Buses arrive at a restaurant as a Poisson process with rate $\alpha$. Each bus contains a random number of hungry customers constituting iid non-negative integer valued random variables. The restaurant has infinite servers. Each customer spends a random amount of time ordering and waiting for his food such that the waiting times are iid and independent of other random variables.
Let $X(t)$ be the number of customers whp arrived in $[0,t]$ and are still waiting for food at time $t$.
- Find $E(X(t))$
- What is the distribution of $X(t)$?
If the random number of hungry customers in each bus was not mentioned, I would have proceeded using thinning with $X(t)$ being a Poisson process with mean measure $$\lambda\int_0^t(1-G(t-s))ds$$ where, the waiting times have the common distribution $G$. However, with this new random variable introduced, I can't figure how to proceed. Each bus has a certain probability of containing some customers, does that mean a factor will be introduced in the above integraln somewhere? Where and how?
For part(2), I need compound processes but a little hint would be great.
To find the expected value of $X(t)$, we can use the fact that $X(t)$ is a counting process, which means that it can be written as a sum of independent indicator random variables.
In this case, each customer arriving at the restaurant corresponds to an indicator random variable that is 1 if the customer is still waiting for food at time $t$ and 0 otherwise. Therefore, we can write:
$$X(t) = \sum_{i=1}^{N(t)} I_i$$
where $N(t)$ is the number of customers who have arrived at the restaurant by time $t$ and $I_i$ is an indicator random variable for the $i$th customer.
Since the number of customers arriving at the restaurant follows a Poisson process with rate $\alpha$, the expected value of $N(t)$ is $\alpha t$. Furthermore, the expected value of each indicator random variable $I_i$ is $1-G(t)$, where $G(t)$ is the distribution function of the waiting times.
Therefore, we can compute the expected value of $X(t)$ as:
$$E(X(t)) = \sum_{i=1}^{\alpha t} E(I_i) = \alpha t (1-G(t))$$
For the distribution of $X(t)$, we can use the fact that $X(t)$ is a compound Poisson process. This means that it can be written as the sum of independent and identically distributed random variables, where the number of random variables is itself a Poisson random variable.
In this case, the random variables are the indicator random variables $I_i$ and the Poisson random variable is $N(t)$. Therefore, the distribution of $X(t)$ is given by:
$$P(X(t) = k) = \sum_{n=k}^\infty P(N(t) = n)P\left(\sum_{i=1}^n I_i = k\right)$$
where $P(N(t) = n)$ is the probability mass function of a Poisson random variable with mean $\alpha t$ and $P\left(\sum_{i=1}^n I_i = k\right)$ is the probability mass function of a sum of $n$ independent and identically distributed Bernoulli random variables with success probability $1-G(t)$.
To compute this distribution explicitly, you can use the probability mass function of the Poisson distribution and the binomial distribution.