How does the equation make sense in case of a Poisson process?

Question

$$P(N_1=2, N_4=6) = P(N_1=2, N_4-N_1=4) = P(N_1=2) \cdot P(N_3=4)$$

How does the above equation make sense in case of a Poisson process?

Can anyone explain?

Answer 1

Hopefully the first equality is clear to you: $$ N_1 = 2 \text{ and } N_4 = 6 \implies N_1 = 2 \text{ and } N_4 - N_1 = 4.$$ Now, a Poisson process (by definition) has independent increments. This means, in particular, that the difference $N_4 - N_1$ is independent of $N_1$. Hence $$ \Pr(N_1 = 2, N_4 - N_1 = 4) = \Pr(N_1 = 2) \Pr(N_4 - N_1 = 4).$$ Furthermore, it has what are called "stationary increments". That means that the distribution of $N_{n+m} - N_n$ depends only on $m$. Hence $N_4 - N_1 \sim N_3 - N_0$, and $N_0 := 0$. Hence $\Pr(N_4 - N_1 = 4) = \Pr(N_3 = 4)$.

Combining these results gives the answer! :)

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You can read up on Poisson processes in Section 1.4 of Perla Sousi's lecture notes on "Applied Probability", available on her website; they're written jointly with Nathanael Berestycki. I highly recommend doing so; they're good notes!

Answer 2

You don't define your notation, but I assume that the carrier space is divided into bins of equal width, the $N_i$ are the numbers of events in the first $i$ bins, and you have a Poisson point process that produces points in that carrier space. Breaking down your equations:

• $P(N_1 = 2, N_4 = 6)$: We want to study the probability that $2$ events land in bin 1 and $6$ events land in the first $4$ bins.
• ${} = P(N_1 = 2, N_4 - N_1 = 4$: If two events land in the first bin, then $6 - 2 = 4$ events land in the second, third, and fourth bins. We have also made the two conditions independent, so ${} = P(N_1 = 2) \cdot P(N_4 - N_1 = 4)$.
• ${} = P(N_1 = 2) P(N_3 = 4)$: The Poisson process is memoryless. The number of events landing in an interval only depends on the width of the interval. The interval represented by $N_4 - N_1$ is three bins wide, so we can replace that with the probability of getting $4$ events in $3$ bins.