From the pdf of the Poisson distribution I would expect $\Pr(x=1)$ to be

$$\lambda dt \cdot \exp(-\lambda dt)$$

I can see that as $dt$ gets very small, $\exp(-\lambda dt)$ becomes close to $1$, and so suggests $\lambda dt$, but I don't see why in the limit as $dt \to 0$ of $\lambda dt \cdot \exp(-\lambda dt)$, doesn't that make the $\lambda dt$ term zero also?

  • 1
    $\begingroup$ Please, use $\LaTeX$ markup for expressions. $\endgroup$
    – user88
    Jun 15, 2011 at 21:57

3 Answers 3


In fact, Leibniz' notation for infinitesimal increments can be confusing.

One has to be careful here to keep all terms of the same order: $e^{-\lambda dt}$ must be approximated to first order of $dt$ (not zeroth order, i.e. without any terms in $dt$), i.e. $e^{-\lambda dt}$ is approximately $1 - \lambda dt$ + (plus terms which are at least quadratic in $dt$ and thus go to zero faster than $dt$ itself)

Then one has:

$$ \lambda dt \cdot e^{-\lambda dt} \approx \lambda dt \cdot (1 - \lambda dt) $$ and then for $dt \rightarrow 0$ one can ignore the non-leading terms ($dt^2$) and is left with $\lambda dt$.


The previous two answer's are I think coming at the problem "backwards" - though they are both correct. They do not start with the postulate and end with the conclusion. If we start from the postulate, then we have:

$$Pr(\text{No event in} [t,t+dt])=1-Pr(\text{1 event in} [t,t+dt])=1-\lambda dt$$

If we define the function $h(t)$ as follows:

$$Pr(\text{No event in} [0,t])=h(t)$$

$$Pr(\text{No event in} [0,t+dt])=h(t+dt)$$

Additionally, we can use the independence of the increments - another postulate of the poisson process and we have:

$$h(t+dt)=h(t)[1-\lambda dt]\implies\frac{h(t+dt)-h(t)}{dt}=-\lambda h(t)$$

Taking the limit as $dt\to 0$ we have $h'(t)=-\lambda h(t)$ which implies $h(t)=K\exp(-\lambda t)$. We can resolve the proportionality constant by noting that $h(0)=1$ - i.e. it is certain to see no events in $[0,0]$. This gives $K=1$. This derivation can be found here (page 4) along with how to extend it to the probability for any number of events (basically by multiplying the zero count probability by $\lambda^n$ where $n$ is the number of events).


Here's an alternative (but basically equivalent) derivation to @Andre Holzner's:

For a Poisson process $N(t)$ with rate $\lambda$,

$Pr(N(t+\tau) - N(t) = 1) = (\tau\lambda)\exp(-\tau\lambda) = Pr(N(\tau) = 1) $

which has Taylor expansion around $\tau=0$

$\tau\lambda - \tau^2\lambda^2 + O(\tau^3)$

and this is approximately $\tau\lambda$ for small $\tau$. You're correct that the actual limit is zero, as one typically assumes $Pr(N(0)=0)=1$ in developing the Poisson process.

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    $\begingroup$ Don't you mean the probability equals $0$, not $1$, in the last sentence? $\endgroup$
    – whuber
    Jun 16, 2011 at 19:44

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