# Distribution of sum of exponentials

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.

• 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.
– whuber
Commented Aug 12, 2019 at 12:28
• 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. Commented Aug 12, 2019 at 21:38
• @whuber I've updated the question specifically mention the more general question. Thank you. Commented Aug 12, 2019 at 21:40
• 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.
– whuber
Commented Aug 12, 2019 at 21:46
– whuber
Commented Aug 12, 2019 at 21:52

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

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.

• You wrote both the question and the answer. What is your point, if I may ask? Commented Oct 14, 2018 at 11:58
• @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. Commented Oct 15, 2018 at 14:31
• @Xi'an, Respectfully, I believe you both asked and answered a question here. Commented Oct 15, 2018 at 17:24