Taking the Zero-Price Limit of Black–Scholes Gamma

Summary

The document considers the Black–Scholes–Merton gamma formula as the underlying stock price approaches zero. Because the formula includes a factor of one over the stock price and an exponential involving d1, it focuses on the limit of that exponential divided by the price. Direct use of L’Hôpital’s rule appears awkward: differentiating the exponential introduces another inverse-price factor through the derivative of d1.

The proposed method is to solve the expression for the stock price in terms of d1, substitute that into the limit, and then take the limit as d1 tends to negative infinity. The questioner states that gamma should tend to zero, but the excerpt provides no worked derivation or proof of the result. It also raises the analogous limit as the stock price tends to infinity without supplying its evaluation. The note is therefore a useful change-of-variable idea for an asymptotic calculation, but readers must carry out the algebra and check assumptions such as fixed maturity and volatility themselves.

Key ideas

• Black–Scholes gamma contains an exponential term divided by the underlying price.
• Differentiating directly with respect to price makes the limit calculation cumbersome.
• Rewriting price as a function of d1 converts the limit to one in d1.
• The excerpt suggests the method but does not show a complete derivation.

Tags

# Limit of BSM Gamma as stock price goes to 0


# Limit of BSM Gamma as stock price goes to 0












BSM gives the following formula for option gamma $$ \Gamma = \frac{e^{-qT-\frac{d_1^2}{2}}}{S\sigma\sqrt{2\pi T}} $$ where $$ d_1=\frac{\ln\frac{S}{K}+(r-q+\frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}} $$ Then, to calculate its limit as $S\rightarrow 0^+$, the key is to calculate $$ \lim_{S\rightarrow 0^+}\frac{e^{-\frac{d_1^2}{2}}}{S} $$ Since $d_1\rightarrow-\infty$ as $S\rightarrow 0^+$, I'm thinking of using L'Hopital's rule. However, everytime one takes a partial derivatve of $e^{-\frac{d_1^2}{2}}$ with respect to $S$, there will be another $S^{-1}$ showing up since $$ \frac{\partial d_1}{\partial S}=\frac{1}{S\sigma\sqrt{T}} $$ Hence, does anyone know how to calculate the limit (I know it should be 0)? A similar problem is how to calculate following limit? $$ \lim_{S\rightarrow +\infty}\frac{e^{-\frac{d_1^2}{2}}}{\frac{1}{S}} $$

## Answer by dm63 (score 1, accepted)

https://quant.stackexchange.com/a/73762

How about expressing $S$ in terms of $d_1$, then plugging this formula into the denominator of your limit, then take limit as $d_1$ tends to negative infinity ( instead of taking limit on S).

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