Saturday, April 29, 2017

black scholes - Understanding the solution of this integral


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(Steμτ12σ2τ+στz)12πez22dz=eμτStN(d1)

with z=lnKSt(μσ2/2)τστ, d1=lnStK+(μ+σ2/2)τστ, and N() cumulative Standardnormal distribution.


Can you please explain how this equality is obtained?



Answer



Let τ=Tt. Then ST=Ste(μ12σ2)τ+στZ,

where Z is a standard normal random variable, independent of Ft. Moreover, E(ST1{ST>K}Ft)=E(Ste(μ12σ2)τ+στZ1{Ste(μ12σ2)τ+στZ>K}Ft)=Ste(μ12σ2)τ+στz1{Ste(μ12σ2)τ+στz>K}12πez22dz=zSte(μ12σ2)τ+στz12πez22dz=z12πSte(μ12σ2)τ+στzz22dz=z12πSte(μ12σ2)τ12(zστ)2+12σ2τdz=Steuτzστ12πex22dx=SteuτN(στz)=SteuτN(d1),
where d1=στz=lnStK+(μ+σ2/2)τστ.


No comments:

Post a Comment

technique - How credible is wikipedia?

I understand that this question relates more to wikipedia than it does writing but... If I was going to use wikipedia for a source for a res...