Assume Wt is a standard Brownian Motion, calculate the the probability that Wt∗W2t is negative, i.e., P(Wt∗W2t<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,W2t−Wt>−Wt)=E(E(1{Wt<0}1{W2t−Wt>−Wt}∣Wt))=E(1{Wt<0}Φ(Wt/√t))=E(1{Wt/√t<0}Φ(Wt/√t))=∫0−∞ϕ(x)Φ(x)dx=12Φ(x)2∣0−∞=18, where ϕ(x)=1√2πe−x22 is the density of a standard normal random variable, and Φ(x) is the cumulative distribution function.
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