I've always been confused with Delta hedging. It is well-known that for a (smooth enough) function of (S,t) we have, due to Ito's lemma, that: dC=(∂C∂t(S,t)+μS∂C∂S(S,t)+12σ2S2∂2C∂S2(S,t))dt+σS∂C∂S(S,t)dX
I understand the usual procedure of Δ, but I never understand how Δ can be defined as −∂C∂S(S,t). By definition a hedging strategy must be a predictable process, is there any justification that Δ given as above is predictable, not merely adapted, to the filtration generated by the stock process?
Even if we accept the definition of Δ, the author also doesn't explain how to compute d(C+ΔS). But, as far as I know, if Δ is really the symbol for −∂C∂S(S,t), then we will have d(ΔS)=Sd(Δ)+ΔdS+(dS)⋅(dΔ).
Wherein lies the problem?
Answer
This question has been asked many times and some clarifications appear needed.
As pointed out in an answer to this question, the portfolio Δ1tSt+Δ2tC,
where Δ1t=−∂C∂S and Δ2t=1, is, generally, neither self-financing nor locally risk-free.To derive the Black-Scholes' PDE, we seek a portfolio Δ1tSt+Δ2tC such that it is self-financing and locally risk-free. As is shown in answer to the above question, we derived the PDE ∂C∂t+rSt∂C∂S+12∂2C∂S2σ2S2t−rC=0.
and the portfolio weights Δ1t=−∂C∂SBtCt−∂C∂SS,Δ2t=BtCt−∂C∂SS,where Bt=ert is the money-market account value. Note that Δ1tSt+Δ2tC=Bt.Based on PDE (1) and the portfolio weights give by (2), it is easy to verify that Δ1tdSt+Δ2tdC=dBt.
That is, the portfolio is indeed self-financing and locally risk-free.Though messy, based on PDE (1) and the portfolio weights give by (2), it is possible to show directly that d(Δ1tSt+Δ2tC)=Δ1tdSt+Δ2tdC
using Ito's lemma. Here, we assume that the respective partial derivatives, such as ∂2C∂t∂S and ∂3C∂S3, exist.
Below, we show the self-financing property using Ito's lemma, that is, SdΔ1t+CdΔ2t+d⟨Δ1t,S⟩t+d⟨Δ2t,C⟩t=0.
For weight Δ1t, ∂Δ1t∂t=−(∂2C∂t∂SBt+r∂C∂SBt)(Ct−∂C∂SS)−(∂C∂t−∂2C∂t∂SS)∂C∂SBt(Ct−∂C∂SS)2=−∂2C∂t∂SBtCt+r∂C∂SBtCt−r(∂C∂S)2BtSt−∂C∂t∂C∂SBt(Ct−∂C∂SS)2=−∂2C∂t∂SBtCt+12∂2C∂S2∂C∂Sσ2S2tBt(Ct−∂C∂SS)2(From Eqn (1))=−∂2C∂t∂SBtC2t+12∂2C∂S2∂C∂Sσ2S2tBtCt−∂2C∂t∂S∂C∂SBtCtS−12∂2C∂S2(∂C∂S)2σ2S3tBt(Ct−∂C∂SS)3,∂Δ1t∂S=−∂2C∂S2Bt(Ct−∂C∂SS)−(∂C∂S−∂2C∂S2S−∂C∂S)∂C∂SBt(Ct−∂C∂SS)2=−∂2C∂S2BtCt(Ct−∂C∂SS)2,∂2Δ1t∂S2=−(∂3C∂S3BtCt+∂2C∂S2∂C∂SBt)(Ct−∂C∂SS)2+2(Ct−∂C∂SS)(∂2C∂S2)2SBtCt(Ct−∂C∂SS)4=−∂3C∂S3BtC2t−∂3C∂S3∂C∂SBtCtS+∂2C∂S2∂C∂SBtCt−∂2C∂S2(∂C∂S)2BtS+2(∂2C∂S2)2SBtCt(Ct−∂C∂SS)3
For weight Δ2t, ∂Δ2t∂t=rBt(Ct−∂C∂SS)−(∂C∂t−∂2C∂t∂SS)Bt(Ct−∂C∂SS)2=rBtCt−r∂C∂SBtS−Bt∂C∂t+∂2C∂t∂SSBt(Ct−∂C∂SS)2=12∂2C∂S2σ2S2tBt+∂2C∂t∂SSBt(Ct−∂C∂SS)2(From Eqn (1))=12∂2C∂S2σ2S2tBtCt+∂2C∂t∂SSBtCt−12∂2C∂S2∂C∂Sσ2S3tBt−∂2C∂t∂S∂C∂SS2Bt(Ct−∂C∂SS)2,∂Δ2t∂S=−(∂C∂S−∂2C∂S2S−∂C∂S)Bt(Ct−∂C∂SS)2=∂2C∂S2BtS(Ct−∂C∂SS)2,∂2Δ2t∂S2=(∂3C∂S3BtS+∂2C∂S2Bt)(Ct−∂C∂SS)2+2(Ct−∂C∂SS)(∂2C∂S2)2S2Bt(Ct−∂C∂SS)4=∂3C∂S3BtCtS−∂3C∂S3∂C∂SBtS2+∂2C∂S2BtCt−∂2C∂S2∂C∂SBtS+2(∂2C∂S2)2S2Bt(Ct−∂C∂SS)3.
Then d⟨Δ1t,S⟩t+d⟨Δ2t,C⟩t=(−∂2C∂S2BtCtσ2S2(Ct−∂C∂SS)2+∂2C∂S2∂C∂Sσ2BtS3(Ct−∂C∂SS)2)dt=(−∂2C∂S2BtC2tσ2S2+∂2C∂S2∂C∂SBtCtσ2S3+∂2C∂S2∂C∂Sσ2BtS3Ct−∂2C∂S2(∂C∂S)2σ2BtS4(Ct−∂C∂SS)3)dt.
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