# How sum of exponential variables is a gamma variable [duplicate]

I have the task to calculate $P(S_{100}\geq 200)$ where $S_{100}=\sum^{100}_{i=1} X_i$ and $X_i$, $i=1,2, \cdots, 100$ are independent $exp(\lambda)$ random variables. One method is to use the fact that a sum of exponential variables make a gamma random variable. So far I've got $P(S_{100} \geq 200)= P(\sum^{100}_{i=1} X_i \geq 200)= \sum^{100}_{i=1} e^{-\lambda200} \frac{(\lambda200)^j}{j!} \geq 200$

## marked as duplicate by heropup, user91500, choco_addicted, Leucippus, hardmathJun 2 '16 at 16:36

• The probabiilty in question should be an integral, not a sum. – Hans Engler Jun 2 '16 at 12:04
• Ok. That just makes it more confusing for me. – Orongo Jun 2 '16 at 12:13
• Your last equality is wrong. It is not true that $P(\sum X_i)=\sum P(X_i)$. The RHS could be larger than $1$, for starters. – MPW Jun 2 '16 at 12:30
• Find the mean and variance and approximate using the normal distribution, use the central limit theorem. – Aravind Jun 2 '16 at 12:46

If $X_n\stackrel{i.i.d.}\sim\mathsf{Exp}(\lambda)$ and $S_n=\sum_{k=1}^n X_k$, then $S_n$ has an Erlang distribution with parameters $n$ and $\lambda$ (this is a special case of the gamma distribution). This distribution is characterized by density $$\frac{(\lambda t)^{n-1}}{(n-1)!}\lambda e^{-\lambda t}\mathsf 1_{(0,\infty)}(t). \tag 1$$ Let $f_n$ be the probability density of $S_n$. For $n=1$ we have $$f_1(t) = \lambda e^{-\lambda t} = \frac{(\lambda t)^0}{(1-1)!}\lambda e^{-\lambda t}.$$ Assume now that the density of $S_n$ is given by $(1)$ for some $n\geqslant 1$. Then \begin{align} f_{n+1}(t) &= f_n\star f(t) \\ &= \int_0^t f_n(s)f(t-s)\ \mathsf ds\\ &= \int_0^t \left(\frac{(\lambda s)^{n-1}}{(n-1)!}\lambda e^{-\lambda s}\right)\lambda e^{-\lambda(t-s)}\ \mathsf ds\\ &= \frac{\lambda^n}{(n-1)!}\lambda e^{-\lambda t} \int_0^t s^{n-1}\ \mathsf ds\\ &= \frac{(\lambda s)^n}{n!}\lambda e^{-\lambda t}\mathsf 1_{(0,\infty)}(t). \end{align} A similar inductive argument may be used to compute the distribution function of $S_n$: $$F_n(t) = 1 - \sum_{k=0}^{n-1}\frac{(\lambda t)^k}{k!}e^{-\lambda t}.$$ It follows that $$\mathbb P(S_{100}\geqslant 200) = 1 - F_{100}(200) = e^{-200 \lambda}\sum_{k=0}^{99}\frac{(200 \lambda)^k}{k!}.$$

• Thank you for a very detailed explanation! It helps a lot! – Orongo Jun 2 '16 at 20:07