Let $X_1$ and $X_2$ be independent and identically distributed exponential random variables with rate $\lambda$. Let $S_2 = X_1 + X_2$.

Q: Show that $S_2$ has PDF $f_{S_2}(x) = \lambda^2 x \text{e}^{-\lambda x},\, x\ge 0$.

Note that if events occurred according to a Poisson Process (PP) with rate $\lambda$, $S_2$ would represent the time of the 2nd event.

Alternate approaches are appreciated. The approaches provided are commonly used when learning queueing theory & stochastic processes.

Recall the Exponential distribution is a special case of the Gamma distribution (with shape parameter $1$). I've learned there is a more general version of this here that can be applied.

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    $\begingroup$ This question is a very special case (and one of the simplest possible examples) of a sum of Gamma distributions. (The Exponential is a Gamma distribution with a shape parameter of $1.$) Thus, you could apply any of the answers at stats.stackexchange.com/questions/72479. $\endgroup$
    – whuber
    Commented Aug 12, 2019 at 12:28
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    $\begingroup$ Thank you. I was unaware of that more general question, though I did know the Exponential is a Gamma distribution with a shape parameter of 1. I hope you'll agree this Q/A is ok as-is and shouldn't be deleted. This is a very frequent question in some engineering disciplines and is certainly more accessible than jumping straight into adding Gamma distributions. $\endgroup$ Commented Aug 12, 2019 at 21:38
  • $\begingroup$ @whuber I've updated the question specifically mention the more general question. Thank you. $\endgroup$ Commented Aug 12, 2019 at 21:40
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    $\begingroup$ For the reasons you gave, and because you have offered a clear account of solutions that work specifically in this case, I have not voted to close this as a duplicate. $\endgroup$
    – whuber
    Commented Aug 12, 2019 at 21:46
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    $\begingroup$ I think the voting on your question and your answer has clearly indicated what the community thinks of this thread. :-) $\endgroup$
    – whuber
    Commented Aug 12, 2019 at 21:52

1 Answer 1


Conditioning Approach
Condition on the value of $X_1$. Start with the cumulative distribution function (CDF) for $S_2$.

$\begin{align} F_{S_2}(x) &= P(S_2\le x) \\ &= P(X_1 + X_2 \le x) \\ &= \int_0^\infty P(X_1+X_2\le x|X_1=x_1)f_{X_1}(x_1)dx_1 \\ &= \int_0^x P(X_1+X_2\le x|X_1=x_1)\lambda \text{e}^{-\lambda x_1}dx_1 \\ &= \int_0^x P(X_2 \le x - x_1)\lambda \text{e}^{-\lambda x_1}dx_1 \\ &= \int_0^x\left(1-\text{e}^{-\lambda(x-x_1)}\right)\lambda \text{e}^{-\lambda x_1}dx_1\\ &=(1-e^{-\lambda x}) - \lambda x e^{-\lambda x}\end{align} $

This is the CDF of the distribution. To get the PDF, differentiate with respect to $x$ (see here).

$$f_{S_2}(x) = \lambda^2 x \text{e}^{-\lambda x} \quad\square$$

This is an Erlang$(2,\lambda)$ distribution (see here).

General Approach
Direct integration relying on the independence of $X_1$ & $X_2$. Again, start with the cumulative distribution function (CDF) for $S_2$.

$\begin{align} F_{S_2}(x) &= P(S_2\le x) \\ &= P(X_1 + X_2 \le x) \\ &= P\left( (X_1,X_2)\in A \right) \quad \quad \text{(See figure below)}\\ &= \int\int_{(x_1,x_2)\in A} f_{X_1,X_2}(x_1,x_2)dx_1 dx_2 \\ &(\text{Joint distribution is the product of marginals by independence}) \\ &= \int_0^{x} \int_0^{x-x_{2}} f_{X_1}(x_1)f_{X_2}(x_2)dx_1 dx_2\\ &= \int_0^{x} \int_0^{x-x_{2}} \lambda \text{e}^{-\lambda x_1}\lambda \text{e}^{-\lambda x_2}dx_1 dx_2\\ \end{align}$

Since this is the CDF, differentiation gives the PDF, $f_{S_2}(x) = \lambda^2 x \text{e}^{-\lambda x} \quad\square$ Figure

MGF Approach
This approach uses the moment generating function (MGF).

$\begin{align} M_{S_2}(t) &= \text{E}\left[\text{e}^{t S_2}\right] \\ &= \text{E}\left[\text{e}^{t(X_1 + X_2)}\right] \\ &= \text{E}\left[\text{e}^{t X_1 + t X_2}\right] \\ &= \text{E}\left[\text{e}^{t X_1} \text{e}^{t X_2}\right] \\ &= \text{E}\left[\text{e}^{t X_1}\right]\text{E}\left[\text{e}^{t X_2}\right] \quad \text{(by independence)} \\ &= M_{X_1}(t)M_{X_2}(t) \\ &= \left(\frac{\lambda}{\lambda-t}\right)\left(\frac{\lambda}{\lambda-t}\right) \quad \quad t<\lambda\\ &= \frac{\lambda^2}{(\lambda-t)^2} \quad \quad t<\lambda \end{align}$

While this may not yield the PDF, once the MGF matches that of a known distribution, the PDF also known.

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    $\begingroup$ You wrote both the question and the answer. What is your point, if I may ask? $\endgroup$
    – Xi'an
    Commented Oct 14, 2018 at 11:58
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    $\begingroup$ @Xi'an, I thought SE encouraged asking the question and answering it...I can screenshot where SE seems to encourage that for you if you want. I've seen a lot of basic questions repeatedly asked and I've been thinking about posting some specific approaches to refer people to. I wasn't able to find something like this and I can refer people to this for a variety of things. If the CV community really hates this post that much, I will voluntarily delete it. $\endgroup$ Commented Oct 15, 2018 at 14:31
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    $\begingroup$ @Xi'an, Respectfully, I believe you both asked and answered a question here. $\endgroup$ Commented Oct 15, 2018 at 17:24
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    $\begingroup$ @Xi'an You may wish to read stats.stackexchange.com/help/self-answer $\endgroup$
    – Sycorax
    Commented Oct 15, 2018 at 17:37
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    $\begingroup$ @Alex good question. I wasn't thinking about getting PDF analytically from MGF. Instead, if you identify the MGF, then you've solved the problem (see my edit). $\endgroup$ Commented May 24, 2020 at 12:59

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