# Mixing average and the Renewal Reward Theorem

I tried to solve the following problem:

Buses arrive to an archeological site according to the discrete renewal process with i.i.d inter-arrival times T1, T2, T3, ... which are distributed Geo(p). Assume k−th bus carries a random number Nk of tourists, and that all Nk -s are independent and, for each n = 1, 2, ... the conditional distribution of Nk given Tk+1 = n is U ni{1, ..., n + 2}. According to the archeological site regulations, tourists enter the site immediately upon arrival and leave it as soon as the next bus arrives. Compute the average time (in a long run) that a tourist spends at the site.

I'm pretty sure I'm not fully understanding some concepts and here is the solution that I tried:

First, I noticed that the time each bus stays at the site is the same as the time between buses' arrivals, so I defined the reward as follows: $$\begin{equation*} R_i = T_{i+1} \end{equation*}$$ Then, I defined the total number of tourists that visit the site as: $$\begin{equation*} m = \sum_{k=1}^{M(t)} N_k \end{equation*}$$ What I believe I need to calculate is the average time a tourist stays at the site, which is the following ($$M(t)$$ is the total number of arrivals and $$C_T$$ the total time needed to for all tourists to visit the site): $$\begin{equation*} \frac{1}{m}\lim_{t\to\infty} \frac{\sum_{i=1}^{M(t)} R_i}{t} = \frac{1}{m} \lim_{t\to\infty} \frac{C_T}{t} \end{equation*}$$

My problem is now concluding the solution with the Renewal Reward Theorem. If I didn't have the average ($$m$$), this would be really straight forward, but given $$m$$ I'm not sure how to proceed.

RRT:

$$\begin{equation*} \lim_{t\to\infty} \frac{C_T}{t}\xrightarrow{t\to\infty} \frac{\mathbb{E}(R_i)}{\mathbb{E}(T)} \end{equation*}$$

I thought I might get the result from RRT and just divide it by $$m$$. Like this, considering that the renewal and the interarrival time follow the same geometric distribution:

$$\begin{equation*} \lim_{t\to\infty} \frac{C_T}{t}\xrightarrow{t\to\infty} \frac{\mathbb{E}(R_i)}{\mathbb{E}(T)} = \frac{p}{p} = 1 \end{equation*}$$

And then the result would just be $$\frac{1}{m}$$. Would that be right?

Define $$A_n$$ - total time, that tourists that arrive on first $$n$$ buses spent on the site. It is easy to see that

$$A_n = \sum_{n=1}^{n}N_kT_{k+1}$$.

Notice that Tk+1 is a time, spent by tourist, who arrived on bus number n.

Define $$B_n$$ - total number of tourists arrived on first n buses. It is also easy to see that

$$B_n \sum_{k=1}^{n}N_k$$.

Define:

$$D_n = \frac{A_n}{B_n}$$,

and average time tourists spent on the bus. We want to find $$\lim_{n\to\infty} D_n$$. This can be done with help of SLLN:

$$\lim_{n\to\infty} D_n = \lim_{n\to\infty} \frac{A_n}{B_n} = \lim_{n\to\infty}\frac{\frac{1}{n}\sum_{n=1}^{n}N_kT_{k+1}}{\frac{1}{n}\sum_{k=1}^{n}N_k} = \frac{\mathbb{E}(N_1T_2)}{\mathbb{E}(N_2)} = \frac{\mathbb{E}(\mathbb{E}(N_1T_2|T_2))}{\mathbb{E}(N_2|T_2)} = \frac{\mathbb{E}(T_2(T_2 +3))/2}{\mathbb{E}(\frac{T2 +3}{2})} = \frac{q/p^2 +3/p}{1/p+3}$$