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!


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

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    $\begingroup$ Dini's theorem won't help; $Y(nu)$ is not a continuous function of $u$, nor is the convergence monotone in general. $\endgroup$ – Nate Eldredge Feb 28 '14 at 17:08
  • $\begingroup$ 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 "$\geq u_0$"). $\endgroup$ – 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:

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

  • $\begingroup$ 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. $\endgroup$ – user52227 Mar 14 '17 at 13:43
  • $\begingroup$ @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 $$\mathbb{P} \left( \sup_{k \geq 1} |S_k| \geq 3\epsilon \right) = \sup_{n \geq 1} \mathbb{P} \left( \sup_{k \leq n} |S_k| \geq 3 \epsilon \right) \leq 3 \sup_k \mathbb{P}(|S_k| \geq \epsilon).$$ $\endgroup$ – saz Mar 14 '17 at 14:16
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    $\begingroup$ 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. $\endgroup$ – saz Mar 14 '17 at 14:18
  • $\begingroup$ Thank you. I think I understand now $\endgroup$ – user52227 Mar 14 '17 at 15:42
  • $\begingroup$ @user52227 You are welcome. $\endgroup$ – saz Mar 14 '17 at 15:45

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