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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.

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  • $\begingroup$ 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 $\frac{W_t}{t}$? $\endgroup$
    – user75514
    Oct 8, 2014 at 18:57
  • $\begingroup$ 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 $\endgroup$ Oct 8, 2014 at 20:46
  • $\begingroup$ @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. $\endgroup$
    – Did
    Oct 9, 2014 at 17:37
  • $\begingroup$ @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. $\endgroup$ Oct 9, 2014 at 17:44
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    $\begingroup$ @NicoDean I guess with $sd(\sin^2 W_s)$ OP means $s\cdot d(\sin^2 W_s)$ $\endgroup$ Oct 9, 2014 at 18:55

1 Answer 1

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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.

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  • $\begingroup$ How can I compute the variance of the compound poisson process to answer the 2nd part of the second question? $\endgroup$
    – user75514
    Oct 9, 2014 at 0:57
  • $\begingroup$ @user75514 this could be found in a description about the CLT of CPP, for example, here or here $\endgroup$ Oct 9, 2014 at 6:31
  • $\begingroup$ 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 $\lambda t E [X_1^2]$? $\endgroup$
    – user75514
    Oct 9, 2014 at 13:29
  • $\begingroup$ @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. $\endgroup$ Oct 9, 2014 at 16:48

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