I tried using the brownian bridge approach to determine the probability P(St<β,t∈[0,T]|S0,ST)
where St is a GBM in the usual Black Scholes setup. We know that for a BM Wt, P(Wt<β,t∈[0,T]|WT=x)=1−exp(−2Tβ(β−x))
and I then tried the following: since St=S0e(μ−1/2σ2)t−σWt
we have P(St<β,t∈[0,T]|S0,ST=s)=P(Wt<ln(β/S0)−(μ−1/2σ2)tσ|S0,WT=ln(s/S0)−(μ−1/2σ2)Tσ)=1−exp(−2T(ln(β/S0)−(μ−1/2σ2)tσ)(ln(β/S0)−ln(s/S0)σ))=1−exp(−2σ2T(ln(β/S0)−(μ−1/2σ2)t)ln(β/s))
Evidently, the correct answer is just1−exp(−2σ2Tln(β/S0)ln(β/s))
so I believe I'm almost there. What am I missing?
Answer
In Inequality Wt<ln(β/S0)−(μ−1/2σ2)tσ,
the right-hand side depends on t, you are not able to use the above result directly. In this case, a Girsanov transformation is usually employed.
Let a=μ−1/2σ2σ, b=ln(β/S0)σ, and c=ln(s/S0)σ. We define the probability measure ˜P such that d˜PdP|t=e−12a2t−aWt,
where P is the original probability measure. Then ˜Wt=Wt+at is a standard Brownian motion under ˜P. Let E and ˜E be expectations with respect to measures P and ˜P. Then, for any Borel set A, E(11{Wt+at<b,t∈[0,T]}11{WT∈A})= ˜E((d˜PdP|T)−111{Wt+at<b,t∈[0,T]}11{WT∈A})= ˜E(e12a2T+aWT11{Wt+at<b,t∈[0,T]}11{WT∈A})= ˜E(e−12a2T+a˜WT11{˜Wt<b,t∈[0,T]}11{˜WT−aT∈A})= ˜E(e−12a2T+a˜WT11{˜WT−aT∈A}˜E(11{˜Wt<b,t∈[0,T]}|˜WT))= ˜E(e−12a2T+a˜WT11{˜WT−aT∈A}[1−exp(−2Tb(b−˜WT))])= E(d˜PdP|Te−12a2T+a˜WT11{˜WT−aT∈A}[1−exp(−2Tb(b−˜WT))])= E(11{WT∈A}[1−exp(−2Tb(b−WT−aT))]).
That is, E(11{Wt+at<b,t∈[0,T]}|WT)=1−exp(−2Tb(b−WT−aT)),
or, equivalently, E(11{Wt+at<b,t∈[0,T]}|WT=x)=1−exp(−2Tb(b−x−aT)).
Therefore, P(Wt<ln(β/S0)−(μ−1/2σ2)tσ,t∈[0,T]|S0,WT=ln(s/S0)−(μ−1/2σ2)Tσ)= E(11{Wt+at<b,t∈[0,T]}|WT=c−aT)= 1−exp(−2Tb(b−c))= 1−exp(−2σ2Tln(β/S0)ln(β/s)).
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