Calculate $P(N_1 = 2 \mid N_3 = 10)$

Question

Assume $(N_t)$ is a non-homogeneous Poisson Process with rate function:

$$\lambda(t) = \begin{cases} 1 & t<1 \\ 2 &1\leq t < 2 \\ 3 & t\geq 2 \\ \end{cases} $$

Calculate $P(N_1 = 2 \mid N_3 = 10)$

I approached this problem thinking I could just make $N_3 = N_1 + (N_3-N_1)$ and then it would be $P(N_1 = 2 \mid N_1 + (N_3 + N_1) = 10)$ and then we can make it $P(N_1 = 2 \mid N_3 - N_1 = 10 -2)$ and so by independence it would just be $P(N_1 = 2)$ which would come out to be $\frac{e^{-2}(2^2)}{2!}$. However, this was the wrong answer and I think where I went wrong was thinking that I could use the fact that $N_1 =2$ and subtracting it from the $10$ but I didn't know how else to try and approach the problem. Any help on figuring it out is appreciated!

Answer 1

You have $N_1\sim \mathrm{Poisson}(1)$ and $N_3-N_1 \sim \mathrm{Poisson}(5)$ and $N_1$ is independent of $N_3-N_2$. So $N_3\sim\mathrm{Poisson}(6)$ and \begin{align} \Pr(N_2=2\mid N_3=10) & = \frac{\Pr(N_1=2\ \&\ N_3=10)}{\Pr(N_3=10)} \\[10pt] & = \frac{\Pr(N_1=2\ \&\ N_3-N_1=8)}{\Pr(N_3=10)} \\[10pt] & = \frac{\dfrac{1^2 e^{-1}}{2!} \cdot \dfrac{5^8 e^{-5}}{8!}}{\left( \dfrac{6^{10} e^{-6}}{10!} \right)} = \frac{10!}{2!8!} \cdot \frac{5^8}{6^{10}} = \frac{10\cdot 9 \cdot 5^8}{2\cdot 6^{10}} = \frac{5^9}{3^2\cdot6^8} = \cdots \end{align}

Answer 2

You have 10 events each distributed independently among three equal-sized subintervals with bias rates $1,2,3$ respectively.   These distributions will be densest where the rate is greatest.

The probability that any particular event occurred within the first interval is : $\frac{1}{1+2+3}$

The probability that exactly two did is : $\binom{10}{2} {\big(\frac 1 6\big)}^{2}{\big(\frac 5 6\big)}^{8}$ ... as this count follows a binomial distribution.

$$\binom{10}{2} \dfrac {5^8} {6^{10}}$$