In a poisson process, getting an exponential distribution from a poisson assumption (for eg. in a given period of time $t$, if number of books that arrive is given by a Poisson, then the time until first email will be given by an Exponential) is pretty straightforward, for eg. see Relationship between poisson and exponential distribution.

I'm trying to get the opposite. Given that each new email arrives following an exponential distribution, how can I retrieve that the number of emails in a time period $t$ is given by a Poisson?

I can get the basic cases. For example, let $X_i$ be time to receive $i$th email after having received $(i-1)$th email, and assume $X_i \sim Expo(\lambda)$. Also let $Y$ be the number of emails received in time period of length $t$.

Then \begin{align*} P(Y=0) &= P(\mbox{no book arrives before time } t) \\ &= P(X_1 > t) = e^{-\lambda t} \end{align*} and \begin{align*} P(Y=1) &= P(\mbox{first book arrives before time }t \mbox{ but then no other book arrives until time }t) \\ &= \int_0^t P(X_0 = x) \cdot P(X_1 > t-x) \,dx \\ &= \int_0^t \lambda e^{-\lambda x} \cdot e^{-\lambda (t-x)} \,dx = \int_0^t \lambda e^{-\lambda t} \,dx = \lambda t e^{-\lambda t} \end{align*}

which are both Poisson. We can continue this process, for example $$ P(Y=2) = \int^t_0 P(X_1=x_1) \left[ \int^t_{x_1} P(X_2=x_2) \cdot P(X_3>t-x_2) \,dy \right] \,dx$$

But this quickly becomes unwieldy. Is there a better method to get this result? For example, I just learned that a sum of exponential r.v. (eg. the $X_i$'s here) can be modeled using a Gamma distribution, but not sure if this is of any help.


3 Answers 3


I'll leave it to you to fill the details of this:

The exponential waiting time implies that an infinitesimal interval of length $dt$ will have probability $\lambda dt$ of seeing a jump. Since the process we generate from these exponentials is memoryless, it resets on every hit. A region $[0,t]$, can be broken into bins of length $t/n$. As $n$ goes to infinity, each interval approximates an infinitesimal. Each bin is independent of all others. Thus the number of jumps in such an interval $[0,t]$ is asymptotically a binomial distribution with $n$ bins and probability $\lambda t/n$ of each bin getting a hit. Thus we get:

$$P(Y=x)=\binom{n}{x}(\lambda t/n)^k(1-\lambda t/n)^{n-k}.$$

Now use your favorite method of showing that a binomial with $n$ bins whose probability scales like $c/n$ approaches a poisson distribution with parameter $c$. You could use Stirling's approximation on the above, or even easier, recall that the characteristic function of a binomial looks like $(1-p+pe^{ix})^n$. The characteristic function of a poisson with rate $\lambda t$ is $\exp(\lambda t (\exp(ix-1))$. Then the definition of $e$ implies:

$$(1-p+pe^{ix})^n=(1-(1-e^{ix})\lambda t/n)^n\rightarrow \exp(\lambda t (\exp(ix-1)).$$

  • $\begingroup$ Why is the probability of seeing a jump in a $dt$ interval $\lambda dt$? Isn't it $\int_0^{dt} \lambda e^{-\lambda x} dx = 1-e^{-\lambda dt}$? $\endgroup$
    – samlaf
    Jul 28, 2017 at 18:29
  • $\begingroup$ Sure, but $dt$ is infinitesimal, so $1-e^{-\lambda dt}=\lambda dt + o(dt)$ (taylor expand the exponential around 0). This reasoning will make the argument rigorous. $\endgroup$
    – Alex R.
    Jul 28, 2017 at 18:34
  • $\begingroup$ Oh wow nice! Is there a more intuitive way to reason/remember that this probability is $\lambda$ (for example using expected value somehow... or...?)? Or did you just do this calculation but left it out of your answer? $\endgroup$
    – samlaf
    Jul 28, 2017 at 18:39
  • $\begingroup$ Maybe the easiest way is to actually derive the exponential waiting time. So we assume that the chance of something occurring in a bin of size $1/n$ is asymptotically $\lambda/n$. We are waiting for the first such occurrence, which gives $P(T=x)=(1-\lambda/n)^x (\lambda/n)$. If we scale $x$ like $cn$, this asymptotically gives you $P(T\in (c,c+dt))\approx\lambda \exp(-\lambda c) dt$, which is precisely the density of the exponential. $\endgroup$
    – Alex R.
    Jul 28, 2017 at 18:49

Here's another way of approaching this problem. Suppose $X_i \sim Exp(\beta)$. For simplicity, define $T_k = \sum_{i=1}^k X_i$. Finally, we define

$$N = \min(n \ \ s.t. \ \ T_{n-1} < 1 \text{ and } T_n \geq 1)$$

Then it follows that $N-1$ is a Poisson random variable with mean $\frac{1}{\beta}$.

