Let $Z=X/(X+Y),$ find the pdf of $Z$.

Question

Let $X$ and $Y$ be independent and $\text{exp}(\lambda)$, and let $Z=X/(X+Y),$ find the pdf of $Z$.

1) How can one draw the conclusion that the range of $Z$ is $[0,1]$?

Let's take a $z$ in this range and consider the cdf

$$F_Z(z)=P(Z\le z)=P\left(\frac{X}{X+Y}\le z\right)=P\left(\frac{X}{X+Y}\le z\right)=P\left(Y\geq\left(\frac{1}{z}-1\right)X\right).$$

With $a=1/z-1$ we have that $F_Z(z)=P(Y\ge aX).$

2) Can someone intuitively and otherwise, in detail, explain to me why computing $P(Y\ge aX)$ is equivalent to integrating the joint pdf over the region $B=\{(x,y):y\ge ax\}$?

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EDIT 1: With the help of the comments and answers thus far, I now draw the region on the xy-plane:

So obviously $-\infty \le x\le \infty $ and $ax\le y \le \infty,$ so

$$P(Y\ge aX)=\int_{-\infty}^{\infty}\int_{ax}^{\infty}\lambda e^{-\lambda y}\lambda e^{-\lambda x} \ dy \ dx.$$

I can't evaluate this integral and the book says that the lower bound for the outer integral is $0$, I don't agree because clearly $x$ can be negative.

Answer 1

1) $\mathbb{P}(Z<0)=\mathbb{P}(X<0,X+Y>0)+\mathbb{P}(X>0,X+Y<0)=0+0=0$.

$\mathbb{P}(Z>1)=\mathbb{P}(X>X+Y)=\mathbb{P}(0>Y)=0$.

Then $\mathbb{P}(Z\in[0,1])=1$.

2) Is by definition. Note that $(X,Y)\in B \iff Y\geq aX$

Answer 2

1) The samples of $\exp(\lambda)$ are almost surely positive: $X \leq X+Y \Leftrightarrow \frac{X}{X+Y} \leq 1$. 2) Let $f(x,y)$ be the joint pdf of $(X,Y)$. Then: $$P(Y \geq aX) = \iint_{\{Y \geq aX\}} f(X,Y) \mathrm{d}X\mathrm{d}Y.$$

We define this set by ${\{Y \geq aX\}}:= \{(X,Y): Y \geq a X\}$, since we are dealing with the joint pdf of $(X,Y)$ directly.

Answer 3

In response to your edit:

Why can $x$ be negative?

$X$, just like $Y$, is an exponential random variable so it take values $\ge 0$, i.e., you should only be looking at the first quadrant (that is where $x,y$ are both nonnegative).

Now you're being asked to compute the PDF of Z, so like you did start with its CDF: $F_Z(z)=P(Z \le z)=P(\frac{X}{X+Y} \le z)=P(Y \ge (\frac{1}{z}-1)X) $
Now just integrate over the correct region, $\int_0^\infty \int_0^{\frac{y}{a}}\lambda e^{-\lambda x}\lambda e^{-\lambda y}dxdy$ (if you integrate first with respect to the $x$-axis), or $\int_0^\infty \int_{ax}^{\infty}\lambda e^{-\lambda x}\lambda e^{-\lambda y}dydx$, (if you integrate first with the respect to the $y$-axis).

After integrating you would get the CDF, so then differentiate with respect to z to get the random variable Z's PDF.