I'm reading Dupire's "Pricing and Hedging with smiles" (1993). After arriving $$\frac12 b^2 \frac{\partial^2 C}{\partial x^2}=\frac{\partial C}{\partial t}$$

(note: here $C$ is the value of a call option, $t$ refers to its maturity, while $x$ refers to its strike)

it says

Both derivatives are positive by arbitrage (butterfly for the convexity and conversion for the maturity).

Sure, a butterfly option's positive value means $\frac{\partial^2 C}{\partial x^2} > 0$, but I'm a bit confused here on the conversion part.

If I'm not wrong, a conversion, is to long the underlying stock and offset it with an equivalent synthetic short stock (long put + short call) position.

How is a conversion related to $$\frac{\partial C}{\partial t}>0$$?

  • $\begingroup$ I agree that the term 'conversion' can cause some confusion. I'd ignore that, and assume that you see/understand why $\frac{\partial C}{\partial T} > 0$? Or are you unsure how to derive the inequality? $\endgroup$ Dec 27, 2021 at 14:43
  • $\begingroup$ @FridoRolloos I'm fine with $\frac{\partial C}{\partial t}>0$, as the model is $dx = a(x,t)dt + b(x,t) dW$, and in risk-neutral measure $a=0$ so $dx = b(x,t) dW^Q$ so $X(T) \sim N(0,\sigma^2(T))$ where $\sigma^2(T) = \mathbb{E} X^2(T) = \mathbb{E} (\int_0^T b(x,t) dW^Q(t))^2 = \mathbb{E} \int_0^T b^2(x,t) dt$, so $\sigma(T)$ is monotonic increasing. $\endgroup$
    – athos
    Dec 27, 2021 at 15:58
  • 1
    $\begingroup$ Ok, another way to see this is \begin{align} E_t \left(S_{T + \Delta T} - K \right)_+ &= E_t \left[ E_T \left(S_{T + \Delta T} - K \right)_+ \right] \\ &\geq E_t \left(E_T(S_{T + \Delta T}) - K \right)_+ \\ &= E_t \left(S_T - K \right)_+ \end{align} $\endgroup$ Dec 27, 2021 at 16:32
  • $\begingroup$ But, the result is model independent I believe. Shouldn’t have to rely on any model. $\endgroup$
    – dm63
    Dec 27, 2021 at 16:42
  • $\begingroup$ @dm63 Yes, my understanding is also that the result should be model independent. I think the OPs derivation relies on local vol model. The alternative derivation I gave does not. $\endgroup$ Dec 27, 2021 at 16:46


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