# Integral of Wiener Process and Central Limit Theorem

I am trying to solve the following exercise:

(1) Given $W$ is a Wiener process, find a constant $M$ such that $\lim\limits_{t\to\infty} \frac{1}{t}\int_{0}^{t}\sin^2W_s ds=M$

(2) Then show that $\frac{1}{\sqrt{t}}\int_{0}^{t}(\sin^2W_s-M) ds$ converges to $N(0,\sigma^2)$ and compute $\sigma^2$.

So far I have only started the first part of the question before getting stumped. I started the following: I simply integrated by parts with respect to s and evaluated at the bounds 0 and t

$$\lim_{t\to\infty} \frac{1}{t}\int_{0}^{t}\sin^2W_s ds= \lim_{t\to\infty} \frac{1}{t}\left(\frac{1}{2}\Big((W_t-\sin(W_t)\cos(W_t)-(W_0-\sin(W_0)\cos(W_0)\Big)-\int_{0}^{t}sd(\sin^2W_s)\right)$$

Given $W_0=0$ We have the following:

$$\lim_{t\to\infty} \frac{1}{t}\left(\frac{1}{2}(W_t-\sin(W_t)\cos(W_t))-\int_{0}^{t}sd(\sin^2W_s)\right)$$

I am assuming I have to apply something like Ito's formula to the stochastic integral term on the RHS. However, I am stumped how to show this limit equals a constant M, and how to begin the second part of the problem statement.

• Did I make a mistake in my initial statement with the integration? Should I be integrating by parts? Hence producing a stochastic integral term. How can you go about computing limit of $\frac{W_t}{t}$? Oct 8, 2014 at 18:57
• I've edited my answer. For the limit of $\frac{W_t}{t}$, you may consider the integer part $[t]$ of $t$ and show that $W_{[t]} - W_t$ is negligeable when $t$ goes to infinity Oct 8, 2014 at 20:46
• @NicoDean You modified $$\int_{0}^{t}sd(\sin^2W_s)\quad\text{into}\quad\int_{0}^{t}ds(\sin^2W_s).$$ This is probably not what the OP intended.
– Did
Oct 9, 2014 at 17:37
• @Did oh, oops. Then I missed the point of the OPs question; it looked like a typo to me. i'm not familiar about what "sd" should mean in that case. if you understand what the question should have been, could you please correct it? thx. Oct 9, 2014 at 17:44
• @NicoDean I guess with $sd(\sin^2 W_s)$ OP means $s\cdot d(\sin^2 W_s)$ Oct 9, 2014 at 18:55

Firstly, by the strong law of large number, the limit should exist.

Consider $$\theta_k = \inf\{t\geq \theta_{k-1}, |W_t- W_{\theta_{k-1}}| = \pi\}, \theta_0 = 0$$ Then the integral can be written as the sum of $\int_{\theta_{k}}^{\theta_{k+1}}\sin^2(W_s)ds$, we have

\begin{align} \frac{1}{t}\int_0^t \sin^2(W_s)ds = \dfrac{\sum_{k=1}^{+\infty}1_{\theta_k < t}}{t} \dfrac{\sum_{k=0}^{+\infty}\int_{\theta_{k}}^{\theta_{k+1}}\sin^2(W_s)ds1_{\theta_{k+1} < t} + \int_{\theta_{k+1}}^t \sin^2(W_s)ds1_{\theta_{k+1} < t \leq \theta_{k+2}} }{\sum_{k=1}^{+\infty}1_{\theta_k < t}} \end{align}

Remark that $$\dfrac{\sum_{k=0}^{+\infty}\int_{\theta_{k}}^{\theta_{k+1}}\sin^2(W_s)ds1_{\theta_{k+1} < t} + \int_{\theta_{k+1}}^t \sin^2(W_s)ds1_{\theta_{k+1} < t \leq \theta_{k+2}} }{\sum_{k=1}^{+\infty}1_{\theta_k < t}}\to E\int_0^{\theta_1}\sin^2(W_s)ds$$ alomst surely by the strong law of large number because $\int_{\theta_{k}}^{\theta_{k+1}}\sin^2(W_s)ds$'s are i.i.d.

And that $$\dfrac{\sum_{k=1}^{+\infty}1_{\theta_k < t}}{t} \to \dfrac{1}{E(\theta_1)}$$ alomost sutrly(Imagine a Poisson process which jumps once a new $\theta_k$ is reached)

So the limit is equal to $\dfrac{E\int_0^{\theta_1}\sin^2(W_s)ds}{E(\theta_1)}$. To compute it, let $$I_1 = \frac{1}{t}\int_0^t \sin^2(W_s)ds$$ $$I_2 = \frac{1}{t}\int_0^t \cos^2(W_s)ds$$ then we have $I_1 + I_2 = 1$.

Intuitively, when $t\to \infty$, the limit of $I_1$ and $I_2$ should be the same, since $\sin^2 x$ is just $\cos^2 x$ delayed by $\frac{\pi}{2}$, so we get $M = \frac{1}{2}$.

To prove it rigorously, let $\tau = \inf\{t \geq 0, W_t = \frac{\pi}{2}\}$, then \begin{align} I_2 &= \frac{1}{t} \left(\int_0^\tau \cos^2(W_s)ds + \int_{\tau}^t\cos^2(W_s)ds\right) \\ & = \frac{1}{t} \left(\int_0^\tau \cos^2(W_s)ds + \int_{\tau}^t\sin^2(W_s - \frac{\pi}{2})ds\right) \end{align}

Since $\tau$ is finite almost surely: $$\lim I_2 = \lim \frac{1}{t} \int_{\tau}^t\sin^2(W_s - \frac{\pi}{2})ds = \lim \frac{1}{t} \int_{0}^{t-\tau}\sin^2(W_s')ds = \lim I_1$$ where $W_s' = W_{s+ \tau} - \frac{\pi}{2}$

So the limit of $I_1$ and $I_2$ are both equal to $\frac{1}{2}$

Once the first part is finished, we can realize that we can treat the problem as a renewal process, which jumps once a new $\theta_k$ is reached and with the jump size $\int_{\theta_{k-1}}^{\theta_k}\sin^2(W_s)ds$. Therefore we can apply the central limit theorem for renewal process to resolve the second question.

• How can I compute the variance of the compound poisson process to answer the 2nd part of the second question? Oct 9, 2014 at 0:57
• @user75514 this could be found in a description about the CLT of CPP, for example, here or here Oct 9, 2014 at 6:31
• Still a bit lost. From the Columbia reference, let $N(t)$ be a Poisson process with rate $\lambda$ so that $E[N(t)]=\lambda t$. Let $X_i$ be iid random variables independent of N, then $D(t)=\sum\limits_{i=1}^{N(t)} X_i, t\ge 0$ is a compound Poisson process. This process has $E[D(t)]=\lambda t E[X_1]$ and $Var[D(T)]=\lambda t E [X_1^2]$. From part (1) we calculated $E[D(t)]$. So is $\lambda t= \frac{1}{E(\theta_1)}$? We used the relationship between sine and cosine to compute the entire quantity $\lambda t E[X_1]$. Do I follow a similar strategy for calculating $\lambda t E [X_1^2]$? Oct 9, 2014 at 13:29
• @user75514 by the first part we have $\lambda = \dfrac{1}{E\theta_1}$, so the variance we are looking for is $\lambda E(X_1^2)$. I spent a while to figure it out but didn't succeed. Oct 9, 2014 at 16:48