Why conversion shows $\frac{\partial C}{\partial T} > 0$?

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 $$\frac{\partial C}{\partial t}>0$$?

• 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? Dec 27, 2021 at 14:43
• @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. Dec 27, 2021 at 15:58
• 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} Dec 27, 2021 at 16:32
• But, the result is model independent I believe. Shouldn’t have to rely on any model.
– dm63
Dec 27, 2021 at 16:42
• @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. Dec 27, 2021 at 16:46