Recall that if \(N\) is a Poisson counting process, then it has exponential inter-arrival times all with the same rate; specifically, we may think of \[ N(t) = \sum_{k=1} ^{\infty} \mathbf{1}[T_k \leq t],\] where \(T_0= 0\) and \[T_k = \sum_{i=1} ^{k} X_i,\] where \(X_i\) are i.i.d. exponential random variables.

We now allow for the possibility that the *positive* inter-arrival times \(X_i\) are i.i.d. random variables that are *not* necessary exponential, and we refer to \(N\) as a **renewal proceess**. We will always assume that \(\mathbb{P}(X_1 = 0) =0\) and \(\mathbb{E} X_1= \mu <\infty\).

Just from our basic knowledge of the law of large numbers and the central limit theorem, we have the following results.

**Theorem (Law of large numbers):** If \(N\) is a renewal process with mean inter-arrival time \(\mu\) then, \[N(t)/ t \to \mu^{-1}\] almost surely, as \(n \to \infty\).

**Theorem (Central limit theorem):** If \(N\) is a renewal process with inter-arrival times with finite mean \(\mu\) and variance \(\sigma^2\), then

\[ \frac{ N(t) - t/\mu}{ \sqrt{t \sigma^2 / \mu^3}}\] converges in distribution to a standard normal, as \(t \to \infty\).

*Proof (Law of large numbers):* We make use of the observation that \[ T_{N(t)} \leq t < T_{N(t) +1}.\] Hence

\[ \frac{N(t)}{T_{N(t) +1}} \leq \frac{N(t)}{ t} \leq \frac{N(t)}{ T_{N(t)}}.\] We also know that if \(X_i\) are inter-arrival times, then \[T_{N(t)} = \sum_{i=1} ^{N(t)} X_i\]

so that the regular strong law of large numbers gives that \[T_{N(t)} /N(t) \to \mu\] almost-surely, as \(t \to \infty\), since \(N(t) \to \infty\), as \(t \to \infty\). Hence the desired result follows.

We will not give a proof the central limit theorem for renewal processes, but we can see why \(var(N(t)) \approx t \sigma^2 / \mu^3\). The law of large numbers tells us that \[ N(t) \approx \frac{ T_{N(t)}}{\mu}\] so that \[ N(t) \approx \frac{1}{\mu} \sum_{i=1} ^{{N(t)}} X_i\] where the \(X_i\) are the inter-arrival times. We also have by the law of large numbers for renewal processes that \[{N(t)} \approx \frac{t}{\mu}\] thus

\[ var(N(t)) \approx var\big( \frac{1}{\mu} \sum_{i=1} ^{ \frac{t}{\mu}} X_i \big) \approx t \sigma^2 / \mu^3,\] where \(var(X_1) = \sigma\).

In this section, we will prove a version of the ergodic theorem for Markov chains by breaking the chain up into independent parts. Let \(X\) be a Markov chain on a state space \(S\). Recall that we say that a state \(s \in S\) is **recurrent** if \[\mathbb{P}( \text{there exists } n\geq 1 \text{ such that } X_n=s \ | \ X_0=s)=1.\] Furthermore, let \(X\) be a Markov chain started at \(s\) and \(T = \inf \{n \geq 1 : X_n=s\}\); we say that a recurrent state \(s\) is **positive-recurrent** if \(\mathbb{E} T < \infty\) and **null-recurrent** if \(\mathbb{E} T = \infty\).

**Theorem (Law of large numbers):** Let \(X\) be an irreducible Markov chain on a countable state space \(S\). Let \(P\) be the transition matrix for \(S\), and assume all the states are positive-recurrent. Fix \(s \in S\). Start the chain \(X\) at \(s\). Let \(T_1 = \inf \{n \geq 1 : X_n=s\}\) and let \[V_n = \sum_{k=0} ^{n} \mathbf{1}[X_k=s].\] Then \[V_n/n \to (\mathbb{E} T_1)^{-1}\] almost surely.

The starting point of the proof of this version of the law of large numbers is the following observation. Let \(T_1 = \inf\{n >0: X_n =s\}\). Set \[T_{n} = \inf\{n >T_{n-1}: X_n=s\}.\] Then \(T_n = \sum_{k=1} ^{n-1} (T_{k+1} - T_k) + T_1\) and \(Y_k = T_{k+1} - T_k\) are i.i.d. random variables. This may appear obvious from the Markov property, but it is a seemingly stronger claim (that turns out is implied by the regular Markov property), since we need to argue that Markov chain *renews* itself each time it reaches the state \(s\), which is a *random* time, rather than a deterministic time.

