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Let $I_t = \int_0^t W_{\sqrt{u}}^2du$. What is the distribution of $I$?

If I recall correctly, if the Brownian motion were instead $W_u$, then it would be $I_t \sim N\left(\frac{t^2}{2},\frac{t^4}{3}\right)$.

I tried to do a simple change of variables with $v = \sqrt{u}$, but then this generates a new term in the equation that I'm not sure how to handle (I believe it turns into $\int_0^\sqrt{t} 2v W_v^2dv$, but because of the leading $2v$ coefficient, I don't know how to relate this back to the previous result).

Radon1337
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1 Answers1

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Integral $I_t$ does not seem to follow the normal distribution.

$\newcommand{\d}{\mathrm{d}}$Define $I_t = \int_0^{t} W_{\sqrt{u}}^2 \d u$ and let $u = v^2$ as you have already suggested.

By integration by parts

$$ I_t = \int_0^{\sqrt{t}}W_v^2 \d v^2 = \underbrace{W_v^2 v^2\big|_0^{\sqrt{t}}}_{tW_{\sqrt{t}}^2} - \underbrace{\int_0^{\sqrt{t}}v^2\d W_v^2}_{J_t}, $$

where $\d W_v^2 = 2W_v \d W_v + \d v$, therefore,

$$ J_t = \underbrace{\int_0^{\sqrt{t}} 2W_v v^2 \d W_v}_{K_t} + \underbrace{\int_0^{\sqrt{t}}v^2 \d v}_{\tfrac{1}{3}t^{3/2}} $$

Now define $f(v, x) = vx^2$ and apply Itô's formula:

\begin{align} &\d(v W_v^2) {}={} W_v^2 \d v + 2vW_v \d W_v + \tfrac{1}{2}(2v)\d v\\ \Rightarrow{}& \sqrt{t}W_{\sqrt{t}}^2{}={}\int_0^{\sqrt{t}} W_v^2 \d v + K_t + \int_0^{\sqrt{t}}v\d v\\ \Rightarrow{}& K_t = \sqrt{t}W_{\sqrt{t}}^2-\tfrac{1}{2}t - \int_0^{\sqrt{t}} W_v^2 \d v, \end{align}

therefore

\begin{align} I_t {}={}& tW_{\sqrt{t}}^2 - K_t - \tfrac{1}{3}t^{3/2} \\ {}={}&(t - \sqrt{t})W_{\sqrt{t}}^2 + \tfrac{1}{2}t - \tfrac{1}{3}t^{3/2} + \int_0^{\sqrt{t}} W_v^2 \d v \end{align}

Note that $W_{\sqrt{t}}^2$ follows a chi-squared distribution, the second and third terms are deterministic and the integrands of the last term, $W_v^2$ follow a chi-squared distribution too.

Regarding the last integral, you can check out this question on QSE.

Pantelis Sopasakis
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