# Expectation of the product of Brownian processes (higher powers)

I have recently sat an exam that had elements of stochastic calculus, but I am now feeling like I might have gone wrong in some questions of it like the following.

I am trying to evaluate $$\mathbb{E}(B^2_t B^2_s)$$, where $$B_s$$ and $$B_t$$ are Browninan processes which are not independent. Moreover, all we know is that $$t,s>0$$.

I have started off by taking an expansion and then simplifying it using the increment rules: $$d(B_s^2B_t^2)=2B_sB_t^2dBs + 2B_tB_s^2dBt + B_t^2(dB_s)^2 + B_s^2(dB_t)^2+4B_tB_sdB_sdB_t=\\ =2B_sB_t^2dBs + 2B_tB_s^2dBt + B_t^2ds + B_s^2dt+4B_tB_sdB_sdB_t$$

Once this was all done, I have integrated this and finally taken the expectation of the integral, which simplifies by the properties of Itô integrals:

$$\mathbb{E}(B^2_t B^2_s)=\mathbb{E}(2\int_0^sB_qB_t^2dB_q)+\mathbb{E}(2\int_0^tB_qB_s^2dB_q)+\mathbb{E}(\int_0^sB_t^2dq)+\mathbb{E}(\int_0^tB_s^2dq)+\mathbb{E}(4\int_0^t(\int_0^sB_qB_rdB_q)dB_r).$$

And by the properties of Itô integrals this should simplify to (under Fubini's theorem):

$$\mathbb{E}(B^2_t B^2_s)=0+0+\int_0^s\mathbb{E}(B_t^2)dq+\int_0^t\mathbb{E}(B_s^2)dq+0= \int_0^st dq+\int_0^tsdq = 2ts$$

Does this make any sense at all? I am not sure if I have assumed anywhere that really they are independent Brownian processes?

I have also solved $$\mathbb{E}(B^3_t B^3_s)$$, but I might have misunderstood the conditions that allow the expansions of $$B_s$$ and $$B_t$$ independently when they are not independent of each other.

Can any kind soul clarify this?

• What exactly do you mean by "[...] $B_s$ and $B_t$ are Brownian processes which are not independent"....? Your notation is suggesting that you are considering just one Brownian motion and evaluate it at different times $s$ and $t$.... that's however not fitting to your description. If you consider two Brownian motions then you should use different symbols for it, e.g $B$ and $W$. In the first case (i.e. if you have only one Brownian motion) then Itô calculus is straight overkill... it's much more straightforward to use the independence of increments of BM and use $B_t \sim N(0,t)$. – saz May 21 at 12:49
• I am in fact considering the same Brownian process taken at two different times. That is why $B_s$ and $B_t$ are not independent of each other, I guess? I think I might have over-killed it then by using Itô. Is it wrong to use it? I am trying to think of what you said instead: If I follow what you suggested I should get: $B_tB_s \sim N(0,ts)$ So then: $\mathbb{E}({B^2_tB^2_s})= Var(B_tB_s)+(\mathbb{E}(B_tB_s))^2=ts+(Min(s,t))^2$ Is this correct? – Luigi Candela May 26 at 20:58