Question

Compute \(E\left[B\left(t_{1}\right) B\left(t_{2}\right) B\left(t_{3}\right)\right]\) for \(t_{1}

Short answer

The expectation of the product of Brownian motion at the three different time points \(t_1, t_2, t_3\) (with \(t_1 < t_2 < t_3\)) is equal to 0: \(E[B(t_1)B(t_2)B(t_3)] = 0\)

Step-by-step solution

Recall properties of Brownian motion increments

Brownian motion is a continuous-time stochastic process, denoted by \(B(t)\), where t represents time. Brownian motion has several important properties, especially the independence and distribution of its increments. Knowing that for any \(0\leq s\leq t\): 1. The increment \(B(t) - B(s) \sim N(0, t - s)\), where N represents the normal distribution with mean 0 and variance \(t - s\). 2. Increments are independent, i.e., \(B(t) - B(s)\) is independent of \(B(s) - B(r)\) for any \(r \leq s \leq t\). These properties will help us compute the expectation of the product of Brownian motion at three different time points.

Write the expression of the product

First, let's write down the product of Brownian motion at three different time points, given by: \[B(t_1)B(t_2)B(t_3)\]

Decompose the product using the increments

Now, we can express each of the Brownian motion terms in terms of the increments. Using the independence property mentioned earlier, we can write: \[B(t_1) = B(t_1) - B(0)\] \[B(t_2) = B(t_2) - B(t_1) + B(t_1)\] \[B(t_3) = B(t_3) - B(t_2) + B(t_2)\] Substitute these expressions into our original product: \[(B(t_1) - B(0))(B(t_2) - B(t_1) + B(t_1))(B(t_3) - B(t_2) + B(t_2))\]

Simplify and find expectations of each term

To compute the expectation, we need to simplify the expression: \begin{align*} &B(t_1)(B(t_2)B(t_3) - B(t_1)B(t_3) + B(t_2)B(t_1)) \\ &- B(0)(B(t_2) - B(t_1) + B(t_1))(B(t_3) - B(t_2) + B(t_2)) \end{align*} Now we need to find the expectations of each term. Notice that \(E(B(t_{1})) = 0\), \(E(B(t_{2})) = 0\), and \(E(B(t_{3})) = 0\), since the expectation of Brownian motion at any time is 0. Similarly, \(E(B(0)) = 0\) as the expectation at the initial time point is 0.

Compute the expectation of the product

Using the previous step's results, we have: \begin{align*} E[B(t_1)B(t_2)B(t_3)] &= E[B(t_1)(B(t_2)B(t_3) - B(t_1)B(t_3) + B(t_2)B(t_1))] \\ &\quad - E[B(0)(B(t_2) - B(t_1) + B(t_1))(B(t_3) - B(t_2) + B(t_2))] \\ &= 0 \cdot (0 - 0 + 0) \\ &= 0 \end{align*} Thus, the expectation of the product of Brownian motion at the three different time points \(t_1, t_2, t_3\) (with \(t_1 < t_2 < t_3\)) is equal to 0.