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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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You could/should try using the general method expanded here. –  Did Dec 9 '11 at 10:50
    
Why did you erase every description of what you had tried? Now your post runs contrary to explicit recommendations about how to ask questions on the site... –  Did Dec 9 '11 at 10:57
    
I think I have worked out the question by using multivariable transfromation. And I just want to check if my answer is correct. –  John Dec 9 '11 at 11:02
    
If that is so, you might want to post your solution as an answer to your own question, then people will be able to check it. This is actually recommended on the site. –  Did Dec 9 '11 at 11:05
1  
Perhaps a further edit is needed. As of right now, the question does not state that $X$ and $Y$ are independent random variables, but the solution included in the question does make the assumption that $X$ and $Y$ are independent. –  Dilip Sarwate Dec 9 '11 at 13:49
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1 Answer

up vote 5 down vote accepted

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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It tooks me half page to work out pdf of Z. But it seems that you can get it directly. Could you explain to me how you do that? –  John Dec 9 '11 at 11:20
    
@ZhouzhouDu: do you know the distribution of $\frac{X}{Y}$ where $X,Y$ are independent standard Gaussian? –  Ilya Dec 9 '11 at 11:31
    
@Ilya Yes. It is standard Cauchy. But X and X+Y are not independent. Is that matters? –  John Dec 9 '11 at 11:35
    
ZhouzhouDu: Quote: If that is so [that is, if you have worked out the question and just want to check if your answer is correct] you might want to post your solution as an answer to your own question, then people will be able to check it. –  Did Dec 9 '11 at 11:43
2  
@ZhouzhouDu: I guess, Didier meant that $X+Y$ and $X-Y$ are independent (since $X+Y$ and $X+Y$ are clearly non-independent), so $$\frac{X}{X+Y} = \frac12\left(1+\frac{X-Y}{X+Y}\right)$$ and $\frac{1}{\sqrt{2}}$ you use in numerator and denominator to normalize them and make standard Guassian and use the fact the quotient of them is Cauchy. Btw, independence of $X-Y$ and $X+Y$ you can just verify by covariation –  Ilya Dec 9 '11 at 11:44
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