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.


\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?


1 Answer 1


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

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


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

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:

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


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