Deriving Gamma from a Central Finite Difference
Summary
The discussion examines how to derive an approximation for an option’s gamma from values at an underlying price above and below the current price. The original question substitutes finite-difference estimates of delta into a difference quotient and obtains an extra factor of one half compared with a textbook formula. The responses identify inconsistent notation and scaling as sources of confusion, including confusing an absolute price shock with a relative shock and using the wrong variable in the denominator.
The accepted explanation starts from Taylor expansions of option value on both sides of the current price. Adding the expansions cancels the first-derivative terms; the second-derivative contributions each carry a half, which combine to give the central second-difference approximation without an extra factor of two. A remainder term involving higher derivatives vanishes in the limit as the shock approaches zero. For a relative shock, the price increment is scaled by the current underlying price, so the denominator must be scaled consistently. A second response emphasizes consistent increments when differencing delta. These are local approximations whose accuracy depends on shock size and smoothness.
Key ideas
- Gamma is the second derivative of contract value with respect to the underlying price.
- A symmetric Taylor expansion cancels the first-derivative terms in the central difference.
- The two half-weighted second-order terms combine, eliminating the apparent extra factor of one half.
- Use either an absolute or relative price shock consistently in both numerator and denominator.
- The finite-difference expression approximates gamma locally, with higher-order remainder effects for nonzero shocks.
Tags
Full text
# Derivation of the Gamma approximation formula from the Delta approximation formula
# Derivation of the Gamma approximation formula from the Delta approximation formula
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 by ir7 (score 3, accepted)
https://quant.stackexchange.com/a/65506
(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 by nbbo2 (score 2)
https://quant.stackexchange.com/a/65509
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".Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.