Since i did not find a good proof of this anywhere on the internet, I want somebody to check my proof:

An experiment runs over a specific time span $[0,t]$. The expected number of arrivals in this time interval is $\lambda$, and the time intervals in between 2 arrivals has exponential distribution (it does not matter when the last arrival took place, the probability distribution until the next one is always the same). E.g.: number of radioactive decays of a sample in a given time interval, number of calls arriving, ...

Let $X$ be the number of arrivals in this time interval $[0,t]$. Then $X$ has the following probability distribution: $$ P_{[0,t]}(X= k) = \frac{\lambda^k}{k!} e^{-\lambda}. $$


If we slice the interval in $n\gg0$ equal parts: $[0,t] = [0,t/n)\cup [t/n, 2\cdot t/n) \cup \dots \cup [(n-1)\cdot t/n, t]$ Let $Y_1, Y_2, \dots, Y_n$ be the random variables describing the number of arrivals in the time intervals $[0,t/n), [t/n, 2t/n), \dots, [(n-1)t/n, t]$.

Since the time from a specific point until the first arrival has exponential distribution, $P(Y_1 = 1) = P(Y_2 = 1) = \dots = P(Y_n = 1)$. Also, the expected number of arrivals in a subinterval is proportional to the length of the interval, therefore: $P(Y_1 = 1)= \dots =P(Y_n = 1) = \frac \lambda n$. Because there can only be one arrival at the same time, $n$ can be chosen so big, that there will always be only zero or one arrivals in every subinterval.

Now one can see, that the probability of $k$ events in the interval $[0,t]$ can be written as a binomial distribution:

\begin{align} P_{[0,t]}(X = k) &= \binom{n}{k} \left(\frac{\lambda}{n}\right)^k \left(1-\frac{\lambda}{n} \right)^{n-k} = \\ &= \frac{n(n-1)\cdots (n-k+1)}{k!} \frac{\lambda^k}{n^k} \left(1-\frac{\lambda}{n}\right)^n\left(1-\frac{\lambda}{n}\right)^{-k}, \end{align}

since, $\lim_{n\rightarrow \infty} \frac{n(n-1)\cdots(n-k+1)}{n^k} = 1$, $\lim_{n\rightarrow \infty} \left( 1-\frac{\lambda}{n}\right)^n = e^{-\lambda}$ and $\lim_{n \rightarrow \infty} \left(1-\frac{\lambda}{n}\right)^{-k} = 1$:

\begin{align} P_{[0,t]}(X=k) = \frac{\lambda^k}{k!} e^{-\lambda}. \end{align}

(I think it should be mostly correct, put please can anyone check this)

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    $\begingroup$ Standard usage is $n\gg0,$ not $n>>0.$ I edited accordingly. $\qquad$ $\endgroup$ Jan 4, 2018 at 15:36
  • $\begingroup$ Oh yeah, so my initial definition of the Poisson distribution was wrong, just corrected it, is the proof now correct? $\endgroup$
    – iqopi
    Jan 4, 2018 at 15:39
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    $\begingroup$ Your proof is correct, but uses some strange or incorrect notation as Michael Hardy points out. I know a different proof of this fact and if you are interested I can post it tomorrow when I have time. $\endgroup$ Jan 4, 2018 at 15:59

1 Answer 1


You wrote:

If $X$ is a random variable and $[0,t]$ a time interval and the expected number of counts of $X$ in the interval $[0,t]$ is $\lambda$, then the probability, of $k$ counts of $X$ in the same time interval $[0,t]$ is:

That is infelicitous phrasing. If $X$ is a random variable, then what is meant by the number of "counts" of $X$ in a time interval $[0,t]$? Would you expect someone who had never heard of Poisson processes to understand that? What if $X$ is a random variable and $X\sim N(0,1)$? What then is the "number of counts of $X$" is a time interval? Saying "If $X$ is a random variable" means that what follows is applicable whenever $X$ is any random variable, whereas here you want $X$ to be a "number of counts"; you don't want to say for any random variable (e.g. a continuously distributed one) "has" a "number of counts".

What you call "events" I'd rather call "arrivals" since "event" is an overworked word in this context.

What you need here is to say that

  • for every subinterval of $[0,t]$ there is a random variable that is the number of arrivals in that time interval, and
  • the numbers of arrivals in disjoint intervals are independent, and
  • the expected number of arrivals in each subinterval is proportional to the length of the subinterval, and
  • the expected number $X$ of arrivals in $[0,t]$ is $\lambda$ (this says $X$ is that number, not that $X$ is just any random variable), and
  • no two arrivals can be simultaneous.

That last assumption does not follow from the others: sometimes one includes a probability distribution of the number of arrivals at a time when arrivals occur.

I also wouldn't say "$X$ has no memory". The thing that "has no memory" is the probability distribution of the time from one arrival to the next, and that is the probability distribution of a continuous random variable, whereas $X$ is a discrete random variable. "Memorylessness" in this context has a precise definition. There is discrete memorylessness of a discrete distribution, but that is a geometric distribution, not a Poisson distribution.

"$\frac\lambda n \ll0$" is not correct, since that implies $\lambda/n$ is negative.

Your way of finding the limit is correct. I would also mention that $\left( 1 - \frac \lambda n \right)^{-k} \to 1$ as $n\to\infty.$

  • $\begingroup$ thanks for your corrections. I have tried my best to rework my proof with your notes. Hope now it is easier to understand. $\endgroup$
    – iqopi
    Jan 4, 2018 at 23:23

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