3

Question:

Suppose buses arrive at a bus stop according to a Poisson process $N_t$ with parameter . Given a fixed $t > 0$. The time of the last bus before t is $S_{N_t}$ , and the time of the next bus after $t$ is $S_{N_{t+1}}$. Show the following identity:$E(S_{N_{t+1}}-S_{N_t})=(2-e^{-\lambda t})/{\lambda}$.

My attempt:

One basic conclusion about Poisson proccess is that: the time interval between two arrivals follows $Expo(\lambda)$.

But here,as t is a fixed time point,it can be any point in the interval $S_{N_t}$ and $S_{N_{t+1}}$.Thus I found it impossible to use the above assumption.So what should I do next?

Math1000
  • 38,041
anmo
  • 85

1 Answers1

2

See my answer here for the derivation of the density of $L_t: = S_{N_{t+1}}-S_{N_t}$:

Deriving the distribution of residual time in a Poisson process.

It follows that \begin{align} \mathbb E[L_t] &= \mathbb E[L_t\mathsf 1_{(0,t)}(x)] + \mathbb E[L_t\mathsf 1_{(t,\infty)}(x)]\\ &= \int_0^t \lambda^2 x^2 e^{-\lambda x} \, dx + \int_t^{\infty } \lambda x (\lambda t+1) e^{-\lambda x} \, dx\\ &= \frac1\lambda e^{-\lambda t} (-\lambda t) (\lambda t+2)-2)+2\lambda + \frac1\lambda e^{-\lambda t} (\lambda t+1)^2 \\ &=\frac{2-e^{-\lambda t}}{\lambda }. \end{align}

Math1000
  • 38,041