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.