Why the Drift Term Vanishes in an Option Pricing PIDE
Summary
The document asks why a drift term obtained after applying Itô's formula to a discounted European option price must vanish in a pricing partial integro-differential equation. The setup models the asset with a Lévy process, so the discounted option value has a finite-variation drift component and a martingale component that includes both Brownian and jump terms. The question focuses on the reasoning that a continuous finite-variation process which is also a martingale must be constant, and on whether the jump integral meets the required continuity and variation conditions.
This is a conceptual question rather than a complete solution: it presents the drift expression and the martingale decomposition but supplies no answer, proof, or worked example. Applying the argument requires suitable integrability and regularity assumptions for Itô's formula and the jump integral. The material therefore highlights a key step in deriving option pricing PIDEs while leaving the precise technical conditions unresolved.
Key ideas
- Itô's formula decomposes a discounted option price into a drift term and a local martingale term.
- If the discounted price is a true martingale, its finite-variation drift component must satisfy additional conditions.
- A continuous finite-variation martingale is constant under standard assumptions.
- The document raises, but does not resolve, the regularity and integrability conditions on the jump integral.
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# Derivation of option pricing PIDE: Why does the drift need to be zero?
# Derivation of option pricing PIDE: Why does the drift need to be zero?
I started studying PIDE methods for option pricing and am struggling to understand or find the necessary theory that shows why the PIDE is obtained by the condition that the drift term has to be zero.
I read this paper where asset models of the form $S_t = s_0 e^{rt + X_t}$ are considered, with $X$ being a Levy process with generating triplet $(\sigma^2,\gamma,\nu)$ and jump measure $J_X$. If $C(t,S_t)$ is the price function of a European option, one can apply Ito's formula to $\hat{C}_t=e^{r(T-t)}C(t,S_t)$ to obtain dynamics of the form $$ d\hat{C}_t = a(t)dt + dM_t $$ where $M_t$ is a martingale consisting of a $dW_t$ and a $\tilde{J_X}(dt\ dx)$ part. As the discounted price process $\hat{C}_t$ is a martingale (by arbitrage theory), $\hat{C}_t - M_t$ is also a martingale. Up to here, everything is clear to me.
But then, the last part is, concluding that $a(t)$ has to be zero since $\hat{C}_t - M_t = \int_0^t a(s)ds$ is also continuous process with finite variation, justified with a reference to almost a whole section of this book, where I don't really see the connection to this reasoning.
So my questions are:
- Why is $\hat{C}_t - M_t = \int_0^t a(s)ds$ a continuous process with finite variation? The process is given by $$ a(t) = e^{r(T-t)} \left[ -rC(t,S_{t-}) + \partial_1 C(t,S_{t-}) + \frac{\sigma^2 S_{t-}^2}{2}\partial_2^2C(t,S_{t-}) + rS_{t-}\partial_2C(t,S_{t-} \\ + \int_\mathbb{R} C(t,S_{t-}e^x) - C(t,S_{t-}) - S_{t-}(e^x-1)\partial_2 C(t,S_{t-})\nu(dx) \right] $$ I don't see immediately that the integral over this should be continuous and of finite variation.
- Now given that this is true, why is a martingale that is continuous and of finite variation equal to zero?
I would be happy for any kind of resources that give a clearer proof of this result than just referencing to a whole range of theorems and propositions in a bookShown 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.