# Poisson Process: Probability of 2 arrivals by time 1 given that there are 6 arrivals by time 4?

Let $(N_t)_{t \ge 0}$ be a Poisson process with parameter $\lambda = 1.5$. Find the following:

(c) $P(N_1 = 2 | N_4 = 6)$

I know how to do it if the times are reversed (it simply uses the independent increments property and reduces to simple mathematics). But I can't apply the same concept to this. At least not in the same form. Any advice on what to do? Thank you in advance!

Edit: The only thought I had was that it should equal,

$$P(N_1=2) P(N_2 \le 6) P(N_3 \le 6) P(N_4=6)$$

But even if this is correct I know it's more complicated that necessary..

• Using the independent increments property does work a little differently in the discrete case. In this case, we know there were 6 arrivals in the first 4 periods, and independent increments implies that each arrival has an equal probability of being in each period (i.e., $1/4$.) So, you have six arrivals, each with a probability of $0.25$ of having occurred in period 1... – jbowman Apr 24 '18 at 22:24
• Hint: Can you compute the probability that there are exactly two arrivals in $(0,1]$ and exactly four arrivals in $(1,4]$? Subhint: $(0,1]$ and $(1,4]$ are disjoint intervals. From this, can you figure out $P((N_1 = 2)\cap (N_4=6))$? – Dilip Sarwate Apr 24 '18 at 23:24

## 1 Answer

Bayes' rule yields \begin{align} \mathbb P(N_1=2\mid N_4=6) &= \frac{\mathbb P(N_4=6\mid N_1=2)\mathbb P(N_1=2)}{\mathbb P(N_4=6)}\\ &= \frac{\frac{(3\lambda)^4}{4!}e^{-3\lambda}\cdot\frac{\lambda^2}{2!}e^{-\lambda}}{\frac{(4\lambda)^6}{6!}e^{-4\lambda}}\\ &= \frac{1215}{4096}. \end{align}