1
$\begingroup$

We know that the exponential distribution characterises the probability distribution for the waiting time between two consecutive Poisson events. Then I think if we fix a time interval $T$ then we should be able to derive the Poisson distribution from the exponential distribution. But some problems arises during my derivation.

Assuming the number of events per unit time is $\lambda$, then we have the Poisson distribution for the time interval $[0,T]$ $$P(k)=\frac{(\lambda T)^ke^{-\lambda T}}{k!}$$ Now we start with the exponential distribution $f(t)=\lambda e^{-\lambda t}$, for $t\ge 0$, and try to calculate, for instance, $P(k=0)$, $P(k=1)$ and $P(k=2)$, to see whether it gives the same result with the general Poisson distribution.

For $k=0$, no events happen in the time interval. Therefore, $$P(k=0)=\int_T^{\infty}f(t)\mathrm{d}t=e^{-\lambda T},$$ which agrees with the standard Poisson form.

For $k=1$, we have to make sure that one event happens at time $t_1 \in [0,T]$ and the the second event happens after $T$, i.e. $0\le t_1\le T \le t_2...$. Therefore, we have $$P(k=1)=\int_0^T\mathrm{d}t_1f(t_1)\int_{T-t_1}^{\infty}\mathrm{d}t'f(t')=\lambda Te^{-\lambda T}$$ where the second integral simply means the waiting time $t''$ between the two events has to be larger than $T-t_1$ to guarantee that the second event happens outside the interval $[0,T]$. Again, this result agrees with the standard Poisson distribution.

For $k=2$, we then have $0\le t_1 \le t_2 \le T \le t_3...$ and similarly we have the following $$P(k=2)=\int_0^T\mathrm{d}t_1f(t_1)\int_{0}^{T-t_1}\mathrm{d}t'f(t')\int_{T-(t_1+t')}^{\infty}\mathrm{d}t''f(t'')=\frac{(\lambda T)^2e^{-\lambda T}}{2}$$ Edited version, thanks to Kevin Zhou. Once the limited is corrected, a correct result will be obtained.

$\endgroup$
  • 1
    $\begingroup$ Would Cross Validated be a better home for this question? $\endgroup$ – Qmechanic Aug 21 '15 at 7:08
  • $\begingroup$ because I think physicist in certain areas might be even better than statisticians at this. $\endgroup$ – M. Zeng Aug 21 '15 at 7:18
1
$\begingroup$

The lower limit of your third integral should be $T - t_2 - t_1$. (I assume that $t' = t_1$ and $t'' = t_2$, otherwise the integrals don't make sense.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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