Suppose $X_1, X_2, X_3, \ldots, X_n$ be independent and identically distributed random variables having an exponential distribution with mean $\frac{1}{\lambda}$.

If $S_n = X_1 + X_2 + \ldots + X_n$ and $N = \inf\{n \geq 1: S_n > 1\}$, then what would be the value of $\mathrm{Var}(N)$ ?

  • 1
    $\begingroup$ You mean $X_N$ and not $X_n$, right? $\endgroup$ Commented Mar 28, 2018 at 10:30
  • 3
    $\begingroup$ Some thoughts/hints: What is the distribution of $S_n$? What is $P(N > n)$? How can you use that to find $\mathbb{E}[N]$ and $\mathbb{E}[N^2]$? $\endgroup$
    – user51547
    Commented Mar 28, 2018 at 10:34
  • $\begingroup$ @Andreas Storvik Strauman: It is $X_n$, according to the question. $\endgroup$ Commented Mar 28, 2018 at 12:20
  • $\begingroup$ Oh. I understand the question now :) $\endgroup$ Commented Mar 28, 2018 at 13:08
  • 2
    $\begingroup$ Looking at the keyword "Poisson process" could possibly help. $\endgroup$
    – Xi'an
    Commented Mar 30, 2018 at 9:35

1 Answer 1


I am going to use the definition $N=\inf \{ n\ge0: S_{n+1}> 1 \}$ which seems more natural.

The event $\{N\le n\}$ is the same as $\{S_{n+1} > 1\}$ (or equivalently, $\{N>n\}$ is the same as $\{S_{n+1} \le 1\}$).

By the tail-sum formula, we can thus get the expectation of $N$ as \[ E[N] = \sum_{n=0}^\infty P(N>n) = \sum_{n=0}^\infty P(S_{n+1}\le 1). \] Now for any fixed $n$, $S_n$ has the $Gamma(n,1/\lambda)$ distribution (because it's a sum of exponential random variables) and by looking up the density of such distribution, we get \[ P(S_{n+1} \le 1) = \int_0^1 \frac{1}{\Gamma(n+1)}\lambda^{n+1}x^{n}e^{-x\lambda}dx = \lambda \int_0^1 \frac{1}{n!}(x\lambda)^{n}e^{-x\lambda}dx \] Since $\exp(x\lambda)=\sum_{n=0}^\infty (x\lambda)^{n}/n!$, we obtain by moving the integral outside of sum by Fubini's theorem \[ E[N] = \lambda \int_0^1 \exp(x\lambda) e^{-x\lambda} dx = \lambda. \]

We can compute $E[N^2]$ similarly using the tail formula, by making "packets" \[ E[N^2] = \sum_{n=0}^\infty P(N^2>n) = \sum_{k=0}^\infty ((k+1)^2-k^2)P(N>k) = \sum_{k=0}^\infty (2k+1)P(S_{k+1}\le 1). \] We use again the density of the $Gamma(n,1/\lambda)$ distribution, and Fubini to take the integral outside of the sum. We have to compute for $x\in[0,1]$ the sum \[ \sum_{n=0}^\infty \frac{2k+1}{\Gamma(k+1)} \lambda^{k+1} x^k e^{-x\lambda} = 2 \lambda^2 x \exp(x\lambda) e^{-x\lambda} + \lambda \exp(x\lambda) e^{-x\lambda} = (2\lambda^{2}x + \lambda). \] We now integrate the above as in $\int_0^1 (...) dx$; we obtain $E[N^2] = \lambda^{2} + \lambda$.

Finally, $Var[N] = E[N^2]-E[N]^2 = \lambda + \lambda^{2} - \lambda^{2} =\lambda$.

Prove that $N$ has Poisson distribution directly

The above method can also show that $N$ has the Poisson distribution. Since $P(N=n) = P(N >n-1) - P(N>n) = P(S_n \le 1 ) - P(S_{n+1} \le 1)$, we have \begin{align} P(N=n) &= \int_0^1 ( \frac{\lambda^n x^{n-1} e^{-\lambda x}}{\Gamma(n)} - \frac{\lambda^{n+1}x^n e^{-x\lambda}}{\Gamma(n+1)} ) dx \\\ &= \frac{1}{(n-1)!} \int_0^1 e^{-x\lambda} (\lambda x)^{n-1}(1-x\lambda/n) (\lambda dx) \\\ & = \frac{1}{(n-1)!} \int_0^\lambda e^{-u} u^{n-1}(1-u/n) (du). \end{align}

A primitive for the rightmost integrand is $e^{-u}u^n/n$, hence $P(N=n) = \lambda^n e^{-\lambda}/n !$ and $N$ has the Poisson distribution.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.