Computing the Variance of Integrated Squared Brownian Motion

# Integral of the square of Brownian motion using definition of variance


# Integral of the square of Brownian motion using definition of variance












Let $B = \{ B(t); t \ge 0\}$ and let $Z = \{ Z(t); t \ge 0 \}$ where $$Z(t) = \int_0^t B^2(s) ds.$$ How do we find $E[Z(t)]$ and $E[Z^2 (t)]$ in order to get the variance $Var [Z^2(t)] = E[Z^2 (t) ] - E[Z(t)]^2$

There have been a number of posts similar to this question Variance of time integral of squared Brownian motion, Distribution of time integral of Brownian motion squared (where the Brownian motion occurs in square root time)? and Integral of Brownian motion w.r.t. time but they all involved using ideas that are not available to me yet, i.e., Ito lemma and symmetry etc.

My question is that is it possible to solve the problem using definition of variance by finding $E[Z^2(t)]$ and $E[Z(t)]^2$ without necessarily using ito? I would be glad if any one helps me out.

## Answer by Christopher K (score 2)

https://quant.stackexchange.com/a/59715

As for the hint, you first write

\begin{align*} \mathbb{E} \left [\left (\int_{0}^{t} W_{s}^{2}\, ds \right )^{2} \right ] &= \mathbb{E} \left [\left (\int_{0}^{t} W_{s}^{2}\, ds \right )\left (\int_{0}^{t} W_{u}^{2}\, du \right ) \right ] \\ &= \mathbb{E} \left [\int_{0}^{t} \int_{0}^{t} W_{s}^{2}W_{u}^{2}\, du \, ds \right ] \\ &= \mathbb{E} \left [\int_{0}^{t} \int_{0}^{s} W_{s}^{2}W_{u}^{2}\, du \, ds \right ] + \mathbb{E} \left [\int_{0}^{t} \int_{s}^{t} W_{s}^{2}W_{u}^{2}\, du \, ds \right ] \\ &= \mathbb{E} \left [\int_{0}^{t} \int_{0}^{s} W_{s}^{2}W_{u}^{2}\, du \, ds \right ] + \mathbb{E} \left [\int_{0}^{t} \int_{0}^{u} W_{s}^{2}W_{u}^{2}\, ds \, du \right ] \\ &= 2\mathbb{E} \left [\int_{0}^{t} \int_{0}^{s} W_{s}^{2}W_{u}^{2}\, du \, ds \right ] \\ &= 2\int_{0}^{t} \int_{0}^{s} \mathbb{E}[W_{s}^{2}W_{u}^{2}]\, du \, ds. \end{align*}

Now calculate (for $u<s$) \begin{align*} \mathbb{E}[W_{u}^{2}W_{s}^{2}] &= \mathbb{E}\left [W_{u}^{2} \left ((W_{s}-W_{u})^{2} + 2W_{u}(W_{s}-W_{u}) + W_{u}^{2} \right )\right ] \\ &= \mathbb{E}[W_{u}^{2}]\mathbb{E}[W_{s-u}^{2}] + 2\mathbb{E}[W_{u}^3]\mathbb{E}[W_{s-u}] + \mathbb{E}[W_{u}^{4}] \\ &= u(s-u)+3u^{2} \\ &= 2u^{2} + us. \end{align*}

Your answer will then be \begin{align*} \mathbb{E} \left [\left (\int_{0}^{t} W_{s}^{2}\, ds \right )^{2} \right ] &= 2\int_{0}^{t} \int_{0}^{s} \mathbb{E}[W_{s}^{2}W_{u}^{2}]\, du \, ds \\ &= 2\int_{0}^{t} \int_{0}^{s} 2u^{2}+us \, du \, ds \\ &= 2\int_{0}^{t} \frac{7}{6}s^{3}\, ds \\ &= \frac{7}{12}t^{4}. \end{align*}

## Answer by holala (score 1)

https://quant.stackexchange.com/a/59717

Finding the variance is due to our benevolent contributors. $$\begin{align*} \mathrm{Var} [Z^2(t)] & = \mathrm{E} \ [Z^2 (t) ] - \mathrm{E}\ [Z(t)]^2 \\ & = \mathrm{E}\ \left [\left (\int_{0}^{t} B_{s}^{2}\, ds \right )^{2} \right ] - \mathrm{E}\ \left[\int_0^t B^2(s) ds\right]^2\\ &= 2\int_{0}^{t} \int_{0}^{s} \mathrm{E} \ [B_{s}^{2}B_{u}^{2}]\, du \, ds - \left(\int_0^t \mathrm{E}\ [B^2(s)] \right)^2 ds\\ &= 2\int_{0}^{t} \int_{0}^{s} (2u^{2}+us) \, du \, ds - \left([0.5s^2]_0^{t} \right)^2\\ &= \frac{7}{12}t^{4} - \frac{1}{4}t^4\\ &= \frac{1}{3}t^{4} . \end {align*} $$