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This is a question on one of final exam at my school...:

X and Y are independent Gaussian random variables with zero mean and unit variance. Z=X/(X+Y). Find the pdf of Z.

After some struggles, I think I can solve this question by letting W = X+Y first and derive the pdf of W. Then I can find the pdf of Z = X/W.

However, after reading these two wikipedia articles:

https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables

https://en.wikipedia.org/wiki/Ratio_distribution.

I realized that the whole answer is going to be very long.

Meanwhile, I came across this posting - Finding the pdf of $(X+Y)^2/(X^2+Y^2)$ where $X$ and $Y$ are independent and normal, which is similar in nature.

It appears that what are shown on the wikipedia articles are the only way to go.

However, considering this is a final exam question, given a short time limit, there should probably a much shorter way to reach my answer. Can anyone please show me the "easier" way?

StubbornAtom
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Amos Ku
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  • Instead of what you have done, you can change variables $(X,Y)\to \left(\frac{X}{X+Y},X+Y\right)$ and hence derive the marginal distribution of $\frac{X}{X+Y}$. Alternatively, note that $\frac{X}{X+Y}=\frac1{1+\frac{Y}{X}}$ where $Y/X$ has a standard Cauchy distribution. – StubbornAtom Apr 08 '20 at 06:51

2 Answers2

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We know that $$\frac{X}{X+Y}=1-\frac{Y}{X+Y}.$$ Note that the last term is simply the equivalent of the first in $Y$. Can you now solve the problem, given this insight?

Also, just to correct a small misconception; for us to calculate the ratio distribution, the variables must be independent. With your proposed method $(Z=X+Y)$, they are not, so we cannot use that method.

gt6989b
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Arunav MAHESHWARI
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  • +1 good eye. Welcome to Math.SE, hope you stay around and contribute to the site. – gt6989b Apr 08 '20 at 02:55
  • If you read the proof, the assumption is "For independent random variables X and Y, the distribution fZ of Z = X + Y equals the convolution of fX and fY: ". Why then they are not independent? Please advise. I am confused. – Amos Ku Apr 08 '20 at 03:57
  • As for your answer, it seems more confusing to me, since Y/(X+Y) is not much different to me as X/(X+Y). Can you show me the details? Thank you. (I am just a little better than a beginner at this point.) – Amos Ku Apr 08 '20 at 03:59
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I think I had missed the given conditions that $\mu_{x}=\mu_{y}=0$ and that $\sigma_{x}=\sigma_{y}=1$.

I realized if I take these into account, I will come out with

$f_{W}(w)=\frac{1}{\sqrt{2\pi }}e^{-\frac{w^{2}}{2}}$

for pdf of W.

Then, continue working with reference on the ratio distribution derivations I will have

$f_{Z}(z)=\frac{1}{\pi (z^{2}+1)}$

Therefore, the simplification comes if I had remembered to substitute in $\mu_{x}=\mu_{y}=0$ and that $\sigma_{x}=\sigma_{y}=1$

Amos Ku
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  • This is incorrect, as you would find out if you take a look at the comments below your post. Your method relies on finding the distribution of $X/W$ where $(X,W)$ is jointly normal (not independent) and $W=X+Y\sim N(0,2)$. But this is not what you have done. – StubbornAtom Apr 08 '20 at 07:10