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...

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    Why do you want to show this? It's not true: en.wikipedia.org/wiki/Erlang_distribution – Brian Tung Jan 21 '16 at 23:20
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    The physically natural thing that is exponentially distributed is the minimum of the two. The sum can't be exponentially distributed, basically because the memoryless property $P(X+Y>t+s \mid X+Y>t)=P(X+Y>s)$ does not hold. Intuitively this is because the stochastic process which corresponds to $X+Y$ is the one where we wait for the event associated to $X$ to occur, then wait for the event associated to $Y$ to occur, then add up the times we spent waiting. So it "knows" whether we are currently waiting for $X$ vs. currently waiting for $Y$. – Ian Jan 21 '16 at 23:23
  • Wierd, I think my teacher gave this as an exercise... Thank you! – Whyka Jan 21 '16 at 23:36
  • Your dark blue expression for the density of the sum looks correct when $\lambda_1 \not = \lambda_2$. As @Ian says, your target expression in light blue is the density of the minimum. – Henry Jan 22 '16 at 7:36
up vote 1 down vote accepted

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.

  • Great answer, thanks a lot! So, in general- the property of the sum having the same distribution as the independent components is equivalent to that distribution being memoryless? Iff? – Whyka Jan 22 '16 at 0:47
  • I also find it useful to think of sum of two i.i.d. exponentials as a waiting time until the first count of a Poisson process with intensity $\lambda -\frac {\lambda}{1+\lambda t}=\lambda\frac {\lambda t}{1+\lambda t}$ which shows transition to a homogeneous Poisson for large $t$. – A.S. Jan 22 '16 at 0:49
  • @Whyka : No: The distribution of either $X$ or $Y$ is memoryless in the example worked above, but the distribution of $X+Y$ is not memoryless. Nor is the distribution of the sum of $20$ of these, as the last paragraph explains. $\qquad$ – Michael Hardy Jan 22 '16 at 1:43
  • So, just to make sure I understood you: "if the sum were exponentially distributed, this distribution would be memoryless" is true only because we are dealing with this specific distribution? – Whyka Jan 22 '16 at 1:50
  • It's true because the exponential distribution is memoryless. $\qquad$ – Michael Hardy Jan 22 '16 at 1:57

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.

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