Expectation of the absolute of the difference between two B.M, $\DeclareMathOperator*{\E}{\mathbb{E}}|B(s)-B(t)|=\sqrt{\frac{2}{\pi}}|t-s|^{1/2}$?

Question

I am going through Bernt Oksendal's (S.D.E an introduction with application sixth edition) book, I can do most of the questions without any issue, but I don't understand how (using my understanding of what I have learned from Bernt up to chapter six) $ Expectation$$$|B(s)-B(t)|=\sqrt{\frac{2}{\pi }}|t-s|^{1/2} $$

Note: I know that I have to use the definition of Brownian motion, as I have found here on wiki, https://en.wikipedia.org/wiki/Brownian_motion

Could someone please show me step by step as to how to work this out? I do not need just hints.

Note that this specific question isn't from Oksendal's book. Thanks.

Answer 1

One can easily show that if $X \sim N(0, \sigma^2)$ then $E(|X|) = \sqrt{\frac{2}{\pi}} \sigma$. Now observe that if $t > s$, $B_t -B_s \sim N(0, t-s)$. By symmetry, one can argue similarly for $s> t$. Result follows immediately.