Let $(B_t, t\ge0)$ be a standard Brownian motion. For $s,t\ge 0$, prove that:
(i) $\Bbb{E}[|B_t-B_s|]=\sqrt{\frac{2|t-s|}{\pi}}$
(ii) $\Bbb{E}[|B_t-B_s|^2]=|s-t|$
The mark scheme is as follows:
(i) $\Bbb{E}[|B_t-B_s|]=\displaystyle\int^∞_02x\frac{1}{\sqrt{2\pi(t-s)}}\exp\biggl(-\frac{x^2}{2(t-s)}\biggr)\mathrm d x=\sqrt{\frac{2|t-s|}{\pi}}$
(ii) $\Bbb{E}[|B_t-B_s|^2]=\Bbb{E}[B_t^2]+\Bbb{E}[B_s^2]-2\Bbb{E}[B_tB_s]=t-s$
For both questions I have been confused as I thought because I read somewhere that as it is standardized, the Brownian motion expectation should be zero.
In part (i) as well I have noticed that they seem to use the pdf for $B_t-B_s\sim N(0,t-s)$ but I was wondering how this integral is solved? I was reading somewhere about the erf function but have never used this before.
Finally, in part 2, I was wondering why $\Bbb{E}[B_t^2]=t$ as my lecture notes have written this out without explanation.
For simplicity set $t>s$. Recall that $$X:=B_t-B_s \sim \mathcal{N}(0,(t-s))$$ So we are just working with a Gaussian random variable with mean zero and variance $\sigma^2:=t-s$. So $$P(|X|\leq z)=P(-z\leq X \leq z)=\int_{-z}^z\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{x^2}{2\sigma^2}}dx=\Phi\bigg(\frac{z}{\sigma}\bigg)-\Phi\bigg(-\frac{z}{\sigma}\bigg)$$ Therefore $$f_{|X|}(z)=\frac{1}{\sigma}\bigg(\phi\bigg(\frac{z}{\sigma}\bigg)+\phi\bigg(-\frac{z}{\sigma}\bigg)\bigg)=\frac{2}{\sigma}\phi\bigg(\frac{z}{\sigma}\bigg), \, \, z \in [0,\infty)$$ where $\phi$ is the standard normal pdf and $\Phi$ is the standard normal cdf. This is a half-normal distribution with expectation and second moment: $$E[|X|]=\sigma \sqrt{\frac{2}{\pi}}=\sqrt{\frac{2(t-s)}{\pi}}$$ $$E[|X|^2]=\sigma^2=t-s$$
Since $B_t \sim \mathcal{N}(0,t)$ then $\textrm{Var}[B_t]=E[B_t^2]=t$.