| Front | Theorem. Suppose that \(\xi\) is non-arithmetic. Let \(E[\xi] = 1/\lambda\). Then\[L(t) \to \hat{\xi}\]in distribution as \(t \to \infty\). Moreover, for all \(y \geq 0\),\[\mathbb{P}(E(t) \leq y) \to \lambda \int_{0}^{y} \mathbb{P}(\xi > x) dx\]as \(t \to \infty\) and the same result holds with \(A(t)\) in place of \(E(t)\). In fact,\[(L(t), E(t)) \to (\hat{\xi}, U\hat{\xi})\]in distribution as \(t \to \infty\), where \(U\) is uniform on \((0, 1)\) and is independent from \(\hat{\xi}\). The same result holds with the pair \((L(t), A(t))\) instead of \((L(t), E(t))\). |
| Back | (\(\xi\) discrete). We prove the theorem, in the case where \(\xi\) is a discrete random variable taking values in \(\mathbb{N}\). We start by proving the second formula which is slightly easier.Then, as suggested by Figure 3, \((E(t), t = 0, 1, \ldots)\) forms a discrete time Markov chain with transitions\[p_{i,i-1} = 1\]for \(i \geq 1\) and\[p_{0,n} = \mathbb{P}(\xi = n + 1)\]for all \( n \geq 0 \). It is clearly irreducible and recurrent, and an invariant measure satisfies:\[\pi_n = \pi_{n+1} + \pi_0\mathbb{P}(\xi = n + 1)\]thus by induction we deduce that\[\pi_n := \sum_{m \geq n+1} \mathbb{P}(\xi = m)\]is an invariant measure for this chain. This can be normalised to be a probability measure if \(\mathbb{E}[\xi] < \infty\) in which case the invariant distribution is\[\pi_n = \lambda\mathbb{P}(\xi > n).\]We recognise the formula (4.2) in the discrete case where \( t \) and \( y \) are restricted to be integers. Using the assumption that \( \xi \) is non-arithmetic gives that the Markov chain \( E \) is aperiodic. Since it is also irreducible, we can apply the convergence to equilibrium theorem to get (4.2).We now consider the slightly more delicate result (4.3), still in the discrete case \( \xi \in \{1, 2, \ldots\} \). Of course, once this is proved, (4.1) follows. Observe that \(\{(L(t), E(t)); t = 0, 1, \ldots\}\) also forms a discrete time Markov chain in the space \( \mathbb{N} \times \mathbb{N} \) and more precisely in the set\[S = \{(n, k) : 0 \leq k \leq n - 1\}.\]The transition probabilities are given by\[p_{(n, k) \to (n, k - 1)} = 1\]if \( k \geq 1 \) and\[p_{(n, 0) \to (k, k - 1)} = \mathbb{P}(\xi = k).\]This is an irreducible recurrent chain for which an invariant measure is given by \(\pi(n, k)\) where:\[\pi(n, k - 1) = \pi(n, k)\]for \( 0 \leq k \leq n - 1 \) and\[\pi(k, k - 1) = \sum_{m=0}^{\infty} \pi(m, 0)\mathbb{P}(\xi = k).\]So taking \(\pi(n, k) = \mathbb{P}(\xi = n)\) works. This can be rewritten as\[\pi(n, k) = n\mathbb{P}(\xi = n) \times \frac{1}{n} \{0 \leq k \leq n - 1\}.\]After normalisation, the first factor becomes \(\mathbb{P}(\xi = n)\) and the second factor tells us that \(E(t)\) is uniformly distributed on \(\{0, \ldots n - 1\}\) given \(L(t) = n\) in the limit. The theorem follows. |