If $X_1$ and $X_2$ are two independent random variables having exponential densities then $f(x_1,x_2)$ is defined as $$f(x_1,x_2)=\exp(-(x_1+x_2))\,{\bf 1}_{(0,\infty)}(x_1){\bf 1}_{(0,\infty)}(x_2).$$ we have to prove the distribution of $z$ is uniform $f(z)=1$, for$\ 0<z<1$ using distribution function technique, where $z=x_1/(x_1+x_2)$.

To solve this I proceed as

$$F(z)=\int_0^{z}\int_0^{x_1(1-z)/z}\exp(-(x_1+x_2))\,dx_2\, dx_1$$ as answer should be $z$, but I cant get this. Can I determine integral limits correctly?

  • 1
    $\begingroup$ what is it exactly you want to prove? you have the product of two iid Exp(1) random variables, so far so good. Then you say the distribution of $z$, what is $z$? $\endgroup$ – user190080 Aug 1 '15 at 14:37
  • $\begingroup$ It is written in the question. z=x1/(x1+x2). $\endgroup$ – Joseph Aug 1 '15 at 14:44
  • $\begingroup$ @Joseph $Z$ is supposed to be a RV and I'd like to know how it's defined. So you say it is actually $Z=X_1/(X_1+X_2)$? $\endgroup$ – user190080 Aug 1 '15 at 15:05

For the sake of my comfort, I will use $X$ and $Y$ instead of $X_1$ and $X_2$. We want the cumulative distribution function of $Z=\frac{X}{X+Y}$.

Let $z$ be between $0$ and $1$. We want $\Pr(Z\le z)$, which is the probability that $X\le z(X+Y)$, or equivalently the probability that $Y\ge \frac{1-z}{z}X$.

So we want to calculate the probability of landing in the part of the first quadrant that is above the line $y=\frac{1-z}{z}x$. This is the key geometric fact.

Integrate first with respect to $y$, where $y$ goes from $\frac{1-z}{z}x$ to $\infty$. There is some cancellation, and we get $e^{-x/z}$.

Now integrate with respect to $x$, from $0$ to $\infty$. An antiderivative is $-ze^{-x/z}$, so the integral is $z$.


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