I came across this example, and I was completly thrown off! Hopefully someone will be able to help me start it or even explain it all to me! Thanks for any help!

X and Y are two independent random variables with exponential distribution and parameter $\lambda = 1$

Suppose that $Z=\frac{X}{X+Y}$ .

a)Explain why ImZ = (0,1)

b) By computing Fz(z) = P(Z ≤ z), show that the random variable Z has a uniform distribution on the interval (0, 1)

I think maybe I have to use something like the Convolution Formula for b) but not sure how to

  • $\begingroup$ Have you done (a), and only are looking for help for (b)? $\endgroup$ – Clement C. Nov 11 '15 at 15:39
  • $\begingroup$ I am confused on both! so help on either is appreciated! $\endgroup$ – boabysands123 Nov 11 '15 at 15:41
  • $\begingroup$ Hint for a: given $x>0$ and $0<z<1$ can you find $y>0$ such that $\frac{x}{x+y}=z$? (This is really just algebra.) For b, call the $y$ that you just solved for $h(x,z)$. Then because of monotonicity, $\frac{X}{X+Y} \leq z$ if and only if $Y \geq h(X,z)$. So that tells you how to set up the integration, as $\int_0^\infty \int_{h(x,z)}^\infty f(x,y) dy dx$, where $f$ is the joint pdf. The rest is calculus. $\endgroup$ – Ian Nov 11 '15 at 15:42

For (a), it suffices to prove that for $x,y > 0$, you have $0 < \frac{x}{x+y} < 1$. This should be straightforward.

For (b): after (a), you can assume $z\in(0,1)$. Rewrite $$\begin{align} \mathbb{P}\{Z \leq z\} &= \mathbb{P}\left\{X \leq \frac{z}{1-z} Y\right\} = \int_{0}^\infty\int_{0}^\infty \mathbb{1}_{\{x \leq \frac{z}{1-z}y\}} dydx e^{-y}e^{-x} \\ &= \int_{0}^\infty dy e^{-y} \int_{0}^{\frac{z}{1-z}y}dx e^{-x} \end{align}$$ and you can compute this last quantity by direct integration: also, to conclude recall that for a uniform distribution, $F(z) = z$ for $z\in(0,1)$.


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