Help~Expectation of Brownian Motion Processes

[mckammer]

Hello everybody! I'm working on a Stochastic Processes course based on Ross' Introduction to Probability Models.
I hope you can help me work through this problem.

Homework Statement

{B(t), t >= 0} is a standard Brownian motion process
Compute E[B(t1)B(t2)B(t3)] for t1 < t2 < t3.

Homework Equations

See my attempt at a solution?

The Attempt at a Solution

E[B(t1)B(t2)B(t3)] =E[E[B(t1)B(t2)B(t3)|B(t1)]]
=E[B(t1)E[B(t2)B(t3)|B(t1)]
Then, I name another Brownian Process C(t), such that C(t) = B(t) - B(t1)
=E[B(t1)E[(C(t2)-B(t1))(C(t3)-B(t1))|B(t1)]

Expanding, I get
E[B(t1)^3 - B(t1)^2(E[C(t2)+C(t3)]) + B(t1)E[C(t2)C(t3)]]

The second term evaluates to 0, since we have the sum of two Brownian motion processes with mean 0

The third term evaluates to
E[C(t2)[C(t3) + C(t2) - C(t2)]] = E[C(t2)(C(t3)-C(t2))] + E[C(t2)^2]
= 0 + t2? (Because the first term here is multiplied by E[C(t2)], and the second term is the second moment, which is equal to the variance due to the mean being 0?)

My final answer is
B(t1)^3 + (t2)B(t1)

Does this look reasonable?

 

[mighty2000]

Thank you for posting your question. I am happy to help you work through this problem.

First, let's review the definition of a standard Brownian motion process. It is a continuous-time stochastic process with the following properties:
1. B(0) = 0
2. B(t) has independent increments
3. B(t) has normally distributed increments with mean 0 and variance t

Now, let's tackle the problem at hand. We want to compute E[B(t1)B(t2)B(t3)] for t1 < t2 < t3.

Using the definition of a standard Brownian motion process, we can rewrite this expectation as:
E[B(t1)(B(t2) - B(t1))(B(t3) - B(t2))] = E[B(t1)C(t2)C(t3)], where C(t) = B(t) - B(t1).

Now, we can use the property of independent increments to rewrite this as:
E[B(t1)C(t2)C(t3)] = E[B(t1)]E[C(t2)]E[C(t3)] = 0, since B(t1) and C(t2) are independent and have mean 0.

Therefore, our final answer is 0.

I hope this helps you understand the solution to this problem. If you have any further questions, please don't hesitate to ask.