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

John
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  • You could/should try using the general method expanded here. – Did Dec 09 '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 09 '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 09 '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 09 '11 at 11:05
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    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 09 '11 at 13:49
  • 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. – Dilip Sarwate Dec 09 '11 at 15:28
  • Related: https://math.stackexchange.com/q/3769136/321264. – StubbornAtom Oct 25 '20 at 07:41

1 Answers1

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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.

Community
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Did
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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 09 '11 at 11:20
  • @ZhouzhouDu: do you know the distribution of $\frac{X}{Y}$ where $X,Y$ are independent standard Gaussian? – SBF Dec 09 '11 at 11:31
  • @Ilya Yes. It is standard Cauchy. But X and X+Y are not independent. Is that matters? – John Dec 09 '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 09 '11 at 11:43
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    @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 – SBF Dec 09 '11 at 11:44
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    @Ilya It seems to me that since $X-Y$ and $X+Y$ are independent Gaussian random variables with the same variance, their ratio is a standard Cauchy random variable: it is not necessary to "normalize" them to be standard Gaussian random variables. What am I missing? In fact, I wish Didier would re-word his argument to say that $Z=(1/2)\left(1+\frac{X-Y}{X+Y}\right)=(1/2)(1+T)$ and hence $f_Z(z)=1/(0.5)f_T((z-0.5)/0.5)=(1/\pi)(z^2+(z-1)^2)^{-1}$ which is more obvious (to me, at least) than simply stating the density of $Z$ and saying hence $Z=(1/2)(1+T)$ where $T$ is Cauchy. – Dilip Sarwate Dec 09 '11 at 14:12
  • @Dilip: you're missing nothing. However, Zhouzhou asked for explanations and I explained Didier's answer to him (where $\frac{1}{\sqrt2}$ was mentioned and might be unclear for Zhouzhou); I don't know Didier's reasoning, but the question has [tag:homework] so giving a complete solution is not considered to be appropriate. – SBF Dec 09 '11 at 14:25
  • @Ilya "but the question has homework so giving a complete solution is not considered to be appropriate." OK. In this case, though, Zhouzhou has provided a correct solution in the question itself (after prodding by Didier), and I feel there is no harm in helping him understand Didier's more elegant approach. – Dilip Sarwate Dec 09 '11 at 15:32
  • @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. – SBF Dec 09 '11 at 15:37