# Poisson process, expected value

Arrivals of passengers at a bus stop form a Poisson process N with rate $$\lambda = 1/3$$ per minute. Assume that a bus has left at time $$t = 0$$ leaving no customers behind. Let $$T$$ denote the time of arrival for the next bus; then the number of passengers present when it arrives will be $$N_T$$. We suppose that $$T$$ is independent of $$N$$ and has the distribution $$\phi$$.

a) Compute $$E[N_T | T]$$ and $$E[N_T^2 | T]$$.

b) Compute $$E[N_T]$$ and $$Var(N_T)$$ for

$$d\phi(t) = 1/2 dt$$ if $$9 \leq t \leq 11, 0$$ otherwise.

Background:

A typical Poisson process denoted by $$N_t$$ refers to the number of arrivals (discrete) at time $$t$$, with rate $$\lambda$$, where

$$P(N_t = k) = \frac{e^{\lambda t} (\lambda t)^k}{k!}$$

It is also given (and can be computed) that $$E[N_t] = Var(N_t) = \lambda t$$, but I don't know if those are applicable here because part a has a conditional and part b has a density function.

I'm confused about part a because it seems that $$N_T$$ already implies that time $$T$$ is met, thus $$E[N_T | T] = E[N_T]$$.

In part b I would think I should integrate from $$t =$$ 9 to 11, using the probability I provided with the given value of $$\lambda$$. However, what should I use for k?

Please provide feedback on both parts (sorry, I don't know if there is a rule about only asking one question per post).

It is also given (and can be computed) that $$E[N_t] = Var(N_t) = \lambda t$$, but I don't know if those are applicable here because part a has a conditional...

What you know is that for a Poisson arrival process with rate $$\lambda$$, if $$N_t$$ is the number of arrivals during some (fixed!) time $$t$$, then the rv $$N_t$$ follows the Poisson distribution you wrote (with parameter $$\lambda t$$), and hence $$E[N_t]=\lambda t$$, etc.

But here we have a time $$T$$ which is not fixed, but it's a random variable.

This is our problem. We know $$N_t$$ (its law) if we "are given" the value of $$t$$. But to say this, is equivalent to say that we know what it "if conditioned" on $$T=t$$.

That is, what we actually know is this: $$P(N_T | T=t) = \frac{e^{\lambda t} (\lambda t)^k}{k!}$$

and hence $$E[N_T | T=t) = \lambda t$$, etc. Or, in terms of the random variable: $$E[N_T | T ] = \lambda T$$.

To compute the raw expectation $$E[N_T]$$ we use the tower property $$E[X] = E[E[X|Y]]$$, so

$$E[N_T] = E[ E[N_T | T ] ] = E[\lambda T]$$

I guess you can go on from here.

• No I'm not sure how to proceed. I don't know how to set up the integral for the expectation unless I can use the probability function, and even so I don't know what to use for k. Commented Jul 20, 2019 at 4:32