# How to calculate the covariance involving Stochastic process

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.

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 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 $$t$$

• 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. – Bogaso Aug 9 at 9:33
• That's probably best explained by a picture (added above) - let me know if its not clear – StackG Aug 9 at 10:12
• great. Many thanks – Bogaso Aug 9 at 10:34