Covariance of squares of Brownian motion

Question

$B(t)$ is a Brownian motion. Calculate $Cov[B(s)^2,B(t)^2]$ for $s,t\ge0$.

For this I would need $E[B(s)^2B(t)^2]$. Without the squares that wouldn't be a problem but I have no idea how to calculate this expected value. Any suggestions?

Answer 1

$B_s^{2}=(B_s-B_t)^{2}+2(B_s-B_t)B_t+B_t^{2}$ for $t<s$. Multiply by $B_t^{2}$ and take expectation.

Answer 2

Suppose $s > t.$ Then $$E[(B(s)-B(t))^2B(t)^2] = E[B(s)^2B(t)^2] - 2E[B(s)B(t)^3] + E[B(t)^4]$$ Use the fact $B(\cdot)$ has independent increments to evaluate the left-hand side. On the right side, we have the expression you need to evaluate and two others. For $E[B(s)B(t)^3]$ you can apply the same method of independent increments again, and $E[B(t)^4]$ is just the fourth moment of a normal variable.

Answer 3

From $dB^2_t=2B_tdB_t+dt$, write

$$B_t^2=2z_t+t$$

where

$$z_t=\int_0^tB_\tau dB_{\tau}$$

Then,

$$E[B_s^2B_t^2]=E[(2z_s+s)(2z_t+t)]=4E[z_sz_t]+ts $$

Assume $t>s$ and evaluate

$$E[z_sz_t]=E\left[\int_0^s B_u dB_u\int_0^t B_v dB_v\right]= E\left[ \int_0^s B_u^2du\right]=\int_0^s u du= \frac 12s^2 $$

Thus,

$$E[B_s^2B_t^2]=2s^2+st=s(2s+t) $$