expected time of a Poisson process exponential time A kid counts the cars passing a certain street according to a Poisson process of rate λ. When a given α time units passes without any car arriving, he gets bored and leaves.
(a) How much time does he spend there on average? 
Hint: Condition on the time of the first car.
(b) How many cars did he count on average?
I'm not sure how to approach question a), so far i have something like this: $s_i$ is the time it takes for the $i$-th event to happen and has a memoryless property. $\alpha P(s_1>\alpha) + E[(s_1+\alpha \mid s_1<\alpha)]+E[(s_2+\alpha \mid s_2<\alpha)]+\cdots$ and i think this will converge to some finite value, however I'm not sure if this approach is the right way of doing it.
for part b) I did: $\sum^\infty_{k=0} k P(T_{k+1}>\alpha)P(T_k<\alpha)\cdots P(T_1<\alpha)$, which will be $k$ times a geometric series times $P(T_{k+1}>\alpha)$ but the problem I encountered is that after writing the geometric series and did the multiplication my answer to the question is $k.$
I did this: $P(T_{k+1}>\alpha)\frac{k}{e^{-\lambda \alpha}} = e^{-\lambda \alpha} \frac{k}{e^{-\lambda \alpha}}$, $e^{-\lambda \alpha}$ cancels out and leaves me $k$, which I know it is definitely wrong but I'm not sure how I would approach this question differently.
 A: We will need to know the conditional expected time of the first car's arrival given that at least one car arrives before time $\alpha.$
Let $S$ be the time of the first car's arrival, so $\operatorname E S = 1/\lambda.$
We have $\Pr(S>t) = e^{-\lambda t}$ (for $t\ge0$).
Therefore $\Pr(S>t\mid S<\alpha) = \dfrac{\Pr(t<S<\alpha)}{\Pr(S<\alpha)} = \dfrac{e^{-\lambda t} - e^{-\lambda\alpha}}{1-e^{-\lambda\alpha}}.$
From this we get the conditional distribution
$$
f_{S\,\mid\,S\,<\,\alpha} (t) \, dt = \left(-\frac d {dt}\, \frac{e^{-\lambda t} - e^{-\lambda\alpha}}{1-e^{-\lambda\alpha}} \right) dt = \frac{e^{-\lambda t}}{1-e^{-\lambda\alpha}} (\lambda\,dt) \quad \text{for } 0<t<\alpha,
$$
and from this we get the conditional expected value:
$$
\operatorname E(S\mid S<\alpha) = \int_0^\alpha t \, \frac{e^{-\lambda t}}{1-e^{-\lambda\alpha}} (\lambda \, dt) = \frac 1 \lambda \cdot \frac{1-e^{-\lambda\alpha}- \lambda\alpha e^{-\lambda\alpha}}{1-e^{-\lambda\alpha}}.
$$
The boy counts cars until an interval of length $\alpha$ appears with no cars.
Let $\beta$ be the average time the boy is there. Let $T$ be the total time the boy is there.
Then
\begin{align}
\beta & = \operatorname ET = \operatorname E(T\mid S>\alpha) \Pr(S>\alpha) + \operatorname E(T\mid S<\alpha) \Pr(S<\alpha) \\[10pt]
& = \alpha \Pr(S>\alpha) + \operatorname E(T + S\mid S<\alpha) \Pr(S<\alpha) \\[10pt]
& = \alpha\Pr(S>\alpha) + \Big( \operatorname E(T) + \operatorname E(S\mid S<\alpha) \Big) \Pr(S<\alpha) \\[10pt]
& = \alpha\Pr(S>\alpha) + \Big( \beta + \operatorname E(S\mid S<\alpha) \Big) \Pr(S<\alpha).
\end{align}
Since we know how to express $\Pr(S>\alpha),$ $\Pr(S>\alpha),$ and $\operatorname E(S\mid S<\alpha)$ as functions of $\alpha$ and $\lambda,$ we can substitute those functions of $\alpha$ and $\lambda$ into that last line, and then we have an algebraic equation that can be solved for $\beta$ as a function of $\alpha$ and $\lambda.$
I got $\displaystyle \beta = \alpha + \frac 1 \lambda\left( 1 - e^{-\lambda\alpha} - \lambda\alpha e^{-\lambda\alpha} \right).$
