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Let $(E_i)_{i\in\mathbb{N}_0}$ be independent exp(1) random variables. Show that for $$N:=\sup\left\{n\in\mathbb{N}:\sum_{i=0}^n E_i \leq \lambda\right\},$$ we have $N \sim \text{Poisson}(\lambda)$.

What I have so far:

Sum of exp(1) = gamma(k,1). My problem with this problem is I am unsure how to eliminate the supermum and condition it such that it will become

a busy cat

2 Answers 2

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Hints:

  1. Show that $$\begin{align*} \mathbb{P}(N=k) &= \mathbb{P} \left( \sum_{i=0}^k E_i \leq \lambda, \sum_{i=0}^{k+1} E_i>\lambda \right) \\ &= \int_{(0,\infty)}\! \ldots\! \int_{(0,\infty)} e^{-\sum_{j=0}^{k} x_j} 1_{[0,\lambda]}(x_0+\ldots+x_{k-1}) 1_{(\lambda,\infty]}(x_0+\ldots+x_k) \, dx_k \ldots dx_0. \end{align*}$$
  2. Conclude that $$\mathbb{P}(N=k) = e^{-\lambda} \int_{(0,\infty)} \ldots \int_{(0,\infty)} 1_{[0,\lambda]}(x_0+\ldots+x_{k-1}) \, dx_{k-1}\, \ldots \, dx_0. \tag{1}$$
  3. Find either a geometric interpretation for the integral on the right-hand side of $(1)$ or use a proof by induction to show that $$\int_{(0,\infty)} \ldots \int_{(0,\infty)} 1_{[0,\lambda]}(x_0+\ldots+x_{k-1}) \, dx_{k-1} \, \ldots \, dx_0 = \frac{\lambda^k}{k!}.$$
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    Mmmhm, at the moment this is still a bit too high level for me. Is there anyway you can further elaborate on the hints? Thank you for your time! – 2017-02-15
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    Ah never mind, I figured it out. Thanks! – 2017-02-15
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    @user3225528 You are welcome. – 2017-02-16
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Define a sequence of random variables $$S_n = \sum_{i=1}^n E_i, \quad n = 1, 2, \ldots. $$ Then the event $N = n$ is equivalent to saying $$S_{n} \le \lambda < S_{n+1},$$ which is in turn equal to $$S_n \le \lambda < S_n + E_{n+1}.$$ Since $S_n$ and $E_{n+1}$ are independent, it follows from the law of total probability that $$\Pr[N = n] = \int_{s=0}^\lambda \Pr[E_{n+1} > \lambda - s] f_{S_n}(s) \, ds.$$ Then since $S_n \sim \operatorname{Gamma}(n,1)$, we have $$\Pr[N = n] = \int_{s=0}^\lambda e^{-(\lambda-s)} \frac{s^{n-1} e^{-s}}{\Gamma(n)} \, ds = e^{-\lambda} \frac{\lambda^n}{n!}.$$ The details of this last computation is left as a straightforward exercise for the reader.