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Consider a standard Brownian motion $\{B(t); t\ge 0\}$ (zero mean and $\sigma^{2} = 1 $) at times

$0 < u < u+v < u+v+w < u+v+w+x $, where $u, v, w, x > 0$

Evaluate the product moment:

$$E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)] $$

I tried using the same trick employed when solving $E[B(t)B(s)] = \sigma^{2} min\{s,t\}$.

Namely for $0<s \le t$,

$$E[B(t)B(s)] = E[B(s)\{B(t)-B(s)+B(s)]=E[B(s)^{2}]+E[B(s)]E[B(t))-B(s)]=\sigma^{2}s$$

Using the same method on the above product moment and starting with the 4th term I got

$$E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)]\\ = \\ E[B(u)B(u+v)B(u+v+w)\{B(u+v+w+x)-B(u+v+w)+B(u+v+w)\}]$$ $$\\ = \\ E[B(u)B(u+v)B(u+v+w)^{2}] + E[B(u)B(u+v)B(u+v+w)\{B(u+v+w+x)-B(u+v+w)\}]$$

I then used the same trick with the other elements eventually I managed to end up with

$$E[B(u)^{2}B(u+v+w)^{2}] + E[B(u)\{B(u+v)-B(u)\}B(u+v+w)^{2}] \\+E[B(u)B(u+v)^{2}\{B(u+v+w+x)-B(u+v+w)\}] \\+ E[B(u)^{2}\{B(u+v+w) -B(u+v)\}\{ B(u+v+w+x)-B(u+v+w)\}] \\+ E[B(u)\{B(u+v)-B(u)\}\{B(u+v+w) -B(u+v)\}\{ B(u+v+w+x)-B(u+v+w)\}]$$

But I've gotten lost and not sure if I am going the right way, And I don't know how I would evaluate elements such as $E[B(u)^2B(u+v+w)^{2}]$ or $E[B(u)B(u+v)^{2}\{B(u+v+w+x)-B(u+v+w)\}]$. I Believe the very last term is equal to zero by the definition of Brownian motion?

Any help would be appreciated.

seraphimk
  • 573

1 Answers1

1

So I managed to figure it out so I thought I would post the solution in case anyone else is having difficulty with the same question.

Firstly one needs to use the law of total Expectation i.e. $E[E[X|Y]]=E[X]$ and apply it iteratively. So starting from the top.

$$E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)] = E[B(u)B(u+v)B(u+v+w)E[B(u+v+w+x)|\mathcal{F}_{u+v+w}]]$$ Where $\mathcal{F}_{u}:= \sigma (\{B(t);t \le u\}) $.

Using $E[B(u+v+w+x)|\mathcal{F}_{u+v+w}]]=B(u+v+w)$ we get,

$$E[B(u)B(u+v)B(u+v+w)E[B(u+v+w+x)|\mathcal{F}_{u+v+w}]]= E[B(u)B(u+v)B^{2}(u+v+w)]$$

Applying the law of total probability again we have

$$E[B(u)B(u+v)B^{2}(u+v+w)] = E[B(u)B(u+v)E[B^{2}(u+v+w)|\mathcal{F}_{u+v}]]$$
Where $E[B^{2}(u+v+w)|\mathcal{F}_{u+v}] = B^2(u+v)+(u+v+w)-(u+v) = B^2(u+v)+w$ , so we have,


$$E[B(u)B(u+v)E[B^{2}(u+v+w)|\mathcal{F}_{u+v}]]= E[B(u)B(u+v)[B^2(u+v)+w]]$$

Expanding the right hand side we have

$$ E[B(u)B(u+v)[B^2(u+v)+w]] = E[B(u)B^3(u+v)] + wE[B(u)B(u+v)]$$

Applying the Law of total probability again to both parts of the RHS we end up with


$$E[B(u)B^3(u+v)] + wE[B(u)B(u+v)] = E[B(u)E[B^3(u+v)| \mathcal{F}_{u}]] + wE[B(u)E[B(u+v)| \mathcal{F}_{u}]] $$

where, $E[B^3(u+v)| \mathcal{F}_{u}] = B^3(u) +3B(u)((u+v)-u))=B^3(u)+3B(u)v$ insertiing this into the first term on the RHS we get


$$E[B(u)E[B^3(u+v)| \mathcal{F}_{u}]] =E[B(u)(B^3(u)+3B(u)v)] = E[B^4(u)]+3vE[B^2(u)] = 3u^2+3uv$$

Looking at the second term on the RHS we have


$$ wE[B(u)E[B(u+v)| \mathcal{F}_{u}]]=wE[B^2(u)]=wu$$

Where we used the result $E[B(u+v)| \mathcal{F}_{u}]=B(u)$


Hence we end up with the full answer as


$$E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)] = 3u^2+3uv+uw $$

seraphimk
  • 573