<s>[INST] Variance of a Brownian motion Let $\{X(t), t \geq 0\}$ be a Brownian motion with drift parameter $\mu = 3$ and variance parameter $\sigma^2 = 9$. If $X(0) = 10$, find $P(X(0.5) > 10)$.
First, I calculated the expectation and variance of $X(0.5)).$ Since $X(0.5) - X(0)$ is normal with mean $1.5$ and variance $4.5$, it follows that $E[X(0.5)] 
= 1.5 + 10 = 11.5$. 
Likewise, we have $\text{Var}(X(0.5) - X(0)) = 4.5$. So I thought that $\text{Var}(X(0.5) - X(0)) = \text{Var}(X(0.5) - 10) = \text{Var}(X(0.5)) = 4.5$. But the answer key says the answer should be $14.5$. 
I also tried writing
$$X(t) = 10 + 3t + 3B(t),$$
so 
$$\text{Var}(X(0.5)) = \text{Var}(3 B(0.5)) = 9 \cdot \text{Var}(B(0.5))$$
 [/INST] It should be $4.5$. There may be an error in the answer key. The variance of the deterministic part is $0$ and doesn't add to the variance of the increment like you wrote. 
At $X(0.5)$, we have $X(0.5) = 10 + 3 * 0.5 + 3\sqrt{0.5}Z$, where $Z$ is a standard normal random variable ($m = 0$, $s^2 = 1$). And so to calculate $P(X(0.5) > 10)$, we are calculating $P(10 + 3 * 0.5 + 3\sqrt{0.5}Z > 10)$.
Simplifying we have 
\begin{align}
P(11.5 + 3\sqrt{0.5}Z > 10) & = P(3\sqrt{0.5}Z > -1.5) \\
& = P(Z > -\frac{\sqrt{2}}{2})\\
& = 1 - P(Z < -\frac{\sqrt{2}}{2}) \\
& = 1 - \Phi(-\frac{\sqrt{2}}{2})\\
& \approx 0.7602
\end{align}
That should be the answer but be sure to check! Hope this helps.
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