Question

Let \( \{W(t)\}_{t \geq 0} \) be a standard Brownian motion. Which one of the following statements is NOT true?

• A. \( E[W(7)] = 0 \)
• B. \( E[W(5)W(9)] = 7 \)
• C. \( 2W(1) \) is normally distributed with mean 0 and variance 4
• D. \( E[W(5) \mid W(3) = 3] = 3 \)

Hint:

For a standard Brownian motion, \( E[W(t)] = 0 \) and \( E[W(t_1)W(t_2)] = \min(t_1, t_2) \).

Answer 1

The Correct Option is B

To determine which of the given statements about a standard Brownian motion \( \{W(t)\}_{t \geq 0} \) is NOT true, let's examine each option:

1. \( E[W(7)] = 0 \)  
   For a standard Brownian motion, the expected value at any time \( t \), \( E[W(t)] \), is zero. Therefore, \( E[W(7)] = 0 \) is true.
2. \( E[W(5)W(9)] = 7 \) 
   The covariance of a standard Brownian motion at times \( s \) and \( t \), where \( s < t \), is given by \( \text{Cov}(W(s), W(t)) = \min(s, t) \). Therefore, for \( W(5) \) and \( W(9) \), the covariance is \( \min(5, 9) = 5 \). Thus, the expected product \( E[W(5)W(9)] = \text{Cov}(W(5), W(9)) = 5 \), not 7. Hence, this statement is NOT true.
3. \( 2W(1) \) is normally distributed with mean 0 and variance 4 
   A multiple of a normal distribution is still normally distributed. Since \( W(1) \sim N(0, 1) \), then \( 2W(1) \sim N(0, 4) \), as both the mean is 0 and the variance is \( 4 = 2^2 \times 1 \). This statement is true.
4. \( E[W(5) \mid W(3) = 3] = 3 \) 
   For a standard Brownian motion, given \( W(a) = x \), the conditional expectation \( E[W(b) \mid W(a) = x] = x \) for any \( b \geq a \). Therefore, \( E[W(5) \mid W(3) = 3] = 3 \). This statement is true.

Therefore, the statement that is NOT true is \(E[W(5)W(9)] = 7\), as the correct value should be 5 based on the properties of the covariance of Brownian motion.

Answer 2

Step 1: Understanding Brownian motion.
A standard Brownian motion \( \{W(t)\}_{t \geq 0} \) has the following properties:
- \( E[W(t)] = 0 \), for all \( t \geq 0 \) (the expected value of \( W(t) \) is 0).
- \( \text{Var}(W(t)) = t \), for all \( t \geq 0 \) (the variance of \( W(t) \) is \( t \)).
- For \( t_1 \neq t_2 \), \( \text{Cov}(W(t_1), W(t_2)) = \min(t_1, t_2) \) (the covariance between two Brownian motions is the minimum of the two times).
Step 2: Analyzing the options.
(A) \( E[W(7)] = 0 \):
This is true. The expected value of a Brownian motion at any time \( t \) is 0, so \( E[W(7)] = 0 \).
(B) \( E[W(5)W(9)] = 7 \): This statement is NOT true. The correct formula for the expectation of the product of two Brownian motions \( W(t_1) \) and \( W(t_2) \) is: \[ E[W(t_1) W(t_2)] = \min(t_1, t_2) \] For \( t_1 = 5 \) and \( t_2 = 9 \), we get: \[ E[W(5) W(9)] = \min(5, 9) = 5 \] So, the correct value should be 5, not 7. Therefore, option (B) is false.
(C) \( 2W(1) \) is normally distributed with mean 0 and variance 4:
This is true. Since \( W(1) \) is a standard Brownian motion, it is normally distributed with mean 0 and variance 1. Therefore, \( 2W(1) \) will be normally distributed with mean \( 2 \times 0 = 0 \) and variance \( 2^2 \times 1 = 4 \).
(D) \( E[W(5) \mid W(3) = 3] = 3 \):
This is true. Given that \( W(3) = 3 \), the conditional expectation \( E[W(5) \mid W(3) = 3] \) is equal to \( W(3) \) because for Brownian motion, the conditional expectation of \( W(t_2) \) given \( W(t_1) \) (with \( t_2>t_1 \)) is simply \( W(t_1) \). Thus: \[ E[W(5) \mid W(3) = 3] = 3 \] Step 3: Conclusion.
The false statement is (B), where the expectation of \( W(5)W(9) \) is incorrectly stated as 7. The correct value is 5.