Poisson with customer arrival to the shop rate given by $\lambda (t)=16-(t-4)^2$ Calculate $P(N(5)-N(3)=40|N(4)=70)$ where $N(i)$ means the number of arrivals in the first $i$ hours. The shop is open for 8 hours.
I thought about using the formula:
$$P(N(5)-N(3)=40)=\frac{e^{-(\Lambda (5)-\Lambda (3)}(\Lambda (5)-\Lambda (3))^{40}}{40!} $$ where $\Lambda (t)=\int_0^t 16-(t-4)^2 dt$.
However, I'm strugling to use the conditional probability.
$P(N(5)-N(3)=40|N(4)=70)=\frac{P(N(5)-N(3)=40,N(4)=70)}{P(N(4)=70)}$
Does it equal?: $$\frac{P(N(5)-N(3)=40)P (N(4-5+3)=70-40)}{P(N(4)=70)}$$
I think I should use independent increments somehow, but I don't see how.
$$P(N(5)-N(3)=40|N(4)=70)$$ $$=\sum_{i=0}^{40}P(N(5)-N(4)=i,N(4)-N(3)=40-i|N(4)=70)$$ $$=\sum_{i=0}^{40}P(N(5)-N(4)=i | N(4)-N(3)=40-i, N(4)=70) \times P(N(4)-N(3)=40-i|N(4)=70)$$ $$=\sum_{i=0}^{40}P(N(5)-N(4)=i | N(4)=70) \times P(N(4)-N(3)=40-i|N(4)=70)$$
But we also know that $$P(N(4)-N(3)=40-i|N(4)=70)=\frac{P(N(3)=i+30\cap N(4)=70)}{P(N(4)=70)}=\frac{P(N(3)=i+30\cap N(4)-N(3)=40-i)}{P(N(4)=70)}=\frac{P(N(3)=i+30)P(N(4)-N(3)=40-i)}{P(N(4)=70)}=\frac{P_{Poi}(i+30;\Lambda(0,3))P_{Poi}(40-i;\Lambda(3,4))}{P_{Poi}(70;\Lambda(0,4))}$$
where $\Lambda(t_1,t_2)=\int_{t_1}^{t_2} \lambda(t) dt$, and $P_{Poi}(i;\mu)=e^{-\mu } \mu ^i / i!$.
And we know that: $P(N(5)-N(4)=i | N(4)=70)=P_{Poi}(i;\Lambda(4,5))$. So we can conclude that the total is:
$$=\sum_{i=0}^{40} \frac{P_{Poi}(i+30;\Lambda(0,3))P_{Poi}(40-i;\Lambda(3,4))P_{Poi}(i;\Lambda(4,5))}{P_{Poi}(70;\Lambda(0,4))}\approx0.0693744$$