The following integral represents an expected value of a geometric brownian motion for ST>K (i.e. part of the Black-Scholes call option price): ∫z∗(Steμτ−12σ2τ+σ√τz)1√2πe−z22dz=eμτStN(d∗1)
with z∗=lnKSt−(μ−σ2/2)τσ√τ, d∗1=lnStK+(μ+σ2/2)τσ√τ, and N(⋅) cumulative Standardnormal distribution.
Can you please explain how this equality is obtained?
Answer
Let τ=T−t. Then ST=Ste(μ−12σ2)τ+σ√τZ,
where Z is a standard normal random variable, independent of Ft. Moreover, E(ST1{ST>K}∣Ft)=E(Ste(μ−12σ2)τ+σ√τZ1{Ste(μ−12σ2)τ+σ√τZ>K}∣Ft)=∫∞−∞Ste(μ−12σ2)τ+σ√τz1{Ste(μ−12σ2)τ+σ√τz>K}1√2πe−z22dz=∫∞z∗Ste(μ−12σ2)τ+σ√τz1√2πe−z22dz=∫∞z∗1√2πSte(μ−12σ2)τ+σ√τz−z22dz=∫∞z∗1√2πSte(μ−12σ2)τ−12(z−σ√τ)2+12σ2τdz=Steuτ∫∞z∗−σ√τ1√2πe−x22dx=SteuτN(σ√τ−z∗)=SteuτN(d∗1),
where d∗1=σ√τ−z∗=lnStK+(μ+σ2/2)τσ√τ.
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