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Let $W_t$ be a standard Brownian motion, i.e., $W_t \sim N(0,t)$. Define the random variable

$$X=\int_0^1\cos(W_t)dt$$

A similiar process, $Y_t=\cos(\omega t+\sigma W_t+\theta)$, with the uniform random variable $\theta\sim U(-\pi,\pi)$ and $\omega\in\Bbb R$, was analyzed by Cai and Wu [1], where the probability density was defined as

$$p_Y(y)=\frac{1}{\pi\sqrt{1-y^2}}$$

Seeing that the process $Y_t$ has a density, how can I obtain the density of the simplified process $\cos(W_t)$ ? I expanded the integral as a Riemann sum,

$$X\approx\frac{1}{n}\sum_{i=1}^n{\cos\left( W_\frac{i}{n}\right)}=\frac{1}{n}\left(n\cos\left(W_\frac{1}{n}\right)+ (n-1)\left(\cos\left(W_\frac{2}{n}\right)-\cos\left(W_\frac{1}{n}\right)\right)+ (n-2)\left(\cos\left(W_\frac{3}{n}\right)-\cos\left(W_\frac{2}{n}\right)\right)+ \cdots + 2\left(\cos\left(W_\frac{n-1}{n}\right)-\cos\left(W_\frac{n-2}{n}\right)\right)+ \cos\left(W_1\right)-\cos\left(W_\frac{n-1}{n}\right) \right)$$

and after rearranging the terms I got the sum $$\frac{1}{n}\sum_{i=1}^n{\cos\left( W_\frac{i}{n}\right)}= \frac{1}{n}\sum_{i=1}^n{-2(n-i+1) \sin\left(\frac{W_\frac{i}{n}+W_\frac{i-1}{n}}{2}\right) \sin\left(\frac{W_\frac{i}{n}-W_\frac{i-1}{n}}{2}\right)}$$

How can I progress from here?


[1] CAI, G. Q.; WU, C. Modeling of bounded stochastic processes. Probabilistic Engineering Mechanics, v. 19, n. 3, p. 197–203, 2004.

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  • $\begingroup$ I find your question somewhat misleading. The process considered in the article is of the form $Y_t = \cos(\sigma W_t+\theta)$ for a uniformly distributed random variable $\theta$... your process $X_t$ is not exactly "similar" to $Y_t$ . In particular, it is not to be expected that $p$ is the density of $X_t$. Please feel free to correct me if I missed something about the article. $\endgroup$ – saz Jan 28 at 15:24
  • $\begingroup$ sure, but the random variable $\theta$ is constant in time. The cosine can them be partitioned in $\cos(\sigma W_t)\cos(\theta)-\sin(\sigma W_t)\sin(\theta)$. In order to understand the final form, I thought that I needed to break the integral into fundamental blocks, which is why I'm specifically asking about this one. But you are right. The process don't need to have the same distributions. I will edit it. $\endgroup$ – Kaio César Borges Benedetti Jan 28 at 15:30
  • $\begingroup$ I'm just saying that $p$ is certainly not going to be the density of $X$. Using that $W_t \sim N(0,t)$ you can verify that $\mathbb{E}(X)>0$. If $Y$ is a random variable with density $p(y)=\frac{1}{\pi \sqrt{1-y^2}} 1_{(-1,1)}(y)$ then it has expectation zero... this shows that something doesn't fit here (or I'm misunderstanding your question). I think it's highly non-trivial to determine the distribution of $X$. $\endgroup$ – saz Jan 28 at 16:03
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    $\begingroup$ If $\theta$ is uniform on $(0,2\pi)$, then, for every real number $z$, $\theta+z$ mod $2\pi$ is also uniform on $(0,2\pi)$, thus, for every $Z$ independent of $\theta$, $\cos(\theta+Z)\stackrel d=\cos\theta$. Now, surely you can compute the PDF of $\cos\theta$... $\endgroup$ – Did Jan 28 at 17:10
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    $\begingroup$ Related: this and this. $\endgroup$ – Rodrigo de Azevedo Jan 29 at 18:08
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I edited my response many times to correct some mistakes. Sorry.

We can compute the density of $U_t=\cos(W_t)$ by appropriately separating the integral and a change of variable. For any measurable $f$

\begin{align} \mathbb{E}(f(U_t)) &= \mathbb{E}(f(\cos(W_t))) = \int_\mathbb{R} f(\cos(w)) \mathbb{P}_{W_t}(w) \mathrm{d} w = \sum_{k \in \mathbb{Z}} \int_{k\pi}^{(k+1)\pi} f(\cos(w)) \mathbb{P}_{W_t}(w) \mathrm{d} w \\ & = \sum_{k \in \mathbb{Z}} \int_{-1}^{1} f(u) \left( \mathbb{P}_{W_t}(\arccos(u)+2k\pi)+\mathbb{P}_{W_t}(\arccos(-u)+(2k+1)\pi) \right)\frac{ \mathrm{d} u}{\sqrt{1-u^2}} \\ &= \int_{-1}^1 f(u) \frac{\sum_{k\in \mathbb{Z}} \mathbb{P}_{W_t}(\arccos(u)+2k\pi)+\mathbb{P}_{W_t}(\arccos(-u)+(2k+1)\pi)}{\sqrt{1-u^2}} \mathrm{d}u \end{align}

From the first to the second line we need to separate odd and even integers but we can group the sums together after that.

So the density of $U_t=\cos(W_t)$ is \begin{equation} \mathbb{P}_{U_t}(u) = \frac{\sum_{k\in \mathbb{Z}} \mathbb{P}_{W_t}(\arccos(u)+2k\pi)+\mathbb{P}_{W_t}(\arccos(-u)+(2k+1)\pi)}{\sqrt{1-u^2}} \end{equation}

Numerical simulations for $\mathbb{P}_{U_t}$:

Histogram and density of U_t

The numerator is an infinite sum as it appears in the wrapped Gaussian. But there is this $\arccos$ and the denominator. I don't know if it can help to evaluate the density of the integral.

The problem is that $U_t$ is neither Gaussian nor independent at different time step which makes difficult to evaluate the density of the Riemann sum.

Numerical simulations give the following histogram for X:

Histogram of X

I don't have any idea of what function it could be.

Interestingly, using a sinus ie $U_t = \sin(W_t)$ seems tractable. It's probably because $U_t$ is centered. To get the density, just replace $\arccos$ by $\arcsin$ above. See the figures. For $\mathbb{P}_{U_t}$

Histogram and density of U_t

And for the density of $X$, I'm able to fit a density $ \propto \exp\left( -\frac{s}{1-x^2} \right) $:

Histogram of X

I hope it could be helpful.

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  • $\begingroup$ I edited my answer many times because of calculus errors and wrong simulations. Now it should be correct. $\endgroup$ – Jonathan Vacher Jan 30 at 4:45

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