Finding distribution functions of exponential random variables Find the distribution functions of X+Y/X and X+Y/Z, given that X, Y, and Z have a common exponential distribution.
I think the main thing is that I wanted to confirm the distribution I got for X+Y. I'm doing the integral, and my calculus is a little rusty. I'm getting -e^-ax - ae^-as with parameters x from -infinity to infinity.
From there presumably I can just treat X+Y like one variable and then divide by z.
Thanks so much!
 A: For $t>1$, the distribution function of $(X+Y)/X$ is given by
\begin{eqnarray*}
P((X+Y)/X < t)
&=&P(Y/(t-1)< X)\cr
&=&\int_0^\infty P(y/(t-1)< X)\ ae^{-ay}\ dy\cr
&=&\int_0^\infty e^{-ay/(t-1)} \ ae^{-ay}\ dy\cr
&=&(t-1)/t.
\end{eqnarray*}
Here, $S:=X+Y$ and $Z$ are independent and the density of 
$S$ is $g(s)=a^2 s e^{-as}$ for $s>0$ and zero otherwise (gamma density). 
Thus, 
for $t>0$, the distribution function of $(X+Y)/Z$ is given by
\begin{eqnarray*}
P(S/Z  < t)
&=&P(S/t< Z)\cr
&=&\int_0^\infty P(s/t< X) \ a^2 s e^{-as}\ ds\cr
&=&\int_0^\infty e^{-as/t} \ a^2se^{-as}\ ds\cr
&=&[t/(t+1)]^2.
\end{eqnarray*}

I want to add a comment on Michael's answer  and your response. I guess that your book gave you a formula for the density of the sum of independent random variables that looks like this: $$f_{X+Y}(s)=\int_{-\infty}^{\infty} f_X(s-y)\ f_Y(y)\  dy.$$
You thought, "both my random  variables are exponential, so I should plug in $a e^{-a(s-y)}\ ae^{-ay}$. "
But the density of an exponential random variable is not $f(y)=a e^{-ay}$, it is $$f(y)=\cases{ae^{-ay} & \text{ if }y>0\cr 0 & \text{ otherwise}.}$$
I often have a hard time convincing students that these two formulas are not the same. The "otherwise zero" part of the formula is crucial.
So the expression $f_X(s-y)f_Y(y)$, to be integrated over all $y$ values, is $a e^{-a(s-y)} a e^{-ay}$ only for those $y$ that satisfy $y>0$ and  $s-y>0$. Otherwise,  $f_X(s-y)f_Y(y)$ is zero.
In particular, the whole integral becomes zero unless $s>0$.
When $s>0$, the terms with exponent $y$ cancel each other, so the integral is quite easy
$$f_{X+Y}(s)=\int_0^s a^2 e^{-as} dy=a^2 s e^{-as}.$$ 
A: If $X$ and $Y$ are independent exponential random variables, then $X+Y$ has a gamma distribution.
Also, your integral shouldn't go from $-\infty$ to $\infty$, because exponentials can't be negative.
A: To answer your question about $X+Y$:  If $X$ and $Y$ are independent exponential random variables, you can check that $X+Y $ has the Gamma distribution by computing the characteristic function of $X+Y$:
$\mathbb{E}[e^{it(X+Y)}] = \mathbb{E}[e^{itX}e^{itY}] = \mathbb{E}[e^{itX}]\mathbb{E}[e^{itY}]$ by independence.  
Then if $X$ and $Y$ are exponential with parameter $\lambda$, you have that
$\mathbb{E}[e^{it(X+Y)}] = (1-\frac{it}{\lambda})^{-1}(1-\frac{it}{\lambda})^{-1} = (1-\frac{it}{\lambda})^{-2}$, and this is the characteristic function of $Gamma(2, \lambda^{-1})$.
A: Thanks so much for your help. I'm still having some trouble, however. Currently for (X+Y)/X I have denoted X+Y = t, adn the distribution of T is \int \boldsymbol{\alpha e^{-\alpha x}(1+\alpha x + \frac{(\alpha x)^2}{2})}. Then doing the integral caculations for the distribution, I get \int \boldsymbol{\alpha e^{-\alpha t x}(1+\alpha t x + \frac{(\alpha t x)^2}{2})}. This seems right but for some reason I'm not getting the answer I'm supposed to. Is it because X+Y and X are related? Would this work for (X+Y)/Z?
Thanks!
