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 n−m times V(S(m)n,t+nΔt)=e−rΔt[pV(uS(m)n,t+(n+1)Δt+(1−p)V(uS(m)n,t+(n+1)Δt]
p=e(r−D)Δt−du−d - risk-neutral probability, r - risk free rate, D continuous dividend yield
u=eσ√Δt, d=e−σ√Δt, 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(u−1)2+(1−p)(d−1)2)∗e−rΔt
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
Now, substitute ˆμ=q(logd+log(u/d)),ˆσ=q(1−q)(log(u/d))2,q=erΔt−e−σ√Δteσ√Δt−e−σ√Δtin f(Δt)=log(K/S0)−ˆμN−εlog(u/d)ˆσ√N
To derive the PDE, note in the Binomial model it holds erΔtCall(S,t)=qCallu(Su,t+Δt)+(1−q)Calld(Sd,t+Δt)
Plug in the value of q=erΔt−e−σ√Δteσ√Δt−e−σ√Δt:
erΔtCall(S,t)=erΔt−e−σ√Δteσ√Δt−e−σ√ΔtCall(Seσ√Δt,t+Δt)+eσ√Δt−erΔteσ√Δt−e−σ√ΔtCall(Se−σ√Δt,t+Δt)
Expand Call(Seσ√Δt,t+Δt) about (S,t):
Call(Seσ√Δt,t+Δt)=Call(S,t)+(eσΔt−1)∂Call∂S+12(eσΔt−1)2S2∂2Call∂S2+Δt(∂Call∂t+(eσΔt−1)S∂2Call∂S∂t+12(eσΔt−1)2S2∂3Call∂S2∂t)+O((S)3)
erΔtCall(S,t)=erΔt−e−σ√Δteσ√Δt−e−σ√ΔtCall(Seσ√Δt,t+Δt)−eσ√Δt−erΔteσ√Δt−e−σ√ΔtCall(Se−σ√Δt,t+Δt)
Expand eq. (2) in Δt to get
0=(rCall(S,t)+∂Call∂t+rS∂Call∂S+12rS2∂2Call∂S2)Δt+O((Δt)3/2)
As Δt→0 we have the Black-Scholes PDE.
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