Question

Let \( \{ W(t) : t \geq 0 \} \) be a standard Brownian motion. Then \[ E\left( (W(2) + W(3))^2 \right) \] equals _______ (answer in integer).

Hint:

For Brownian motion, use the properties \( E(W(t)) = 0 \), \( {Var}(W(t)) = t \), and \( {Cov}(W(t), W(s)) = \min(t, s) \) to compute expectations and variances of linear combinations of Brownian motions.

Answer 1

We use the properties of Brownian motion to solve the problem. Specifically, for a standard Brownian motion:
\( E(W(t)) = 0 \) for all \( t \).
\( \text{Var}(W(t)) = t \) for all \( t \).
\( \text{Cov}(W(t), W(s)) = \min(t, s) \) for \( t, s \geq 0 \).

We need to find \( E\left( (W(2) + W(3))^2 \right) \). Expanding the square:
\[ E\left( (W(2) + W(3))^2 \right) = E(W(2)^2) + E(W(3)^2) + 2E(W(2)W(3)). \]
Using the properties of Brownian motion:
\( E(W(2)^2) = \text{Var}(W(2)) = 2 \),
\( E(W(3)^2) = \text{Var}(W(3)) = 3 \),
\( E(W(2)W(3)) = \text{Cov}(W(2), W(3)) = \min(2, 3) = 2 \).

Thus:
\[ E\left( (W(2) + W(3))^2 \right) = 2 + 3 + 2 \times 2 = 2 + 3 + 4 = 9. \]

Thus, the answer is \( \boxed{9} \).