# Strong law of large numbers for Poisson process

My question regards the strong law of large numbers as stated, e.g., in Ethier and Kurtz (1986, p. 456 Eq. (2.5)), as follows:

If $Y$ is a unit Poisson process, then for each $u_0>0$, \begin{eqnarray*} \lim_{n \to \infty} \sup_{u \geq u_0} \vert Y(nu)/n - u \vert = 0 \quad a.s. \end{eqnarray*}

My question regards the uniform convergence. More precisely, I see how to apply the strong law of large numbers in order to get \begin{eqnarray*} \lim_{n \to \infty} \vert Y(nu)/n - u \vert = 0 \quad a.s. \end{eqnarray*} for each $u>0$ (just note that $Y(nu) = \sum_{k=1}^n Y(ku)-Y((k-1)u)$ is a sum of i.i.d. random variables with mean $u$ since $Y$ is a unit rate Poisson process). But I don't know how to show that the convergence is not only pointwise, but also uniform. I suspect that if I could show that $Y(nu)/n$ is an increasing sequence of functions, i.e., $Y(nu)/n \leq Y((n+1)u)/(n+1)$, I could use Dini's theorem to show the uniform convergence. I don't see, however, how to show this monotonicity either.

I've been thinking about the proof now for a while and would be happy if somebody could help!

Reference:

Ethier, S. N. and Kurtz, T. (1986). Markov processes: Characterization and Convergence. John Wiley & Sons, Inc.

• Dini's theorem won't help; $Y(nu)$ is not a continuous function of $u$, nor is the convergence monotone in general. – Nate Eldredge Feb 28 '14 at 17:08
• I believe it should read $$\lim_{n \to \infty} \sup_{u \leq u_0} |Y(nu)/n-u|=0$$ (so "$\leq u_0$" instead of "$\geq u_0$"). – saz Mar 1 '14 at 8:45

In my edition of the book it reads

$$\lim_{n \to \infty} \sup_{\color{red}{u \leq u_0} } \left| \frac{Y(nu)}{n}-u \right|=0 \quad \text{a.s.}$$

So let's prove this. Fix $\varepsilon>0$. For any $n \in \mathbb{N}$ we have by Etemadi's inequality

$$p_n := \mathbb{P} \left( \sup_{u \leq u_0} \left| \frac{Y(nu)}{n}-u\right| > 3\varepsilon \right) \leq 3\sup_{u \leq u_0} \mathbb{P} \left( \left| \frac{Y(nu)}{n}-u\right|>\varepsilon \right). \tag{1}$$

The idea is to show that $$\sum_{n \in \mathbb{N}} p_n<\infty; \tag{2}$$ the claim then follows from the Borel-Cantelli lemma. In order to prove $(2)$ we note that we can choose a constant $C>0$ such that for any $|\lambda| \leq 1$

$$\mathbb{E}e^{\lambda \tilde{Y}_t} \leq e^{Ct \lambda^2}, \tag{3}$$

where $\tilde{Y}_t :=Y_t-t$ denotes the compensated Poisson process; see the lemma below. The (exponential) Markov inequality and $(1)$ then shows

\begin{align*} p_n &\leq 3\sup_{u \leq u_0} \mathbb{P} \bigg( Y(nu)-nu>\varepsilon n\bigg)+3\sup_{u \leq u_0} \mathbb{P} \bigg( -(Y(nu)-nu)>\varepsilon n \bigg) \\ &\leq 3 \sup_{u \leq u_0} \bigg[ \exp \left(\lambda \tilde{Y}(nu)-\varepsilon n \lambda\right)+ \exp \left(-\lambda \tilde{Y}(nu)-\varepsilon n \lambda\right) \bigg]. \end{align*}

If we choose $\lambda=\frac{1}{\sqrt{n}}$ and apply $(2)$, then we get

$$p_n \leq 6 \exp \left( C u_0-\varepsilon \sqrt{n} \right).$$

Obviously, this entails $(2)$.

Lemma Let $(Y_t)_{t \geq 0}$ be a Poisson process (with rate $1$) and $\tilde{Y}:=Y_t-t$ the compensated Poisson process. Then $(3)$ holds.

Proof: Since $Y_t \sim \text{Poi}(t)$, the exponential moments can be calculated explicitely: $$\mathbb{E}e^{\lambda Y_t} = e^{t \cdot (e^{\lambda}-1)}.$$ Hence, $$\mathbb{E}e^{\lambda \tilde{Y}_t} = e^{t \cdot (e^{\lambda}-1-\lambda)}.$$ For $\lambda \in [-1,1]$, we have $$|e^{\lambda}-1-\lambda| \leq C \cdot \lambda^2$$ and this proves $(3)$.

Remark The claim holds for any Lévy process $(Y_t)_{t \geq 0}$ with finite exponential moments:

$$\lim_{n \to \infty} \sup_{u \leq u_0} \left| \frac{Y(nu)}{n}-\mathbb{E}Y_1 \cdot u \right|=0 \quad \text{a.s.}$$

• Etemadi's inequality as stated in the reference given above applies to a finite sequence of random variables. Here we have the sup over an uncountable set and not a max over a finite set. So I don't understand its application. – user52227 Mar 14 '17 at 13:43
• @user52227 Etemadi's inequality gives $$\mathbb{P}(\sup_{k \leq n} |S_k| \geq 3 \epsilon) \leq 3 \sup_{k \leq n} \mathbb{P}(|S_k| \geq \epsilon),$$ right? Obviously this implies $$\mathbb{P} \left( \sup_{k \leq n} |S_k| \geq 3 \epsilon \right) \leq 3 \sup_{k \geq 1} \mathbb{P}(|S_k| \geq \epsilon).$$ Using the monotone convergence theorem (... or the continuity of the probability measure), we get $$\mathbb{P} \left( \sup_{k \geq 1} |S_k| \geq 3\epsilon \right) = \sup_{n \geq 1} \mathbb{P} \left( \sup_{k \leq n} |S_k| \geq 3 \epsilon \right) \leq 3 \sup_k \mathbb{P}(|S_k| \geq \epsilon).$$ – saz Mar 14 '17 at 14:16
• This proves Etemadi's inequality for countable many random variables. Finally, note that if $(X_t)_{t \geq 0}$ is a process with càdlàg sample paths, then $$\sup_{u \leq u_0} |X_u| = \sup_{u \leq u_0, u \in \mathbb{Q}} |X_u|$$ and so everything boils down to the countable setting. Consequently, we get $$\mathbb{P} \left( \sup_{u \leq u_0} |X_u| \geq 3\epsilon \right) \leq 3 \sup_{u \leq u_0} \mathbb{P}(|X_u| \geq \epsilon)$$ for any càdlàg process $(X_t)_t$ with independent increments. – saz Mar 14 '17 at 14:18
• Thank you. I think I understand now – user52227 Mar 14 '17 at 15:42
• @user52227 You are welcome. – saz Mar 14 '17 at 15:45