Superposition of processes Considering the traffic on the road pictured in Figure 4.8.2, the following is known. The number of vehicles passing the point A in an hour follows the Poisson

distribution with mean 60; 20% of these vehicles are trucks. The number of vehicles passing B in an hour is also Poisson distributed with mean 80: 30% of these are trucks. In general, 10% of all vehicles stop at the restaurant. The number of persons in a truck is one; the number of passengers in a car is equal to 1, 2, 3, 4, or 5 with respective probabilities 0.30, 0.30, 0.20, 0.10, and 0.10.

Find the expected value E[Z] of the number of persons Z arriving at the restaurant within that one hour
Compute $E[\alpha^{Z}]$ for $\alpha \in [0, 1]$.

My attempt
Taking the superposition of the processes $N_T(t)$ and $N_c(t)$, then
$\delta_T = ((60)(0.2)+(0.8)(0.3))0.1 = 3.6$ trucks per hour
$\delta_c = ((60)(0.8)+(0.8)(0.7))0.1 = 10.4$ cars per hour
From here, we have $E(N_T(1))=3.6$ and
$E(paseengers\ \ cars) = 1(0.4)(0.3)+2(10.4)(0.3)+3(10.4)(0.2)+4(10.4)(0.1)+5(10.4)(0.1) = 24.96$
Then, $E(Z)=28.56$. Is it correct?
Coud someone can helo for $b)$, I really can understand how to compute $E(\alpha^Z)$ for $\alpha \in [0,1]$
Thanks for your time and help.
 A: Question 1
Note: I admit that my way of resolving this question may not necessarily be standard. But it is just to check the validity of OP's results.
Well, first things first. Let's calculate the average number of people coming from a vehicle coming from each lane. We'll call $C$ the random variable containing the number of passengers coming out of a vehicle at the restaurant.


*

*When $k = 1$ we have both trucks and cars to consider. In that case, we have $P(C=1 \cap A) = 0.2 + 0.8 \times 0.3 = 0.44$ and $P(C=1 \cap B) = 0.3 + 0.7 \times 0.3 = 0.51$.

*When $k \geq 2$ we have only cars to consider. In that case, and only for $k = 2$ we have $P(C=2 \cap A) = 0.8 \times 0.3 = 0.24$ and $P(C=1 \cap B) = 0.7 \times 0.3 = 0.21$. I'll leave the rest of calculations to you.
So, we can now calculate $E[C\cap A]$ and $E[C\cap B]$. I'll leave the detail to you, but $E[C\cap A] = 2.12$ and $E[C\cap B] = 1.98$.
So, now that we've got that, we can say that the average number of people getting out of the vehicle when it comes from $A$ is 0.212 (Because of the 10% probability that the vehicle stops at the restaurant in the first place) and 0.198 for $B$.
But that's for one vehicle. In one hour, we've got on average 60 vehicles coming from $A$ and 80 from $B$. So, in that hour, that means that, on average, $60 \times 0.212 = 12.72 $ people will come from $A$ to the restaurant and $80 \times 0.198 = 15.84$ will come from $B$ to the restaurant.
Knowing vehicles can't come from another direction, it means that, on average, the restaurant will receive $12.72 + 15.84 = 28.56$ people.
My process is different, but it's still the same.
Question 2
The transfer theorem indicates that $ E[\alpha^Z] = \sum_{k = 0}^{\infty} \alpha^k P(Z=k)$. The most straightforward way to compute this sum would be to find a simple way to express $P(Z=k)$, and to bring ourselves to an expression that can be reduced.
I'll keep digging at this question.
