conditional probability poisson process

Question

I am stuck on how to find the conditional probability of a poisson process. I know generally, if you have a poisson process with intensity parameter $\lambda$, then the conditional probability of having $m$ events in the first $t$ hours given $n$ events in the first $T$ hours is $${n\choose m}\frac{(T-t)^{n-m}t^m}{T^n}.$$ Now, my question is how to find the conditional probability of having $m$ events in the last $t$ hours given $n$ events in the first $T$ hours with $0\le m\le n$ and $0<t<T$. Any help is appreciated. Thanks.

Answer 1

Both answers are found in the same way.

Let's denote the number of events in the first t hours(resp. in the last t hours) by $N_t$, that in the first $T$ hours by $N_T$. Then we know that $N_t$ is a Poisson variable with parameter $\lambda t$ and $N_T$ with $\lambda T$. What's more $N_T - N_t$ is Poisson with parameter $\lambda (T-t)$ and it is independent of $N_t$ since the increments of Poisson process are independent.

\begin{align} P(N_t = m | N_T = n) &= \frac{P(N_t = m, N_T = n)}{P(N_T = n)}\\ & = \frac{P(N_t = m, N_T - N_t = n - m)}{P(N_T = n)} \\ & = \frac{P(N_t = m) P( N_T - N_t = n - m)}{P(N_T = n)} \\ & = \frac{e^{-\lambda t}\frac{(\lambda t)^m}{m!}e^{-\lambda (T-t)}\frac{(\lambda (T-t))^{n-m}}{(n-m)!}}{e^{-\lambda T}\frac{(\lambda T)^n}{n!}} \\ & = \frac{n!}{m!(n-m!)} (t/T)^m(1-t/T)^{n-m} \end{align}

More generally, if $S = X_1 + X_2 + \cdots + X_n$ where $X_i$'s are independent Poisson variables with parameters $\mu_i$, then $S$ is Poisson with paramter $\sigma = \sum \mu_i$ and

\begin{align} P(X_1 = r_1, X_2 = r_2, \cdots, X_n = r_n | S = s) = \frac{s!}{r_1 ! r_2 !\cdots r_n !}(\frac{\mu_1}{\sigma})^{r_1}(\frac{\mu_2}{\sigma})^{r_2}\cdots (\frac{\mu_n}{\sigma})^{r_n} \end{align}