Delta of an option derived from the binomial model

Question

I have the following function $V=V(S,t)$, $V^- = V(vS,t+\delta t)$, $V^+ = V(uS, t +\delta t)$. The book proceeds to explain that if we use Taylor series expansion on the above we will confirm that $\frac{\partial V}{\partial S} \sim \Delta$ by substituting the results into $\Delta = \frac{V^+ - V^-}{S(u-v)}$.

I get the following results for the expansion: $V^- = V^- (vS,t)+ \frac{\partial V^-}{\partial t} \delta t$

$V^+ = V^+ (uS,t)+\frac{\partial V^+}{\partial t} \delta t$

Not sure how I proceed, subbing right now does not yield any sensible result. Not sure where we get partial derivative of $V$ with respect to $S$ either.. Is that from expanding $V(S,t)$, in that case, how do expand it. Basically, how do I arrive at the result that shows that partial derivative of $V$ with respect to $S$ is delta?

Answer 1

This is not the Taylor expansion with respect to $t$, instead, it is the Taylor expansion with respect to $S$. Moreover, the prices at time $t+\delta t$ is used for approximation. That is, \begin{align*} \frac{\partial V}{\partial S}\big|_t &\approx \frac{V(uS, t) - V(vS, t)}{uS - vS}\\ &\approx \frac{V(uS, t+\delta t) - V(vS, t+\delta t)}{S(u-v)}\\ &= \frac{V^+-V^-}{S(u-v)}. \end{align*} The reason is that, at time $t=0$, there is only a single price available, and, to compute the delta and gamma respective hedge ratios, other trees are needed. For computational efficiency, we use the prices at time $0+\delta t$, where multiple prices are available.