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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.

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    $\begingroup$ Let me have a shot at half of it: $$\mathbb{E}[Z_t]=\mathbb{E}\left[\int_0^t B^2(s) ds\right]=\int_0^t \mathbb{E}[B^2(s)] ds=\int_0^t s ds=[0.5s^2]_0^{t}=0.5t^2$$ Above, I used the fact that expectation is an integral and the order of integration can be changed (the conditions for being able to change the order of integration are in Fubini's Theorem: we don't need to go there). $$Z_t^2=\int_0^tB^2(s)ds\int_0^tB^2(h)dh$$ Now with $\mathbb{E}[Z_t^2]$, I am not sure how to proceed. Let me have a think about it (and I will happily take an advice from more experienced users here...). $\endgroup$ Dec 1 '20 at 8:18
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    $\begingroup$ Note that $$Z_t^2 = \int_0^tB^2_sds\int_0^tB^2_hdh = \int_0^t\int_0^t B_s^2B_h^2dsdh.$$ You can use Fubini again. $\endgroup$
    – Gordon
    Dec 1 '20 at 15:39
  • $\begingroup$ @JanStuller. Whiles you think about the second piece, could you throw more light on $E[B^2(s)] = s ?$ $\endgroup$
    – holala
    Dec 1 '20 at 18:36
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    $\begingroup$ @JanStuller: For a given sample $\omega$, the integrals are for continuous functions and are thus can be changed to a double integral. Only when you are taking the expectation, the random part is involved. There are no two-dimensional Brownian motions involved here, given that W_s and W_h, for given $\omega$, are from the same continuous function, but evaluated at different time point. $\endgroup$
    – Gordon
    Dec 2 '20 at 13:32
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    $\begingroup$ @JanStuller: You can treat it as a volume, but it is not an area squared. In general, $$\int_a^b dy \int_c^d f(x, y) dx = \int_a^b \int_c^d f(x, y) dx dy,$$ for which you can treat it as the volume over the rectangle $[a, b]\times[c, d]$, but with the top surface $f(x, y)$. $\endgroup$
    – Gordon
    Dec 2 '20 at 16:56
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As per 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}]\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*}

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  • $\begingroup$ Out of curiosity, what is your rational for rewriting $$\mathbb{E}\left[\int_0^t \int_0^t W_s^2 W_h^2 du ds\right]$$ as $$2\mathbb{E}\left[\int_0^t \int_0^s W_s^2 W_h^2 du ds\right]$$ ? The expectation can already be put inside the integral at the first step... $\endgroup$ Dec 3 '20 at 10:31
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    $\begingroup$ You can split it up into $$\int_{0}^{t} \int_{0}^{s} W_{s}^{2}W_{u}^{2}\, du \, ds$$ and $$\int_{0}^{t} \int_{s}^{t} W_{s}^{2}W_{u}^{2}\, du\, ds$$ if you'd like and integrate (you would verify that they are the same). But your $$\mathbb{E}[W_{s}^{2}W_{u}^{2}] = 2\min\{s,u\}^{2} + su$$ so you still need to break up into two triangles to integrate. $\endgroup$ Dec 3 '20 at 17:40
  • $\begingroup$ True, good point. $\endgroup$ Dec 3 '20 at 18:00
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Finding the variance is due to our benevolent contributors. $$\begin{align*} \mathrm{Var} [Z^2(t)] & = \mathrm{E} \ [Z^2 (t) ] - \mathrm{E}\ [Z(t)]^2 \\ & = \mathrm{E}\ \left [\left (\int_{0}^{t} B_{s}^{2}\, ds \right )^{2} \right ] - \mathrm{E}\ \left[\int_0^t B^2(s) ds\right]^2\\ &= 2\int_{0}^{t} \int_{0}^{s} \mathrm{E} \ [B_{s}^{2}B_{u}^{2}]\, du \, ds - \left(\int_0^t \mathrm{E}\ [B^2(s)] \right)^2 ds\\ &= 2\int_{0}^{t} \int_{0}^{s} (2u^{2}+us) \, du \, ds - \left([0.5s^2]_0^{t} \right)^2\\ &= \frac{7}{12}t^{4} - \frac{1}{4}t^4\\ &= \frac{1}{3}t^{4} . \end {align*} $$

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