The time between accidents on the Riverfront Bridge follows a Poisson process with a mean time of 40 days between accidents. The time between accidents on the Overview Bridge follows a Poisson process with a mean time of 20 days between accidents. These two bridges are in separate states. Let Z = the average of the time until the next accident on the Riverfront Bridge and the time until the next accident on the Overview Bridge. The moment generating function of Z is then given by:
My approach:
It very clear that the following two identical distribution are independent, therefore we can use the linear combination technique to solve the problem.
Given that the moment generating function of poisson distribution is $e^{\lambda(e^t - 1)}$. Hence by linear combination we can define the following:
$$M_{x_i}(t) = \prod_{i=1}^{n} M_{x_i}(t) = \prod_{i=1}^{n}e^{\lambda(e^t - 1)} = e^{\sum \lambda(e^t - 1)}$$
Hence, I got my answer for the following problem as $$e^{60(e^t - 1)}$$
Can anyone check the answer for me?
There's one key misunderstanding in your approach. A Poisson process with mean time $T$ between accidents has a mean of $\lambda=1/T$ events per day. Times between accidents are actually exponentially distributed for Poisson processes, not Poisson-distributed. The process is named for the distribution of the event count per unit time.
The time until the next accident on the Riverfront bridge has moment-generating function $\frac{1}{1-40t}$ (if we work with the day as our time unit). Now using $M_{(X+Y)/2}(t)=M_X(\frac{t}{2})M_Y(\frac{t}{2})$, the answer to the question is actually $\frac{1}{(1-10t)(1-20t)}$. The MGF's convergence condition is $t<\frac{1}{20}$.