# 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

$$\text{Var}(X(0.5)) = \text{Var}(3 B(0.5)) = 9 \cdot \text{Var}(B(0.5))$$

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}