Distribution of the Brownian Motion

brownian motioncovarianceexpected valuenormal distributionstochastic-processes

Let $(B_t)_{t\geq 0}$ be the standard wiener process and define:

$$X_{t} = \frac{B_{t+h} – B_t}{h} \quad , \quad h > 0.$$

Now I'm asked to prove that $(X_t)_{t \geq 0}$ is a Gaussian Process and calculate its mean $\mathbb{E}[X_t]$ and covariance $Cov(X_t, X_0)$.

Best Answer

HINT

Well, the sum of any two Normals is a Normal. Note that $$ \mathbb{E}[X_t] = \frac{\mathbb{E}[B_{t+h}] - \mathbb{E}[B_t]}{h} $$ and can you now compute $\mathbb{E}\left[X_t^2\right]$ and from that compute the variance?