# Gamma Distribution out of sum of exponential random variables

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?

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

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 $$b = \lambda$$.
• 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? Jan 29, 2014 at 21:01
• @heropup Could you please let me know how to find the PDF of the sum of $n$ (independent non-identically distributed) Exponential random variables? Sep 7, 2016 at 21:49