Question

Let \( W(t) \) be a standard Brownian motion. Then the variance of \( W(1)W(2) \) equals

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

When computing the variance of the product of two Brownian motions, use the covariance properties and Isserlis' theorem to simplify the calculation.

Answer 1

The Correct Option is C

We are asked to find the variance of \( W(1)W(2) \), where \( W(t) \) is a standard Brownian motion.
Step 1: Understanding the properties of Brownian motion.
A standard Brownian motion \( W(t) \) has the following properties:
1. \( W(0) = 0 \).
2. The increments \( W(t) - W(s) \) are independent for \( t>s \), and \( W(t) - W(s) \sim \mathcal{N}(0, t - s) \).
3. The covariance \( \text{Cov}(W(t), W(s)) = \min(t, s) \).
Step 2: Apply the formula for variance.
The variance of the product of two random variables can be written as: \[ \text{Var}(W(1)W(2)) = \mathbb{E}[(W(1)W(2))^2] - (\mathbb{E}[W(1)W(2)])^2. \] First, we compute the expectation \( \mathbb{E}[W(1)W(2)] \). Since \( W(1) \) and \( W(2) \) are both part of the same Brownian motion, we use the covariance property: \[ \mathbb{E}[W(1)W(2)] = \text{Cov}(W(1), W(2)) = 1. \] Next, we compute \( \mathbb{E}[(W(1)W(2))^2] \): \[ \mathbb{E}[(W(1)W(2))^2] = \mathbb{E}[W(1)^2 W(2)^2]. \] Since \( W(1) \) and \( W(2) \) are jointly Gaussian, we can use Isserlis' theorem to express this expectation as: \[ \mathbb{E}[W(1)^2 W(2)^2] = \mathbb{E}[W(1)^2] \mathbb{E}[W(2)^2] + 2(\mathbb{E}[W(1)W(2)])^2. \] We know that \( \mathbb{E}[W(1)^2] = 1 \) and \( \mathbb{E}[W(2)^2] = 2 \), so: \[ \mathbb{E}[W(1)^2 W(2)^2] = 1 \cdot 2 + 2 \cdot 1^2 = 2 + 2 = 4. \] Step 3: Calculate the variance.
Now, we can compute the variance: \[ \text{Var}(W(1)W(2)) = 4 - 1^2 = 4 - 1 = 3. \] Final Answer: \[ \boxed{3}. \]