We notice that $T_{n-1} \sim Gamma(n-1, \beta)$ and that $T_{n-1}$ and $X_n$ are independent and thus the joint pdf is easily attainable. In order to leverage this fact, we write:

$$\begin{align*} P(N=n) &= P(T_{n-1} < 1, \ T_n \geq 1)\\ &= P(T_{n-1} < 1, \ T_{n-1} \geq 1 - X_n) \end{align*}$$

We can obtain this probability by integrating the joint pdf $f(x, t)$ over the appropriate domain.

$$\begin{align*} P(N=n) &= \int_0^1\int_{1-t}^\infty f(x,t)\ dx\ dt \\ &= \int_0^1\int_{1-t}^\infty \frac{e^{-t/\beta}t^{n-2}}{(n-2)!\beta^{n-1}}\cdot\frac{e^{-x/\beta}}{\beta} \ dx \ dt \\ &= \frac{e^{-1/\beta}}{(n-2)!\beta^{n-1}}\int_0^1 t^{n-2} \ dt \\ &= \frac{e^{-1/\beta}(1/\beta)^{n-1}}{(n-1)!} \end{align*}$$

If we make the simple transformation $Y = N-1$, and define $\lambda = 1/\beta$ we obtain:

$$P(Y=y) = P(N=y+1) = \frac{e^{-\lambda}\lambda^y}{y!}$$

Hence $Y = N-1$ is Poisson as desired.

  • $\begingroup$ $T_k$ I'm guessing is the sum of $X_i$'s? But $X_i$ and $\beta$ seem to be circularly defined, so I'm a bit loss. $\endgroup$
    – samlaf
    Jul 28, 2017 at 18:22
  • $\begingroup$ Yes, I just use $T_k$ to avoid having to type out so many sums. And $\beta$ is just the parameter of $X_i$'s distribution. There are two common ways to parameterize the exponential distribution, by stating $\beta = E(X_i)$, i was just trying to be explicit with which parameterization I was using. (: $\endgroup$
    – knrumsey
    Jul 28, 2017 at 18:58

For $P(Y=2)$ and beyond, I used the result for the previous integer and constructed a proof by induction. followed the same method

To motivate the proof, for $Y=2$ $$ \begin{align*} P(Y=2) &= \int_0^t P(\text{First arrival at $t_1 < t$ and exactly one arrival between $t_1$ and $t$}) dt_1 \\ &= \int_0^t \lambda \exp(-\lambda t_1)\lambda(t-t_1)\exp(-\lambda(t-t_1)) dt_1 \\ &= \lambda^2 \exp(-\lambda t) \int_0^t (t-t_1) dt_1 \\ &= \lambda^2 \exp(-\lambda t) t^2/2 \\ &= \frac{(\lambda t)^2 \exp(-\lambda t)}{2!} \end{align*} $$

In the above, $\lambda \exp(-\lambda t_1)$ is the pdf for the first arrival time, and $\lambda(t-t_1)\exp(-\lambda(t-t_1))$ is the probability of having exactly one arrival between $t_1$ and $t$. The total probability of having exactly two arrivals is given by integrating over $t_1$ from $0$ to $t$.

Proof by Induction

Let $P(Y=k) = \frac{(\lambda t)^k \exp(-\lambda t)}{k!}$, the result of the Poisson process. Then

$$ \begin{align*} P(Y=k+1) &= \int_0^t P(\text{First arrival at $t_1 < t$ and exactly $k$ arrivals between $t_1$ and $t$}) dt_1 \\ &= \int_0^t \lambda \exp(-\lambda t_1)(\lambda(t-t_1))^k\frac{\exp(-\lambda(t-t_1))}{k!} dt_1 \\ &= \frac{\lambda^{k+1} \exp(-\lambda t)}{k!} \int_0^t (t-t_1)^k dt_1 \\ &= \frac{(\lambda t)^{k+1} \exp(-\lambda t)}{(k+1)!} \end{align*} $$


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