c**a
发帖数: 316 | 1 Please help!
X(t) is the stochastic exponential of a standard Brownian motion,
i.e. X(t) = exp(w(t) - 0.5 t), w(t) being a standard Brownian motion.
What is the probability that X(T) is greater than H (>1)?
My solution:
log(X(T)) > log (H)
=>
w(T) - 0.5 T > log(H)
=>
w(T) > log(H) + 0.5 T
=>
y > log(H)/sqrt(T) + 0.5*sqrt(T) where y is a standard normal r.v.
The PROBLEM is as T approaches +inf
[log(H)/sqrt(T) + 0.5*sqrt(T)] approaches +inf as well.
Hence, we have y > +inf.
Hence, the probability that X(T) is greater than H is zero when T->\inf
Which does not make ANY sense. | Q***5
发帖数: 994 | 2 So it converges to 0 by probability,why it does not make sense? | b****d
发帖数: 1311 | 3 As T->infinity, the probability that w(T) > 0.5T definitely approaches 0. |
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