# 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.

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 \\ &= \lambda^2 x \text{e}^{-\lambda x} \quad\square \end{align}

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\\ &= \lambda^2 x \text{e}^{-\lambda x} \quad\square \end{align}

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}

• You wrote both the question and the answer. What is your point, if I may ask? – Xi'an Oct 14 '18 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. – SecretAgentMan Oct 15 '18 at 14:31
• @Xi'an, Respectfully, I believe you both asked and answered a question here. – SecretAgentMan Oct 15 '18 at 17:24
• @Xi'an You may wish to read stats.stackexchange.com/help/self-answer – Sycorax Oct 15 '18 at 17:37
• An easier solution would be to use moment generating functions – kjetil b halvorsen Nov 19 '18 at 20:10