Let \(X = (X_i)_{i \in \mathbb{N}}\) be a random process taking values in \(S\). If \(T\) is a nonnegative integer-valued random variable with the property that for all \(n \in \mathbb{N}\) there exists a deterministic function \(\phi_n: S^n \to \{0,1\}\) such that \(\mathbf{1}[T=n] = \phi_n(X_0, X_1, \ldots, X_n)\), then we say that \(T\) is a **stopping time**. Thus \(T\) does not look into the future, to determine whether to *stop* or not. We will usually assume that \(\mathbb{P}(T < \infty) = 1\).

**Theorem (Strong Markov property:** Let \(X\) be a Markov chain taking values in a state space \(S\) with a transition matrix \(P\). If \(T\) is a stopping time, then conditional on the event \(\{X_T=s\}\) we have that \(Y=(X_{T+k})_{k=0} ^{\infty}\) is a Markov chain started at \(s\) with transition matrix \(P\) that is independent of \(Z=(X_k)_{k=0}^{T}\).

*Proof:* We will show that \[\mathbb{P}(Y \in A, Z \in C \ | \ X_T=s) = \mathbb{P}(X \in A \ | \ X_0 =s)\mathbb{P}(Z \in C \ | \ X_T=s),\]

for sets \(A\) of the form \[A= \{a \in S^{\mathbb{N}}: a_0=z_0, \ldots, a_k=z_k\}.\] The event \(\{Z \in C\}\) that we need to check are given by the disjoint union of \[\{Z \in C\} \cap \{T=n\}.\]

Thus we will check events of the form \(B_n \cap \{T=n\}\), where \[B_n =\{ X_0 = b_0, X_1=b_1, \ldots, X_{n-1}=b_{n-1}, X_n=b_n\}.\] Recall that the event \(\{T=n\}\) depends on only on the the Markov chain up to time \(n\). The (regular) Markov property gives that \[\mathbb{P}(X_T=z_0, \ldots, X_{T+k} = z_{k}, \ B_n, T=n, X_T=s) =\] \[\mathbb{P}(X_0=z_0, \ldots, X_k = z_k \ | X_0=s)\mathbb{P}(B_n, T=n, X_T=s).\] Summing over all \(n\geq 0\), we have \[\mathbb{P}(X_T=z_0, \ldots, X_{T+k} = z_{k}, Z \in C, X_T=s) =\] \[\mathbb{P}(X_0=z_0, \ldots, X_k = z_k \ | X_0=s)\mathbb{P}(Z \in C, X_T=s),\] and dividing by \(\mathbb{P}({X_T=s})\), we obtain the required result.

*Proof of law of large numbers:*

By the Strong Markov property the random variables \(Y_k = T_{k+1} - T_k\) are i.i.d. random variables. The law of large numbers gives that \(T_n/ n \to \mathbb{E} T_1\) almost surely. Observe that \(V_{T_n} = n+1\), and thus \(V_{T_n} /T_n \to (\mathbb{E} T_1)^{-1}\). Hence \(V_n / n \to (\mathbb{E} T_1)^{-1}\), since \(T_n \to \infty\).

**Corollary** Let \(\pi\) is the stationary distribution and \(T^s = \inf \{ n \geq 1: X_n =s | X_0 =s\}\). Then \[(\mathbb{E} T^s)^{-1} = \pi(s).\]

*Proof (aperiodic irreducible finite state space)* We proved that \(V_n/ n \to \pi(s)\) in probability. We also proved that \(V_n/ n \to (\mathbb{E} T^s)^{-1}\) almost surely, and thus also in probability. thus \((\mathbb{E} T^s)^{-1} = \pi(s)\).

Recall that the holding times a continuous time Markov chain were chosen to be exponential, as to preserve the Markov property. When we allow for the possibility that the holding times are *not* exponential, then resulting process is often referred to as a *Semi-Markov* or **Markov-renewal** process and the jump chain a **hidden** Markov chain.

We introduced varies relaxations of models that we previously studied.

In particular, we generalized the Poisson counting processes by allowing non-exponential inter-arrival times.

We observed basic limit theorems that follow from the law of the large numbers and the central limit theorem.

We also applied renewal type arguments to the study of Markov chains