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

Step 1: Recall key facts about standard Brownian motion. 
Let $\{W(t)\}_{t\ge0}$ denote a standard Brownian motion. It satisfies the following properties:

• The mean is zero at every time: $E[W(t)] = 0$.
• The variance grows linearly with time: $\mathrm{Var}(W(t)) = t$.
• The covariance between two times is given by $\mathrm{Cov}(W(t_1), W(t_2)) = \min(t_1,t_2)$.

Using these properties, we evaluate each statement.

Step 2: Check statement (A).
The expected value of Brownian motion at any fixed time is zero.
Therefore:

\[ E[W(7)] = 0 \]

This statement is correct.

Step 3: Check statement (B).
For Brownian motion, the expected value of the product is:

\[ E[W(t_1)W(t_2)] = \min(t_1,t_2) \]

Substituting $t_1=5$ and $t_2=9$:

\[ E[W(5)W(9)] = \min(5,9) = 5 \]

Since the statement claims the value is 7, it is incorrect.

Step 4: Check statement (C).
The random variable $W(1)$ follows a normal distribution with mean 0 and variance 1.
Multiplying by 2 scales the mean and variance as follows:

\[ E[2W(1)] = 0, \quad \mathrm{Var}(2W(1)) = 2^2 \cdot 1 = 4 \]

Hence, $2W(1)$ is normally distributed with mean 0 and variance 4, making this statement correct.

Step 5: Check statement (D).
For Brownian motion, the conditional expectation satisfies:

\[ E[W(t_2)\mid W(t_1)] = W(t_1), \quad \text{for } t_2>t_1 \]

Given $W(3)=3$, we obtain:

\[ E[W(5)\mid W(3)=3] = 3 \]

Thus, this statement is also correct.

Step 6: Final conclusion.
Among the given options, the only incorrect statement is:

\[ \boxed{(B)} \]