a.s convergence of $\frac{W_{t}}{N_{t}}$, $\frac{W_{t}}{\sqrt{N_{t}}}$, $\frac{|W_{t}|}{\sqrt{t \ln \ln t^{2}}}$

Question

Let $W$ be a standard Wiener process in $\mathbb{R}$, and let $N$ be a Poisson process with parameter $\lambda$. Decide if those limits exist a.e and calculate them:

$\lim_{t \rightarrow \infty}\frac{W_{t}}{N_{t}}$

$\lim_{t \rightarrow \infty}\frac{W_{t}}{\sqrt{N_{t}}}$,

$\lim_{t \rightarrow \infty}\frac{|W_{t}|}{\sqrt{t \ln \ln t^{2}}}$

I know that the last one can be solved by the law of the iterated logarithm, but I'm not sure how, because of the absolute value.

I was trying to solve the first one by using the large number's law, like this:

$\lim_{t \rightarrow \infty}\frac{W_{t}}{N_{t}}$ = $\lim_{t \rightarrow \infty} \frac{W_{t}}{t} \cdot \frac{t}{N_{t}} = \lim_{M \rightarrow \infty} \frac{W_{0} + W_{1} - W_{0} + \ldots + W_{M} - W_{M-1}}{M+1} \cdot \frac{M+1}{N_{0} + N_{1} - N_{0} + \ldots + N_{M} - N_{M-1}}$ , where $M \in \mathbb{N}$. The problem is, I can't split it into a product of two limits, because each of them would equal $0$...

Answer 1

It follows from the strong law of large numbers that $$\frac{N_t}{t} \xrightarrow[]{t \to \infty} \mathbb{E}(N_1)= \lambda \quad \text{a.e.} \tag{1}$$ and $$\frac{W_t}{t} \xrightarrow[]{t \to \infty} \mathbb{E}(W_1)=0 \quad \text{a.e.} \tag{2}$$ Combining $(1)$ and $(2)$, yields $$\lim_{t \to \infty} \frac{W_t}{N_t} = \lim_{t \to \infty} \frac{W_t}{t} \frac{t}{N_t} = 0 \cdot \lambda = 0. \tag{3}$$

For the second limit, use the law of iterated logarithm to show that $$\frac{W_t}{\sqrt{t}}$$ does not converge a.e. as $t \to \infty$, and then use the same trick as in $(3)$ to conclude that the pointwise $\lim_{t \to \infty} W_t/\sqrt{N_t}$ does not exist.

For the third limit, note that by the law of iterated logarithm $$\limsup_{t \to \infty} \frac{W_t}{\sqrt{2t \log \log t}}=1.$$ By the symmetry of Brownian motion, i.e. the fact that $-W_t$ is also a Brownian motion, we also have $$\limsup_{t \to \infty} \frac{-W_t}{\sqrt{2t \log \log t}}=1,$$ i.e. $$\liminf_{t \to \infty} \frac{W_t}{\sqrt{2t \log \log t}}=-1.$$ Because of the continuity of the sample paths, this implies $$0 = \liminf_{t \to \infty} \frac{|W_t|}{\sqrt{2t \log \log t}} < \limsup_{t \to \infty} \frac{|W_t|}{\sqrt{2 t\log \log t}} = \infty.$$ Study the behaviour of $$\frac{\sqrt{2t \log \log t}}{\sqrt{t \log \log (t^2)}}$$ for $t \to \infty$, to find out whether the third limit exists or not.