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I was looking at some old post : Variance of time integral of squared Brownian motion

I failed to grasp 2 derivations -

  1. $\text{Cov}\left(\int_{0}^{t}W^3_sdW_s\,,\,\int_{0}^{t}W^2_sds\right)$. I know this can eventually be written as $\mathbb{E} \left[ \left( \int_{0}^{t}W^3_sdW_s \right) \left( \int_{0}^{t}W^2_sds\right) \right]$, because $ \mathbb{E} \left[ \int_{0}^{t}W^3_sdW_s \right] = 0 $. But, how to proceed to the final expression from here?

  2. How to calculate the expression $ \int_{0}^{t}\mathbb{E}[W^6_s]ds$

Any pointer will be highly appreciated.

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  1. Let's use the following expression (derived in Quantuple's answer in your link), which will help us tidy up the product using Ito's Isometry

\begin{align} \int^t_0 W^2_s ds = 2 \int^t_0 (t-s)W_s dW_s + {\frac {t^2} 2} \end{align}

Now looking at the expectation \begin{align} {\mathbb E}\Bigl[ \int^t_0 W^3_s dW_s \cdot \int^t_0 W^2_s ds \Bigr] &= {\mathbb E}\Bigl[ \int^t_0 W^3_s dW_s \cdot \Bigl( 2 \int^t_0 (t-s)W_s dW_s + {\frac {t^2} 2} \Bigr) \Bigr] \\ &= {\mathbb E}\Bigl[ {\frac {t^2} 2}\int^t_0 W^3_s dW_s + 2 \int^t_0 W^3_s dW_s \cdot \int^t_0 (t-s)W_s dW_s \Bigr] \end{align}

As you identified above, the expectation of the first term in the sum is $0$, and we can use Ito's Isometry on the second

\begin{align} {\mathbb E}\Bigl[ \int^t_0 W^3_s dW_s \cdot \int^t_0 W^2_s ds \Bigr] &= 0 + {\mathbb E}\Bigl[ 2 \int^t_0 (t-s) W^4_s ds \Bigr] \\ &= 2 \int^t_0 (t-s) {\mathbb E}\bigl[ W^4_s \bigr] ds \\ &= 2 \int^t_0 (t-s) 3s^2 ds \\ &= {\frac 1 2} t^4 \\ \end{align}

In the initial question, the expression has multiplicative prefactors of $2$, $4$ and $6$, so this multiplies out to $24t^4$

In the above, I used the expression ${\mathbb E}\bigl[ W^4_s \bigr] = 3s^2$, which comes from the step-down formula given in the question you linked, ie. \begin{align} {\mathbb E}\bigl[ W^{2n}_t \bigr] = {\frac {(2n)!} {2^n n!}} t^n \end{align}

  1. This can be solved using the same step-down formula \begin{align} \int^t_0 {\mathbb E} \bigl[ W^6_s \bigr] ds &= \int^t_0 {\frac {6!} {2^3 3!}} s^3 ds\\ &= 15 \Bigl[ {\frac 1 4} s^4 \Bigr]^t_0\\ &= {\frac {15} 4} t^4 \end{align}

  2. Limits of the Fubini double integral

\begin{align} \int_0^t \int^s_0 W_u dW_u ds = \int_0^t \int^t_u W_u ds dW_u \end{align}

This change of limits is required so that the double integral is integrating over the same part of the $(s,u)$ space, as shown in the diagram

integration limits in 2d

Basically, we can parameterise the lower triangle either by letting $u$ run from $0$ to $s$, and then letting $s$ run from $0$ to $t$, or if we switch the order we need to let $u$ run fron $u$ to $t$ and then let $u$ run from $0$ to $t$

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  • $\begingroup$ Thanks. In the calculation of $\int^t_0 W^2_s ds$, the Fubini's theorem is used for $\int_0^t \int_0^s W_u dW_u ds = \int_0^t \int_u^t W_u ds dW_u$. How the limits of the final integration is set using Fubini's theorem? Wkipedia ref. doesnt seem to say anything like this. $\endgroup$ – Bogaso Aug 9 at 9:33
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    $\begingroup$ That's probably best explained by a picture (added above) - let me know if its not clear $\endgroup$ – StackG Aug 9 at 10:12
  • $\begingroup$ great. Many thanks $\endgroup$ – Bogaso Aug 9 at 10:34

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