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 $$\phi(T,s)$$.

1 Answer

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.

• 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 5 '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 5 '18 at 14:32