Why an Option Value SDE Usually Depends on the Underlying
Summary
The document asks whether an option’s value can be simulated directly from an SDE whose drift and volatility depend only on the option value and time. Applying Ito’s lemma to an option value expressed as a function of the underlying and time produces an SDE whose coefficients involve the underlying. The response emphasizes that an option price ordinarily depends on the underlying, so its evolution cannot generally be determined without that state information.
A second answer gives a risk-neutral representation: discounted option value is a martingale, and the martingale representation theorem expresses it as an integral against Brownian motion. This yields an SDE with the short rate as drift and a diffusion coefficient expressed through the option’s martingale representation. That coefficient is adapted to the underlying information, however, and need not be a function of option value and time alone. The formulation therefore does not remove the underlying dependence in general; the document provides no conditions for a closed one-dimensional option-value process.
Key ideas
- Ito’s lemma yields option-value dynamics with coefficients that generally depend on the underlying.
- Discounted option value is a martingale under the stated risk-neutral setup.
- Martingale representation provides a Brownian diffusion term for the option value.
- That diffusion coefficient need not depend only on option value and time.
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Full text
# SDE for option value
# SDE for option value
Given an SDE for an underlying:
$$dS(t) = \mu(S,t)dt+\sigma(S,t)dW(t)$$
the SDE for the value of the option $V=V(S,t)$ is given via Ito's lemma as:
$$dV = V_tdt+V_S\mu(S,t)dt+\frac{1}{2}V_{SS}\sigma^2(S,t)dt+V_S\sigma(S,t)dW(t)$$
It seems that this would results in an SDE containing $S(t)$.
How does one then obtain an SDE for the option value so that it can be simulated directly without simulating the underlyings, i.e. something like
$$dV(t) = m(V,t)dt+s(V,t)dW(t)?$$
## Answer by Gordon (score 4, accepted)
https://quant.stackexchange.com/a/42346
This has been indirectly discussed in this question. We assume that $\{\mathcal{F}_t, \, t\ge 0\}$ is the natural filtration generated by the Brownian motion $\{W_t,\, t \ge 0\}$. Moreover, let $B_t= e^{\int_0^t r_sds}$ be the money market account value at time $t\ge0$. Let $V_T$ be the option payoff at maturity $T$. Then \begin{align*} V_t = B_tE_Q\left(\frac{V_T}{B_T}\mid \mathcal{F}_t \right), \end{align*} where $E_Q$ is the expectation operator under the risk-neutral probability measure $Q$. Note that, $\{V_t/B_t, t \ge 0\}$ is a martingale. Therefore, by the martingale representation theorem, there exists an adapted process $\{\kappa_t, 0\le t \le T\}$ such that \begin{align*} \frac{V_t}{B_t} &= V_0 + \int_0^t \kappa_u dW_u. \end{align*} Then \begin{align*} dV_t &= d\left(B_t \frac{V_t}{B_t} \right)\\ &=r_tV_t dt + B_t \kappa_t dW_t\\ &=V_t\Big(r_t dt + s(V,t) dW_t\Big). \end{align*} where $s(V,t)=\frac{B_t \kappa_t}{V_t}$.
## Answer by Daneel Olivaw (score 2)
https://quant.stackexchange.com/a/42343
That's impossible. The value of your option is a function of time and the value of the underlying: $$ V(t) \triangleq V(t,S_t) $$ How can you possibly know the value of the option at time $t$ without knowing the value of its underlying? Equivalently, you need to know how your underlying has evolved in order to know how the value of your option has evolved. You will always have a dependency between the value of an option and its evolution on the one hand, and the value of the underlying on the other hand.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.