Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Premise
Let $W_t$ be the standard Wiener process, and let $X_t = \int_0^t W_s^2 \mathrm{d} s$. I am interested in determining the distribution of $X_t$.

What I did
My line of attack has been to determine moments $\mathbb{E}(X_t^r)$. To this end I set up a system of It$\bar{\text{o}}$ differential equations: $$ \mathrm{d} X_t = Y_t^2 \mathrm{d} t, \qquad \mathrm{d} Y_t = \mathrm{d} W_t, \qquad X_0 = Y_0 = 0 $$ Then $$ \mathbb{E}(X_t^r) = \int_0^t \mathbb{E}( \mathcal{L}_0(X_s^r) ) \mathrm{d} s = \int_0^t \mathrm{d}s_1 \int_0^{s_1} \mathrm{d}s_2 \cdots \int_0^{s_{n-1}} \mathrm{d} s_n \mathbb{E}( \mathcal{L}_0^{ \circ n}(X_{s_n}^r) ) $$ where $n$ is chosen so that $\mathcal{L}_0^{ \circ n}(X_{s_n}^r)$ is constant equal to $a_{n,r}$, then $\mathbb{E}(X_t^r) = \frac{t^r}{n!} a_{n,r}$, where $$ \mathcal{L}_0(f(X_t, Y_t)) = \left(\frac{1}{2} \frac{\partial^2}{\partial Y_t^2} + Y_t^2 \frac{\partial}{\partial X_t} \right)(f(X_t, Y_t)) $$ It is not hard to see that $$ \mathbb{E}(X_t^r) = \frac{t^r}{(2r)!} \left( \frac{1}{2} \frac{\mathrm{d}^2}{\mathrm{d} y^2} + y^2 \frac{\mathrm{d}}{\mathrm{d} x} \right)^{2r} (x^r) $$ Poking around OEIS, I stumbled across A126156, corresponding to $\frac{(2r)!}{r!} \mathbb{E}(X_1^r)$, from where I learnt the moment generating function for $X_t$: $$ \phi(u)= \mathbb{E}(\exp(u X_t)) = \frac{1}{\sqrt{ \cos(\sqrt{2 u} t)}} = 1 + \frac{u t^2}{1!} \cdot \frac{1}{2} + \frac{u^2 t^4}{2!} \cdot \frac{7}{12} + \frac{u^3 t^6}{3!} \cdot \frac{139}{120} + \mathcal{o}(u^3) $$ for $-\infty < u < \frac{\pi^2}{8 t^2}$, i.e. until the first zero of the cosine, which indicates that the main asymptotic term of the probability density is $f_{X_t}(x) \sim \exp\left(- \frac{\pi^2}{8 t^2} x \right)$. Indeed, using Weierstrass representation: $$ \phi(u) = \prod_{n=0}^\infty \frac{1}{\sqrt{1-\frac{8 u t^2}{\pi^2} \frac{1}{(2n+1)^2}}} $$ which suggests that $X_t = \sum_{n=0}^\infty Z_n$, where $Z_n$ are independent and $Z_n \sim \Gamma\left(\frac{1}{2}, 2 \left( \frac{2 t}{(2n+1) \pi} \right)^2 \right)$, which in turn means that $\kappa_r(X_t) = -(-8 t^2)^r (r-1)! (4^r-1) \frac{B_{2r}}{4 (2r)!} $

This unexpected success gives (slim) hope that the distribution function of $X_t$, or its probability density can be found in closed form.

Question
I am looking for references where $X_t$, or $X_1$, has been studied, or possible suggestions on how one would go about proving my guesswork.

share|improve this question
    
An alternative approach is to discretize $(W_s)_{0\leqslant s\leqslant t}$, which yields your characterization of $X_t$ as a sum of Gammas. And you are probably well aware that $X_t$ and $t^2X_1$ have the same generalized chi-squared distribution‌​. –  Did Jan 14 '12 at 16:02
    
@DidierPiau Yes, I considered equal splitting with $s_i = t \frac{i}{n}$. $W_{s_i} = \sqrt{\frac{t}{n}} \sum_{k=1}^n Z_k$, for independ standard normals $Z_k$. This would give $X_1 = \frac{1}{n^2} \sum_{k=1}^n \sum_{m_1=1}^k \sum_{m_2=1}^k Z_{m_1} Z_{m_2}$. Because of cross terms it was difficult to separate that into sums of independent generalized $\chi^2$-distributions. It is natural to expect $\chi^2$ to appear, but conspiracy of scales is that some requires an explanation, while what I now have is a guesswork. –  Sasha Jan 14 '12 at 17:45
add comment

2 Answers 2

up vote 2 down vote accepted

The distribution of $X_t$ was first computed by Cameron & Martin in 1944. A reference to this paper, along with a nifty computation of the Laplace transform of $X_t$, can be found in a paper of Mark Kac: "On distributions of certain Wiener functionals" [Trans. Amer. Math. Soc. vol. 65 (1949) pp. 1–13]. Kac's method is an instance of what we would nowadays call the "Feynman-Kac method". Kac shows that $E[\exp(-uX_t)] = [{\rm sech}((2u)^{1/2}t)]^{1/2}$.

share|improve this answer
    
Thank you. This is the link to the issue containing Mark Kac's article on AMS.org available free of charge. –  Sasha Jan 15 '12 at 0:54
add comment

I haven't checked the details, but I think this approach works.

First off, as Didier Piau notes in a comment, it suffices to find the distribution of $X_1$.

Now, use the Karhunen–Loève expansion of the Wiener process: $$W(s)=\sqrt{2}\,\sum_{k=1}^\infty Z_k {\sin((k-1/2)\pi s)\over(k-1/2)\pi},$$ where $Z_k$ are i.i.d. standard normal random variables. When you square this and integrate, all the cross-product terms disappear and you are left with the representation: $$X_1=\int^1_0 W^2(s)\,ds=\sum_{k=1}^\infty {Z_k^2 \over (k-1/2)^2\pi^2}.$$

share|improve this answer
    
Beautiful! Thank you very much. –  Sasha Jan 14 '12 at 17:50
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.