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Show that the random variable $U=\frac{X}{X+Y}$ has a uniform distribution on the range [0,1] when X and Y are independent random variables with the same exponential distribution.

I'm stuck at

$F_U\left(u\right)=P\left(U\le u\right)=P\left(\frac{X}{X+Y}\le u\right)$

Even if I assume that X=1 I get a divergent integral in the end

And If I assume that

$P\left(0\le \frac{X}{X+Y}\le 1\right)=P\left(0\le \:X\le \:X+Y\right)=P\left(-X\le Y\right)=1-P\left(Y\le X\right)$

Then I still get an divergent integral

So what should I do?

StubbornAtom
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Student
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1 Answers1

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You want to prove $P(U\le u)=u$ for $u\in[0,\,1]$. Indeed,$$\begin{align}P(X\le uX+uY)&=P(X\le\tfrac{uY}{1-u})\\&=\int_0^\infty\lambda e^{-\lambda y}(1-e^{-\lambda uy/(1-u)})dy\\&=1-\int_0^\infty\lambda e^{-\lambda y/(1-u)}dy\\&=1-(1-u)\\&=u.\end{align}$$You can also do it by noting the problem is equivalent to $Z:=Y/X$ having PDF $\frac{d}{dz}[1-(1+z)^{-1}]=(1+z)^{-2}$ on $\Bbb R^+$, because $U=1/(1+Z)$. Indeed, $Z$'s PDF is$$\int_0^\infty x\lambda^2e^{-\lambda(1+z)x}dx=(1+z)^{-2}.$$

J.G.
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  • Why $\int_0^\infty\lambda e^{-\lambda y}(1-e^{-\lambda uy/(1-u)})dy\$? I don't get it. I mean, shouldn't it be $\int :_0^{\frac{uY}{1-u}}\lambda e^{\left(-\lambda x\right)}dx$? – Student Jun 16 '20 at 17:22
  • @Student My approach was to average the probability over all values of $Y$. What you've suggested as an alternative is a random variable, a function of $Y$. If in your integral's limit you replace $Y$ with $y$, you get the bracketed factor I've averaged. – J.G. Jun 16 '20 at 17:26
  • If I calculate $\int :_0^{\frac{uY}{1-u}}\lambda e^{\left(-\lambda x\right)}dx$ I get $1-\mathrm{e}^\frac{\lambda uy}{u-1}$, so what am I supposed to do with this? – Student Jun 16 '20 at 17:28
  • @Student Presumably, $c=-\lambda$. Note you're still confusing the random variable $Y$ with a constant $y$ to which $Y$ can be compared. The correct calculation can be thought of as $\Bbb E(1-e^{-\lambda uY/(1-u)})$, which is the integral I evaluated. – J.G. Jun 16 '20 at 17:30
  • Could you tell me from where did you got this formula for this whole thing? Where can i find more information on this topic? – Student Jun 16 '20 at 17:33
  • @Student You can learn more about it here. But it's just a more complicated variant of the discrete case. – J.G. Jun 16 '20 at 17:35
  • I see it now, it's the expected value you calculate. But why is there this kind of connection? – Student Jun 16 '20 at 17:35
  • @Student Say I toss a dice, and based on its result I toss one of several coins that each have a different bias. Then the probability that I get a head is a mean (weighted if the dice isn't fair) of the coins' head probabilities. What we're doing is a continuous version of that, where the mean comes from an integral. It'll make more sense if you read my links. – J.G. Jun 16 '20 at 17:37
  • Hmm I kind of understand the intuition, because here in this case we have two random variables thus two let's call them "outcomes" (the dice and coin). But why should the expected value be the distribution here in this place? We are looking for the distribution of X <= ..., so if at all shouldn't it be the probability of what number I get when rolling a dice? – Student Jun 16 '20 at 17:43
  • @Student You misunderstand the analogy. Each value of $Y$ gives a different value for $P(X\le uY/(1-u))$, as if a different coin were tossed to determine whether that inequality holds. So it's like $Y$ is a dice with infinitely many sides. – J.G. Jun 16 '20 at 17:49
  • What about: "The random variables X and Y are independent and have the same normal distribution N(0, 1). Find the distribution of random variable Z=sqrt(X^2 + Y^2)". I tried it with your method, so I got $P\left(X\le \sqrt{z^2-Y^2}\right)$ which should get me the integral $\int :_{-\infty }^{\infty }\dfrac{\sqrt{z^2-y^2}\mathrm{e}^{-\frac{y^2}{2}}}{\sqrt{2}\sqrt{{\pi}}}dy$ but it's uncomputable according to WolframAlpha – Student Jun 16 '20 at 18:43
  • @Student You mean $P(-\sqrt{z^2-Y^2}\le X\le\sqrt{z^2-Y^2})$. The surd in your integrand is also wrong; the correct integral is$$\int_{-\infty}^\infty\frac{e^{-y^2/2}}{\sqrt{2\pi}}\int_{-\sqrt{z^2-y^2}}^{\sqrt{z^2-y^2}}\frac{e^{-x^2/2}}{\sqrt{2\pi}}dxdy.$$It is, however, much easier to solve in polar coordinates. See here if you're curious what we get. – J.G. Jun 16 '20 at 19:20