Monday, May 1, 2017

probability question about brownian motion


Assume Wt is a standard Brownian Motion, calculate the the probability that WtW2t is negative, i.e., P(WtW2t<0). I find it tricky to calculate the probability.Thank you.



Answer



Your decomposition is correct. I will show here the computation for one term: P(Wt<0,W2t>0)=P(Wt<0,W2tWt>Wt)=E(E(1{Wt<0}1{W2tWt>Wt}Wt))=E(1{Wt<0}Φ(Wt/t))=E(1{Wt/t<0}Φ(Wt/t))=0ϕ(x)Φ(x)dx=12Φ(x)20=18,

where ϕ(x)=12πex22
is the density of a standard normal random variable, and Φ(x) is the cumulative distribution function.


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