Expected Value and Variance of Poisson Process Bus Stop

Question

Arrivals of passengers at a bus stop form a Poisson process {N(t),t ≥ 0} with rate $\lambda$ = 2 per unit of time. Assume that a bus departed at time t = 0 leaving no customers behind. Let T be the arrival time of the next bus. Then, the number of passengers present when it arrives is N(T). Suppose that the bus arrival time T is independent of the Poisson process and that T ∼ U$_{[0,1]}$ (Uniform distribution over [0, 1]). By 􏰀first conditioning on T, determine the mean and the variance of the number of passengers present when the next bus arrives (i.e., E(N(T)) and Var(N(T))).

$\lambda$ = 2

E(N(T) = ??

Var(N(T)) = ??

I am not sure about the formulas for how to solve for these with the Uniform distribution.

Answer 1

Expectation

Let $f(t)=1$ be the probability distribution function for the uniform distribution.

$\displaystyle\mathbb{E}\left[N(T)\right]=\int_0^1\mathbb{E}\left[N(t)\right]f(t)\mathrm{d}t=\int_0^1\lambda t\mathrm{d}t=\lambda/2$

Variance

Let $\displaystyle g(n)=\frac{e^{-\lambda}\lambda^n}{n!}$ be the probability mass function for the Poisson distribution. $\displaystyle\text{Var}\left[N(T)\right]=\int_0^1 \mathbb{E}\left[ \left(N(t)-\lambda/2\right)^2\right]f(t)\mathrm{d}t$

We already know that $\displaystyle\text{Var}\left[N(t)\right]=\mathbb{E}\left[\left(N(t)-\lambda t\right)^2\right]=\lambda t$.

So using the fact that$\left(N(t)-\lambda/2 \right)^2=\left(N(t)-\lambda t\right)^2+\lambda(2t-1)N(t)+\lambda^2(1/4-t^2)$, we have that $\mathbb{E}\left[ \left(N(t)-\lambda/2\right)^2\right]=\lambda t+(2t-1)\lambda^2t+\lambda^2\left(1/4-t^2\right)$.

So $\displaystyle\text{Var}\left[N(T)\right]=\int_0^1 \mathbb{E}\left[ \left(N(t)-\lambda/2\right)^2\right]\mathrm{d}t=\lambda^2/12+\lambda/2$.