\sqrt2}(X-Y)$ are independent standard gaussian random variables as well.

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    @Dilip: long story short 1) Didier puts an answer 2) Zhouzhou asks for explanations 3) I give them 4) Didier prods Zhouzhou 5) Zhouzhou puts his solution ... n) your first comment :) hope it's clear now. And finally, do you know that writing here in the comments you're pinging Didier? I'll be happy to know why are you so interested in this.2011-12-09