Poisson process, expected value

Question

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).

Answer 1

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.