Friday, January 2, 2015

brownian motion - GBM probability of hitting barrier


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)=1exp(2Tβ(βx))
and I then tried the following: since St=S0e(μ1/2σ2)tσWt
we have P(St<β,t[0,T]|S0,ST=s)=P(Wt<ln(β/S0)(μ1/2σ2)tσ|S0,WT=ln(s/S0)(μ1/2σ2)Tσ)=1exp(2T(ln(β/S0)(μ1/2σ2)tσ)(ln(β/S0)ln(s/S0)σ))=1exp(2σ2T(ln(β/S0)(μ1/2σ2)t)ln(β/s))
Evidently, the correct answer is just1exp(2σ2Tln(β/S0)ln(β/s))
so I believe I'm almost there. What am I missing?



Answer



In Inequality Wt<ln(β/S0)(μ1/2σ2)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=μ1/2σ2σ, b=ln(β/S0)σ, and c=ln(s/S0)σ. We define the probability measure ˜P such that d˜PdP|t=e12a2taWt,

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{WTA})= ˜E((d˜PdP|T)111{Wt+at<b,t[0,T]}11{WTA})= ˜E(e12a2T+aWT11{Wt+at<b,t[0,T]}11{WTA})= ˜E(e12a2T+a˜WT11{˜Wt<b,t[0,T]}11{˜WTaTA})= ˜E(e12a2T+a˜WT11{˜WTaTA}˜E(11{˜Wt<b,t[0,T]}|˜WT))= ˜E(e12a2T+a˜WT11{˜WTaTA}[1exp(2Tb(b˜WT))])= E(d˜PdP|Te12a2T+a˜WT11{˜WTaTA}[1exp(2Tb(b˜WT))])= E(11{WTA}[1exp(2Tb(bWTaT))]).
That is, E(11{Wt+at<b,t[0,T]}|WT)=1exp(2Tb(bWTaT)),
or, equivalently, E(11{Wt+at<b,t[0,T]}|WT=x)=1exp(2Tb(bxaT)).
Therefore,  P(Wt<ln(β/S0)(μ1/2σ2)tσ,t[0,T]|S0,WT=ln(s/S0)(μ1/2σ2)Tσ)= E(11{Wt+at<b,t[0,T]}|WT=caT)= 1exp(2Tb(bc))= 1exp(2σ2Tln(β/S0)ln(β/s)).


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