Derivation of the Gamma approximation formula from the Delta approximation formula

Question

The formula of $Gamma = (Vplus + Vminus - 2V0)/(V0 * dS^2)$, where

• $V$ is the contract value,
• $S$ is the stock price.

We also know that $Gamma = (Dplus-Dminus)/(Splus-Sminus)$, where

• $D$ is the contract Delta,
• $Splus = S0 + dS$,
• $Sminus = S0 - dS$,
• $Dplus = (Vplus - V0)/(V0 * dS)$,
• $Dminus = (V0 - Vminus)/(V0 * dS)$.

Substituting $Dplus$, $Dminus$, and replacing $Splus-Sminus$ with $2dS$ we get:
$(Vplus + Vminus - 2V0)/(V0 * dS)/2dS = (Vplus + Vminus - 2V0)/(V0 * 2 * dS^2)$

I must be doing something wrong, because 1/2 is not present in the book formula.
Could you, please, help me with the derivation?

Answer 1

(Welcome to Quant SE. It looks like you haven't made up your mind on whether your shock is $dS$ or $S_0dS$. Also, $V_0$ doesn't belong in the denominator. You probably mean $S_0$. Please try to use Latex on this site next time you visit.)

It's better to start with Taylor's theorem with remainder to convince yourself of the validity of these finite difference schemes:

$$V(S+dS)= V(S)+V'(S)dS+\boxed{\frac{1}{2}V''(S)(dS)^2}+ \frac{1}{6}V'''(S_1)(dS)^3$$

for some $S_1\in (S, S+dS)$ $$V(S-dS)= V(S)-V'(S)dS+\boxed{\frac{1}{2}V''(S)(dS)^2}-\frac{1}{6}V'''(S_2)(dS)^3$$ for some $S_2\in (S-dS, S)$

We then get (note that the two halves in the boxes will make sure there is no $2$ in the final denominator):

$$ \frac{V(S+dS)+V(S-dS)-2V(S)}{(dS)^2} = V''(S)+ \frac{1}{6}(V'''(S_1) + V'''(S_2))dS$$

and finally we let $dS \rightarrow 0$.

Replacing $dS$ by $SdS$ everywhere, we also get:

$$ \frac{V(S+SdS)+V(S-SdS)-2V(S)}{S^2(dS)^2} \approx V''(S),$$

for small $dS$.

Answer 2

Also, I do not think Gamma = (Dplus-Dminus)/(Splus-Sminus) is correct, rather it is Gamma = (Dplus-Dminus)/(dS). You must use the increment dS throughout, for consistency. By making the denominator be a 2*dS step you are not correctly differentiating with respect to the variable "S".