I have a sequence $T_1,T_2,\ldots$ of independent exponential random variables with paramter $\lambda$. I take the sum $S=\sum_{i=1}^n T_i$ and now I would like to calculate the probability density function.

Well, I know that $P(T_i>t)=e^{-\lambda t}$ and therefore $f_{T_i}(t)=\lambda e^{-\lambda t}$ so I need to find $P(T_1+\cdots+T_n>t)$ and take the derivative. But I cannot expand the probability term, you have any ideas?

  • 1
    $\begingroup$ it's called Erlang distribution $\endgroup$
    – Alex
    Jan 29, 2014 at 0:14
  • $\begingroup$ $P(T_1+...+T_n>t)$ is $1-F_S(t)$ i.e. it relates to the cumulative distribution function, not to the density. Different degrees of computational difficulty. $\endgroup$ Jan 29, 2014 at 0:26
  • $\begingroup$ math.stackexchange.com/q/474775/321264 $\endgroup$ Apr 27, 2020 at 15:50

1 Answer 1


The usual way to do this is to consider the moment generating function (MGF). We note that if $S = \sum_{i=1}^n X_i$ is the sum of independent and identically distributed (IID) random variables $X_i$, each with MGF $M_X(t)$, then the MGF of $S$ is $M_S(t) = (M_X(t))^n$. Applied to the exponential distribution, we can get the gamma distribution as a result.

If you don't go the MGF route, then you can prove it by induction. Consider the simple case of the sum of a gamma-distributed random variable and an exponential-distributed random variable, both of which have the same rate parameter. Suppose $Y \sim {\rm Gamma}(a,b)$ and $X \sim {\rm Exponential}(b)$ are independent. Their probability density functions (PDF) are given, respectively, as $$ f_Y(y) = \frac{b^a y^{a-1} e^{-by}}{\Gamma(a)} \mathbb 1 (y > 0), \quad f_X(x) = be^{-bx} \mathbb 1 (x > 0), $$ where $a, b > 0$ and $\mathbb 1$ is the usual indicator function. Note that if $a = 1$, $Y$ is exponentially distributed (i.e., the exponential distribution is a special case of the Gamma with $a = 1$). The PDF of $Z = X+Y$ is given by $$\begin{align*} f_Z(z) &= \int_{y=0}^z f_Y(y) f_X(z-y) \, dy, \end{align*}$$
hence $$\begin{align*} f_Z(z) &= \int_{y=0}^z \frac{b^a y^{a-1} e^{-by}}{\Gamma(a)} \cdot be^{-b(z-y)} \, dy \\ &= \int_{y=0}^z \frac{b^{a+1} y^{a-1} e^{-by} e^{-b(z-y)}}{\Gamma(a)} \, dy \\ &= \frac{b^{a+1} e^{-bz}}{\Gamma(a)} \int_{y=0}^z y^{a-1} \, dy \\ &= \frac{b^{a+1} e^{-bz}}{\Gamma(a)} \frac{z^a}{a} \\ &= \frac{b^{a+1} z^a e^{-bz}}{\Gamma(a+1)}, \end{align*}$$
which is the PDF of a gamma distributed random variable with shape parameter $a^*= a+1$. By induction, one finds that the sum of $n$ IID exponential-distributed random variables with common rate parameter $\lambda$ is a gamma-distributed random variable with shape parameter $a = n$, and rate parameter $b = \lambda$.

  • $\begingroup$ Thanks for your answer, according to your first part concerning the MGF, I have some difficuties with this, the MGF for each of the $T_i$ is $M_T(x)=(1-\frac{x}{\lambda})^{-1}$, therefore $M_S(x)=(1-\frac{x}{\lambda})^{-n}$, but how to derive $f_S(x)$ now? $\endgroup$
    – TI Jones
    Jan 29, 2014 at 21:01
  • $\begingroup$ Take the PDF of a gamma distribution, and calculate its MGF. When you see it has the same form, it follows from the uniqueness of MGFs that the sum of exponential RVs is therefore gamma distributed. $\endgroup$
    – heropup
    Jan 29, 2014 at 21:10
  • $\begingroup$ @heropup Could you please let me know how to find the PDF of the sum of $n$ (independent non-identically distributed) Exponential random variables? $\endgroup$
    – sky-light
    Sep 7, 2016 at 21:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .