Maximum of Wiener Brownian Motion

Let $W_0$ be the standard Wiener Brownian Motion process, and $M(t)$ defined as $M(t)=\max_{0 \leq s \leq t} W_{0}(t)$ , we then have the following result:

$P(M(t) \geq x)= 2(1-\phi(\frac{x}{\sqrt{t}}))$

where $\phi$ denotes the standard normal cumulative distribution function:

$\phi(z)=\frac{1}{\sqrt{2\pi}}e^{-z^{2}/2}$ $\forall{z}\in\mathbb{R}$

This result is simply a direct application of Theorem 2 from this paper with $\alpha=0$

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May 4, 2009 - Posted by Jean-Philippe | Wiener Brownian Motion

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