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

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

(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?
– user34971
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}
– user34971
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.
– user34971
Dec 27, 2021 at 16:46

## 1 Answer

I'm pretty sure that "by conversion" means "by exercise", that is converting it to stock. Only American options cane be exercised (converted) before maturity and it is well known that a longer maturity American option cannot be less valuable than a shorter maturity option because you can always make a longer maturity American option into a shorter maturity one by throwing it away on the earlier maturity date.

Now if there are no dividends, American and European calls are worth the same so longer maturity European calls are also worth more. However, if there are dividends, then longer maturity European calls are not necessarily more valuable and that partial derivative is not necessarily positive for them.