Find the covariance of a brownian motion.

Question

Given a standard Brownian motion $\{W_t\}_{t\geq0}$, find the value of $\operatorname{Cov}\left(W_t,W_s\right)$.

Is there a way to simplify $$\operatorname{Cov}\left(W_t,W_s\right)?$$

Answer 1

Using the properties and Brownian motion and the linearity of the Covariance, we easily get for $t \geq s$:

\begin{align} \operatorname{Cov}\bigl(W_s, W_t\bigr) &= \operatorname{Cov}\bigl(W_s, W_t - W_s + W_s\bigr) =\operatorname{Cov}\bigl(W_s, W_t-W_s\bigr) + \operatorname{Cov}\bigl(W_s, W_s\bigr) \\ &= 0 + Var(W_s) = s. \end{align}

Similarly if $s \geq t$ you get $$ \operatorname{Cov}\bigl(W_s, W_t\bigr) = t. $$

So in general, we have that $$ \operatorname{Cov}\bigl(W_s, W_t\bigr) = \min\{s,t\}. $$