Tuesday, June 20, 2017

binomial tree - Approximation of CRR as Black Scholes PDE


I have a formula for intermediate european option price calculated at, say, m-th possible tree value.


S(m)n is a price at node after going up n times and down nm times V(S(m)n,t+nΔt)=erΔt[pV(uS(m)n,t+(n+1)Δt+(1p)V(uS(m)n,t+(n+1)Δt]



p=e(rD)Δtdud - risk-neutral probability, r - risk free rate, D continuous dividend yield


u=eσΔt, d=eσΔt, σ is volatility of stock price


I need to figure out using taylor approximations that under limit for Δt>0 this formula becomes BS PDE.




Ok so first of all, I used the identity ex=1+x and applied it everywhere possible. Then I wrote taylor expansion for both functions of the RHS but at this point my results did not really match.


Any value beyond and equal to O(Δt2) I made equal to zero.


So, for instance, for the term 12d2VdS2 I ended with the following:


... + 12d2VdS2S2(p(u1)2+(1p)(d1)2)erΔt


Unfortunatelly after substituting all the variables after rewriting them with taylor approximation, the resulting expression after all the cancellations did not turn into the 12d2VdS2S2σ2


I wrote taylor expansion centered around (S(m)n,t+Δt), respectively.



Could someone let me know where I made the mistake? I did not bother continuing with other parts of the formula mainly due to getting the first one wrong ...



Answer



Assuming continuously compounded returns for a multi-period model with N being the number of periods:


{loguwith probability qlogdwith probability 1-q

given the stock price at maturity log(STS0)=ilogu+(Ni)logd=ilog(ud)+Nlogd
where i is a binomial r.v. under the risk-neutral measure Q. limNlog(STS0)N(μT,σ2T)
with μT=E[i]log(u/d)+Nlogdσ2T=Var[i][log(u/d)]2
By CLT, the distribution of the sum of continuously compounded returns converges to a normal distribution. Since, by assumption u=1/d, we have the equations: u=eΔt,d=eΔT,q=12+μ2σΔt,Δt=tN
Let us re-write the binomial distribution: Bin(log(K/S0dN)log(u/d)+1,N,q) 1Bin(log(K/S0dN)log(u/d)+1,N,q)=Q(ilog(K/S0dN)log(u/d)+11)=Q((iNqNq(1q))=log(ST/S0)NlogdNqlog(u/d)log(u/d)Nq(1q)log(K/S0dN)log(u/d)NqNq(1q)=:A)
log(K/S0dN)log(u/d)=log(K/S0dN)log(u/d)ε,ε[0,1)A=log(K/S0)+N(logd+qlog(u/d))εlog(u/d)log(u/d)q(1q)N


Now, substitute ˆμ=q(logd+log(u/d)),ˆσ=q(1q)(log(u/d))2,q=erΔteσΔteσΔteσΔtin f(Δt)=log(K/S0)ˆμNεlog(u/d)ˆσN

and compute its Taylor expansion: f(Δt)=log(K/S0)σNΔt2εσNσ+12σ2NΔtrNΔtσNΔt+O((Δt)3)
As h0, N and NΔt=T, f(Δt)log(K/S0)(r12σ2)TσTBin(log(K/S0dN)log(u/d)+1,N,q)Φ(log(K/S0)(r12σ2)TσT)
As a result, we have the convergence towards the Black-Scholes model: Call0=S0Φ(d1)KerTΦ(d2)


To derive the PDE, note in the Binomial model it holds erΔtCall(S,t)=qCallu(Su,t+Δt)+(1q)Calld(Sd,t+Δt)


Plug in the value of q=erΔteσΔteσΔteσΔt:


erΔtCall(S,t)=erΔteσΔteσΔteσΔtCall(SeσΔt,t+Δt)+eσΔterΔteσΔteσΔtCall(SeσΔt,t+Δt)


Expand Call(SeσΔt,t+Δt) about (S,t):


Call(SeσΔt,t+Δt)=Call(S,t)+(eσΔt1)CallS+12(eσΔt1)2S22CallS2+Δt(Callt+(eσΔt1)S2CallSt+12(eσΔt1)2S23CallS2t)+O((S)3)

Then, perform the same expansion for Call(SeσΔt,t+Δt) about (S,t), and re-write eq. (1)



erΔtCall(S,t)=erΔteσΔteσΔteσΔtCall(SeσΔt,t+Δt)eσΔterΔteσΔteσΔtCall(SeσΔt,t+Δt)


Expand eq. (2) in Δt to get


0=(rCall(S,t)+Callt+rSCallS+12rS22CallS2)Δt+O((Δt)3/2)


As Δt0 we have the Black-Scholes PDE.


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