Question

Let \( \{W_t\}_{t \geq 0} \) be a standard Brownian motion. Then \( E(W_4^2 \mid W_2 = 2) \) (in integer) equals:

Hint:

- For Brownian motion, the conditional expectation of the square of the process at a future time \( t \) given the value at time \( s \) is \( w^2 + (t - s) \), where \( w \) is the value of the process at time \( s \).

Answer 1

For standard Brownian motion, we know that the conditional expectation \( E(W_t^2 | W_s = w) \) is given by: \[ E(W_t^2 | W_s = w) = w^2 + (t - s) \] In this problem, \( t = 4 \), \( s = 2 \), and \( W_2 = 2 \), so we can directly apply the formula: \[ E(W_4^2 | W_2 = 2) = 2^2 + (4 - 2) = 4 + 2 = 6 \] Thus, the correct answer is 6.