Expectation of squared increment (Brownian Motion)

Question

I'm trying to show that $\mathbb{E}[|X_t - X_s|^2] = |t - s|$ but the absolute value is confusing me.

I know that $X_t - X_s \sim\mathcal{N}(0,t-s)$

so $\mathbb{E}[(X_t - X_s)^2] - \mathbb{E}[X_t - X_s]^2 = t-s $

but how do I bring the absolute value in?

Answer 1

You statements only hold true if $t \ge s$. If $t < s$, you cannot end up with a negative on the RHS since the LHS is all positive.

So then you rename $X_t$ to $X_s$ and vice versa, ending up in $s-t$, but if you want one expression for all cases, it is always $|t-s| = |s-t|$, just a convenient notation not to write the branches out.

Answer 2

\begin{align}\mathbb{E}[\,|X_t-X_s|^2\,]&=\mathbb{E}[\,(X_t-X_s)^2\,]=\mathbb{E}[\,X_t^2\,]+\mathbb{E}[\,X_s^2\,]-2\mathbb{E}[\,X_tX_s\,]\\ \\ &=\text{Var}(X_t)+\text{Var}(X_s)-2\text{Cov}(X_t,X_s)\\ \\ &=t+s-2\min\{t,s\}\\ \\ &=t+s-2\left(\frac{t+s}{2}-\frac{|t-s|}{2}\right)\\ \\ &=|t-s| \end{align}