Given two exponentially distributed random variables, show their sum is also exponentially distributed Given two independent exponentially distributed random variables, I want to show their sum is also exponentially distributed.
This is my try, I used convolution. It didn't get me anywhere...

 A: You proof that $f_{X+Y}(t)=\dfrac{\lambda_1\lambda_2}{\lambda_2-\lambda_1}[e^{-\lambda_2t}-e^{-\lambda_1t}]$ in the case $\lambda_2>\lambda_1$.
If $\lambda_2=\lambda_1$, you will obtain $\Gamma(2,\lambda_1)$, which follows easly using moment generating functions.
A: If $\Pr(X>t) = e^{-t}$ for all $t\ge0$ and $\Pr(Y=2X)=1$, then $X$ and $Y$ are exponentially distributed and so is their sum.
At the opposite extreme, you'd have two independent exponentially distributed random variables.  Their sum will never be exponentially distributed.  The convolution you compute gives the density function of the sum only if they are independent.
The simplest case would be $\Pr(X>t) =\Pr(Y>t) = e^{-\lambda t}$ for all $t\ge0$ and $X,Y$ are independent.  Then they both have the same density, $t\mapsto \lambda e^{-\lambda t}$ for $t\ge0$.  The convolution of densities is
\begin{align}
t\mapsto \int_0^t \lambda e^{-\lambda u} \lambda e^{-\lambda(t-u)}\, du = \lambda^2 t e^{-\lambda t}.
\end{align}
The distribution with this density is not an exponential distribution.
Remember that the exponential distribution is memoryless.  Your standing by the road measuring the time between when a car passes and when the next one passes.  The probability that you need to wait another minute does not depend on how long you've waited.  But not suppose you're measuring the time until the $20$th car passes, and there's an average of one per minute.  After $25$ minutes the $20$th car car hasn't passed yet and you don't know whether the first car or the $19$th has passed yet.  Now the probability that the $20$th car comes in the next minute is higher than if you had waited only two minutes so far, because the probability that the first $19$ cars have already passed is much higher than it would be after just two minutes.  So this is not a memoryless distribution.  If the sum of independent exponentially distributed random variables were exponentially distributed, then this distribution would be memoryless.
