Arrivals of passengers at a bus stop form a poisson process $X = {X(t); t>= 0}$ with the rate of 4 per unit of time. Assume that $T$ denotes the arrival time of the next bus. Then $X(T)$ is the number of passengers present at that time. Assume that the bus arrival time $(T)$ is independent of the Poisson process $(X)$ and $T$ follows the exponential distribution with the intensity $(\theta > 0)$.
Determine the conditional moments of 1st and 2nd order for $X(T)$, given that $T=t$.
Derive the distribution of $X(T)$.
Determine the mean and variance of $X(T)$.
I need someone to check my work for numbers 1 and 3, and help me with number 2.
- $\frac {X(T)}T = t$ ~ $Poi(4t)$ since $T$ follows the exponential distr. and has the memorylessness property.
1st moment = $4t$
2nd moment = $(4t)^2 + 4t$
3.. Mean = $E(E(X(T))) = 4E(T) = \frac 4 \theta$
Variance = $Var(X(T)) = E(Var(X(T))) + Var(E(X(T)))$ = $E(16T^2) + Var(4T)$ = $16(\frac 1 {\theta^2} + \frac 1 {\theta^2}) + 4(\frac 1{\theta^2}) = \frac {36} {\theta^2} $
For every integer $n\geqslant0$, the independence of $(X_t)$ and $T$ yields $$P(X_T=n)=\int_0^\infty P(X_T=n\mid T=t)f_T(t)\mathrm dt=\int_0^\infty P(X_t=n)f_T(t)\mathrm dt, $$ that is, $$ P(X_T=n)=\int_0^\infty \mathrm e^{-4t}\frac{(4t)^n}{n!}\theta\mathrm e^{-\theta t}\mathrm dt.$$ The change of variable $s=(4+\theta)t$ yields $$ P(X_T=n)=\frac{4^n\theta}{(4+\theta)^{n+1}}\int_0^\infty \mathrm e^{-s}\frac{s^n}{n!}\mathrm ds=\frac{\theta}{4+\theta}\cdot\left(\frac{4}{4+\theta}\right)^n.$$