Greeks: Estimate gamma by Monte Carlo finite difference

Question

When I was using Monte Carlo to calculate the gamma of a vanilla call option by finite difference method, I stuck in this weird situation as below.

Consider this, $$ Gamma = \frac{CallPrice(S^{up}_{T}) - 2 * CallPrice(S_{T}) + CallPrice(S^{down}_{T})}{dS^2} $$ And we can choose dS small enough such that when $$ S_{T}>K \text{ then } S^{down}_{T}>K $$ and $$ S_{T}<K \text{ then } S^{up}_{T}<K $$

That is, we can write the above Gamma formula as $$ Gamma = \frac{(S^{up}_{T}-K)I(S_{T}>K) - 2 * (S_{T}-K)*I(S_{T}>K) + (S^{down}_{T}-K)I(S_{T}>K)}{dS^2} $$ $$ = \frac{(S^{up}_{T} - 2 * S_{T} + S^{down}_{T})I(S_{T}>K)}{dS^2} $$ $$ = \frac{(S_{0}+dS)*exp(...) - 2*S_{0}*exp(...) + (S_{0}-dS)*exp(...)}{dS^2} = 0 $$

So every time I run the simulation, I always get a correct delta but wrong gamma. (Not equal to zero maybe because of rounding error?)

I know gamma is nonzero, but I can't find where I did it wrong. Any help?

Note: This question is a bit similar to this one "Greeks: Why does my Monte Carlo give correct delta but incorrect gamma?", but slightly different.

Answer 1

Pathwise finite difference Gamma formula is indeed:

$$\Gamma(S_0,T, dS; Z) = (dS)^{-2} \left[ (S_T^{up} (Z) - K)^+ -2 (S_T (Z) - K)^+ + (S_T^{dn} (Z) - K)^+ \right], $$

where $Z$ is a standard normal rv, and

$$ S_T (Z) = S_0\eta (Z), $$

$$ S_T^{up} (Z) = (S_0+dS)\eta(Z) = (S_0+dS)S_0^{-1} S_T (Z) $$

$$ S_T^{dn} (Z) = (S_0-dS)\eta(Z) = (S_0-dS)S_0^{-1} S_T (Z), $$

and

$$ \eta(Z) = \exp \left( (r-0.5\sigma^2)T + \sigma \sqrt{T} Z \right). $$

For realizations of $Z$, if $ S_T (Z) > K$, then $ S_T^{up} (Z) > K$. We also have $ S_T^{dn} (Z) > K$, but only if:

$$ dS < (S_T(Z) - K)\eta(Z)^{-1}.$$

If $ S_T (Z) <K$, then $ S_T^{dn} (Z) <K$. And $ S_T^{up} (Z) <K$, if

$$ dS < (K -S_T(Z) )\eta(Z)^{-1}. $$

So, for such $Z$ realization and $dS$, we do have:

$$ I(S_T(Z)-K) = I(S_T^{up}(Z)-K) = I(S_T^{up}(Z)-K). $$

If $ S_T (Z) = K$, then $ S_T^{up} (Z) > K$, and $ S_T^{dn} (Z) < K$ no matter what $dS>0$ we choose. So, in this case:

$$ I(S_T(Z)-K) = I(S_T^{dn}(Z)-K) =0 \not= 1 = I(S_T^{up}(Z)-K). $$

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The link that you provided offers answers and resources on how to deal with Gamma calculations in Monte Carlo framework.

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Note that, with notations above,

$$ \lim_{dS\rightarrow 0}\Gamma(S_0,T, dS; Z) = \delta (S_T(Z)-K) S_T(Z)S_0^{-1}, $$

where $\delta$ is the Dirac delta function. So, computing its expectation via Monte Carlo is bound to give meaningless results. The probability of giving a non-zero result is zero, while in reality we know that:

$$ E[\delta (S_T(Z)-K) S_T(Z)S_0^{-1}] = KS_0^{-1}f(K), $$

where $f$ is the pdf of $S_T(Z)$.