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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.

K.defaoite
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user928670
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  • The integral in (i) is pretty easy, you just integrate by parts. If that factor of $x$ weren't there, that's when you'd need erf. (ii) is usually part of the definition of Brownian motion; if you don't see it then you should state the precise definition you're using. – Nate Eldredge May 17 '21 at 16:37

1 Answers1

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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 $\textrm{Var}[B_t]=E[B_t^2]=t$.

Snoop
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