# Normal probability and Brownian motion

Let $X_t$ be a Brownian motion with parameter $\sigma$. Find the probability in terms of $$\Phi(x)= \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^x e^{- \frac{ \alpha ^2}{2}}d\alpha$$ How would I do this for $P(X_2>2)$? I know that $P(X_2>2)=1-P(X_2<2)$ but where do I go from there? $X_2$ is a normal random variable with mean $0$ and variance $2 \sigma$.

• This is $\Phi$ (type \Phi), not $\phi$. – Did Apr 17 '13 at 3:24
• @Did Fixed it, thanks! – Alti Apr 17 '13 at 3:25

A normal random variable $X$ with mean $\mu$ and variance $\sigma^2$ is related to the standard normal random variable $Z$ (zero mean and unit variance) with affine transform: $$X \stackrel{d}{=} \mu + \sigma Z$$ where $\sigma > 0$. Hence $$\Pr(X \leqslant x) = \Pr\left( \mu + \sigma Z \leqslant x \right) = \Pr\left( Z \leqslant \frac{x-\mu}{\sigma} \right) = \Phi\left(\frac{x-\mu}{\sigma} \right)$$