# Poisson Process and Conditional Probability

Let $X= (X(t); t\ge0)$ be a poisson process with the intensity ($\lambda$ per hour)

A) Find the conditional probability of having $m$ events in the first $t$ hours, given that there were $n$ events in first $T$ hours. (here $0\le m \le n$ and $0<t<T$)

I need someone to check if this is right.

$P(m$ events in $t$ hours | $n$ events in $T$ hours) = $P(m$ events in $t$ hours)/$P(n$ events in $T$ hours) =

$${((e^{-\lambda t})(\lambda t)^m) }/m! \over {((e^{-\lambda T})(\lambda T)^n) /n!}$$

Any help would be greatly appreciated.

Hint: In order for you answer to be correct you also need to add the following term in the numerator \begin{align*}&\phantom{\,\=}P(m \text{ events in t hours | n events in T hours})= \\ \\& =\frac{P(m \text{ events in t hours, n-m events in T-t hours})}{P(n \text{ events in T hours})}=\\\\& =\frac{P(m \text{ events in t hours})\cdot P(n-m \text{ events in T-t hours})}{P(n \text{ events in T hours})}=\ldots\end{align*} Then the formula you used for the poisson distribution is correct. The result should reflect the binomial distribution (see second bullet).

• So for P(n-m events in T-t hours) = $$e^{-\lambda T + \lambda t} (\lambda T - \lambda t)^{n-m} / (n-m)!$$ Commented Mar 8, 2014 at 17:43
• @YoungGrasshopper yes, exactly. The result should be free of $e$ terms and be the probability of a binomial distribution. Commented Mar 8, 2014 at 17:45
• And when I multiply all of this together (a headache) it will reduce to something I recognize? Commented Mar 8, 2014 at 17:46
• @YoungGrasshopper Yes, exactly, that is correct! That is the binomial distribution $B(n=5, p=1/3)$... Commented Mar 8, 2014 at 18:12
• @YoungGrasshopper Practically, nothing, the result would be the same. That is due to the independence of events in different periods. You can also see that, from the fact that the binomial coefficients $\dbinom{n}{m}$ and $\dbinom{n}{n-m}$ yield the same result. Commented Mar 8, 2014 at 18:58

Hint: Let $B$ be the event "$n$ events in the first $T$" and $A$ the event $m$ events in the first $t$."

We want $\Pr(A|B)$, which is $\frac{\Pr(A\cap B)}{\Pr(B)}$.

To compute $\Pr(A\cap B)$, note that for this event to occur we need $m$ events in the first $t$ hours and $n-m$ additional events in the remaining $T-t$ hours. Thus you need to multiply your current numerator by a suitable factor.

Then simplify. The answer will look quite familiar.