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I am stuck with a question doing one of my stats tutorial and question is as follows:

Suppose X and Y are two independent exponential random variables with parameter $\theta$, i.e. their joint probability density function is

$f(x,y; \theta) = \frac{1}{\theta^2}e^\frac{-(x+y)}\theta, x\geq0, y\geq0$

where parameter $\theta >0$.

Find the probability density function of $Z = \frac{X}{X+Y}$

Can anyone kindly guide me with this question please? I'm not exactly sure how to begin. Thank you!

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    $\begingroup$ The section 'Dependent variables and change of variables' in the Wikepedia article about probability density functions will be helpful. $\endgroup$ – Nussig Sep 15 '18 at 10:14
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    $\begingroup$ You might want to add the self-study tag and read it's wiki. $\endgroup$ – StubbornAtom Sep 15 '18 at 10:17
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The joint density of $(X,Y)$ is of the form

\begin{align} f_{X,Y}(x,y)&=\frac{1}{\theta^2}\exp\left({-\frac{x+y}{\theta}}\right)\mathbf1_{x,y>0}\quad,\,\theta>0 \\&=\frac{1}{\theta}e^{-x/\theta}\mathbf1_{x>0}\frac{1}{\theta}e^{-y/\theta}\mathbf1_{y>0} \\&=f_X(x)f_Y(y)\quad,\text{ say } \end{align}

So $X$ and $Y$ are independent and identically distributed Exponential variables with mean $\theta$.

You seek the distribution of $\frac{X}{X+Y}=Z$ (say).

Among several ways to find the distribution of $Z$, we could find the distribution function (DF) $P(Z\leqslant z)$ of $Z$, or we could use a change of variables (as mentioned in a comment) along the lines of $(X,Y)\to (Z,W)$ such that $Z=\frac{X}{X+Y}$ and $W=X+Y$.

For the DF, we can proceed using the total probability theorem :

For each $z\in(0,1)$,

\begin{align} F_Z(z)&=P\left(\frac{X}{X+Y}\leqslant z\right) \\&=\int P(X\leqslant zX+y\mid Y=y)f_Y(y)\,dy \\&=\int_0^{\infty}P\left (X\leqslant \frac{y}{1-z}\right)f_Y(y)\,dy \\&=\cdots \end{align}

Differentiating $F_Z$ wrt $z$ would yield the density of $Z$.

If you use the second method with that particular transformation, you would find from the joint density of $(Z,W)$ that $Z$ and $W$ are also independently distributed, and finally identify the distribution of $Z$ from the joint density alone.

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  • $\begingroup$ Thank you very much for your answer! This was the solution I was seeking! @StubbornAtom $\endgroup$ – InvadersMustDie Sep 17 '18 at 16:24
  • $\begingroup$ @InvadersMustDie Let me know your final answer. $\endgroup$ – StubbornAtom Sep 17 '18 at 19:07
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    $\begingroup$ it is a uniform distribution U, since the pdf = 1 $\endgroup$ – InvadersMustDie Sep 18 '18 at 18:28
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A common method is to find CDF, then differentiate: $$F_Z(z)=P(Z \leq z) = P(\frac{X}{X+Y}\leq z) = P(\frac{1-z}{z}X\leq Y)$$ which is 1 outside $0 \leq z \leq 1$.Let $\alpha=\frac{1-z}{z}$, and we seek for $P(\alpha X\leq Y)$, where $\alpha \geq 0$. In the 2D plane, you'll draw the line $y=\alpha x$, and integrate the joint PDF in the area in-between the line and the x-axis; then substitute $z$, and differentiate with respect to it.

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