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 $

Question

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 $\phi(T,s)$.

score 2 · Accepted Answer · answered Oct 05 '18 at 12:59

2

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.

answered Oct 05 '18 at 12:59

Magic is in the chain

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I agree that the transition density goes to zero when the underlying goes to infinity, but we need to know the derivative $ \frac{d}{ds} \left[ \sigma(T,s)^2s^2\phi(T,s)\right] $ before taking the limit right? – user30614 Oct 05 '18 at 13:18
Generally yes as exchanging derivative and limit can lead to different result, but in this case as all are smooth functions so should not be a problem. Note the density would appear in all components if you apply the differentiation and it will take over. – Magic is in the chain Oct 05 '18 at 14:32

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 $

1 Answers1