Poisson process with parameter $\lambda$ and calculation of conditional probability

Question

• Let {$N_t : t ≥ 0$} be a Poisson process of rate λ, and let $S_n$ denote the time of the nth event. Find: $P(S_3 >5|N_2 =1).$

I tried to calculate $$P(S_3>5|N_2=1)=\frac{P(S_3>5, N_2=1)}{P(N_2=1)}$$

and got $$P(S_3>5)=P(N_5<3)=1-P(N_5\geq 3)=1-\frac{(\lambda t)^{n-1}}{(n-1)!}e^{\lambda t}$$

But I struggle to figure out how the top of the fraction relates to each other.

Answer 1

I would have thought you could use the memoryless property to say something like

$$P(S_3>5\mid N_2=1) = P(S_{3-1}> 5-2) \\= P(S_{2}> 3) = P(N_3 < 2) \\= P(N_3=0)+P(N_3=1) \\= e^{-3\lambda}(1+3\lambda)$$

Simulation in R suggests this is reasonable

set.seed(2020)
cases  <- 10^6
lambda <- 0.4321
S1 <-      rexp(cases, rate=lambda)
S2 <- S1 + rexp(cases, rate=lambda)
S3 <- S2 + rexp(cases, rate=lambda)
N2eq1 <- (S1 <= 2 & S2 > 2)
S3gt5 <- (S3 > 5)
mean(S3gt5[N2eq1])
# 0.6283518
exp(-3 * lambda) * (1 + 3 * lambda) 
# 0.6281346