6
$\begingroup$

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?

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

3 Answers 3

6
$\begingroup$

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 $\lambda dt$.

$\endgroup$
3
$\begingroup$

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).

$\endgroup$
2
$\begingroup$

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.

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

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.