91Ó°ÊÓ

Let's first work on the given hint. We need to determine whether \(N(t) + 1 = n\) can be expressed in terms of the information available at time \(t\). From the hint, we can write: $$ N(t)=n \Leftrightarrow X_{1}+\cdots+X_{n} \leqslant t \quad \text { and } \quad X_{1}+\cdots+X_{n+1}>t $$ Using these expressions, we can now rewrite \(N(t) + 1 = n\) as: $$ N(t) + 1 = n \Leftrightarrow X_{1}+\cdots+X_{n-1} \leqslant t \quad \text { and } \quad X_{1}+\cdots+X_{n}>t $$ Since it only includes arrival times up to time t, we can determine whether \(N(t) + 1 = n\) based on information available at time \(t\). Therefore, \(N(t) + 1\) is a stopping time.

Using Wald's equation, for a stopping time \(n\), we have: $$ E\left[\sum_{i=1}^{n} X_{i}\right] = \mu E[n] $$ Since we proved that \(N(t) + 1\) is a stopping time in part (a), we can use Wald's equation for it: $$ E\left[\sum_{i=1}^{N(t) + 1} X_{i}\right] = \mu E[N(t) + 1] $$ Now we need to relate \(E[N(t)+1]\) with \(m(t)\). Since \(N(t)\) is the number of renewals by time \(t\), we can easily relate them as follows: $$ E[N(t) + 1] = E[N(t)] + 1 = m(t) + 1 $$ Therefore, we have: $$ E\left[\sum_{i=1}^{N(t) + 1} X_{i}\right] = \mu[m(t) + 1] $$

Given that interarrival times are bounded by a constant \(M\), we have: $$ P\{X_i < M\} = 1 $$ Now let's consider the sum of the interarrival times up to \(N(t) + 1\): $$ t < \sum_{i=1}^{N(t)+1} X_{i} < t + M $$ The left-hand inequality holds since \(N(t)\) is the maximum number of renewals up to time \(t\), and there hasn't been a renewal after time \(t\). The right-hand inequality holds because the maximum possible value of any single interarrival time is \(M\), and since there has been \(N(t)\) renewals up to time \(t\), the maximum possible extra value would only be \(M\).

To prove the elementary renewal theorem when the interarrival times are bounded, we will use the results we have found in the previous parts. We know that: $$ E\left[\sum_{i=1}^{N(t)+1} X_{i}\right] = \mu[m(t)+1] $$ It's also true that: $$ t < \sum_{i=1}^{N(t) + 1} X_{i} < t + M $$ Now, let's take expectations for all sides of this inequality: $$ E[t] < E\left[\sum_{i=1}^{N(t)+1} X_{i}\right] < E[t + M] $$ Using the result from part (b), we can replace the middle term: $$ E[t] < \mu[m(t) + 1] < E[t + M] $$ Dividing all sides by \(\mu\): $$ \frac{E[t]}{\mu} < m(t) + 1 < \frac{E[t + M]}{\mu} $$ As \(t\to\infty\), we have: $$ \lim_{t\to\infty} \frac{E[t]}{\mu} \leq \lim_{t\to\infty} m(t) \leq \lim_{t\to\infty} \frac{E[t + M]}{\mu} $$ Since \(\lim_{t\to\infty} E[t] = \lim_{t\to\infty} E[t + M] = \infty\), we can conclude that: $$ \lim_{t\to\infty} m(t) = \infty $$ Thus, the elementary renewal theorem is proven when interarrival times are bounded.