Boundary Decay Assumption in Dupire’s Local Volatility Derivation

Summary

The document asks why an integration-by-parts boundary term vanishes when deriving Dupire’s local volatility formula from European call prices. The proposed explanation is that the transition density of the terminal stock price tends to zero as the price grows without bound, and must decay quickly enough that multiplying by the squared price still tends to zero.

This is a tail condition on the density, not a result established for every possible model. The answer gives no proof or specific model conditions guaranteeing that rate of decay, and it notes a possible typo in the derivative in the original question. Researchers applying the derivation should verify the required tail behavior under their chosen assumptions.

Key ideas

• The boundary term arises in an integration-by-parts step in the derivation of Dupire’s formula.
• The explanation relies on the terminal transition density vanishing at high prices.
• The density must decay faster than the squared-price factor grows for the limit to vanish.
• The document does not establish this tail condition for specific asset models.

Tags

# Answer by Magic is in the chain (score 2, accepted)


# Deriving Dupire's Volatility Formula : Why $\lim_{s \rightarrow \infty} (s-K) \frac{d}{ds} \big[ \sigma^2(T,s)s^2\phi(T,s)\big] = 0 $












In deriving Dupire's formula for the local volatility, using European call option, this is used in the integration by part :

$$\lim_{s \rightarrow \infty} (s-K) \frac{d}{ds} \Big[ \sigma^2(T,s) s^2\phi(T,s)\Big] = 0 $$

Why is it the case?

Notation :

$s$ : value of the final stock price $S_T$

$T$: expiry of the call option

$K$: strike of the call option

$\phi(T,s ;t_0,s_0)$: transition density, or probability of going from state $(t_0,s_0)$ to state $(T,s)$. $t_0$ and $s_0$ assumed to be known constant, so it is noted $\phi(T,s)$.

## Answer by Magic is in the chain (score 2, accepted)

https://quant.stackexchange.com/a/42061

As spot goes to infinity, the transition density goes to zero, and hence the result. Underlying assumption being that it goes to zero faster than quadratic($s^2$).

Ps: there seems to be a typo in your derivative but does not matter for the purpose here.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.