Why Longer Maturity Calls Usually Have Higher Value
Summary
The note explains why call value can increase with maturity, clarifying a reference to “conversion” in a derivation of the Dupire equation. The answer interprets the term as early exercise into stock: an American option with a later expiration can be held until an earlier expiration and exercised then, so its value cannot be lower than that of the shorter-dated American option.
For calls without dividends, American and European values coincide, so the maturity monotonicity also applies to European calls. The caveat is that dividends can break this equivalence: a longer-dated European call may be worth less, and the maturity derivative need not be positive. The note offers this as an interpretation of the cited argument, rather than a full derivation or general proof for all option types and market conditions.
Key ideas
- Early exercise lets a longer-dated American option replicate the choice to exercise at an earlier date.
- This flexibility means a longer-maturity American option cannot be worth less than a comparable shorter-maturity one.
- Without dividends, American and European calls have equal value, extending the maturity comparison to European calls.
- With dividends, European call value need not rise with maturity.
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# Why conversion shows $\frac{\partial C}{\partial T} > 0$?
# 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 $\frac{\partial C}{\partial t}>0$?
## Answer by Jon Ingersoll (score 2)
https://quant.stackexchange.com/a/75876
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.