Brownian Motion Finding M(t)

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If I have that {$B(t); t >=0$} is a standard Brownian motion, with $B(0)=0$, and I let $M(t)$ = max{$B(u) ; 0 \leq u \leq t$} and I am supposed to:

a) Evaluate Pr{$M(4) \leq 2$} and

b) Find the number c for which Pr{$M(9) > c$}=0.10

I'm having a hard time coming up with a formula to evaluate these two. I know that the Pr{$max B(t) > x$} = 2Pr{$B(t)>x$}, and I have tried to use the fact that Pr{$M(t) < a$}=Pr{$\tau_a > t$}=2Pr{$B(t)>a$} for (a) but I can't seem to get the answer that the back of the book has (which is 0.6826)

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hint: Use the reflection principle (like for a simple random walk)

$P(M(4)>2) = 2P(B(4)>2)$ $\Rightarrow$ $P(M(4)≤2) = 1- 2P(B(4)>2) = .6826$

$P(M(9)>c) = .1 \Rightarrow P(B(9)>c)= .05$

solve for c