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

$$dV(t) = m(V,t)dt+s(V,t)dW(t)?$$


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 $s(V,t)=\frac{B_t \kappa_t}{V_t}$.

  • $\begingroup$ Thank you. So, the question now is how to find an expression for the adaptive process. But the martingale representation theorem is about the existence, and it doesn't provide a way for us to do that, correct? $\endgroup$ – Confounded Oct 23 '18 at 18:46
  • $\begingroup$ That is indeed for the existence. For the specific computation, I think the underlying dynamics will still be used. $\endgroup$ – Gordon Oct 23 '18 at 19:02
  • $\begingroup$ Would we have: $E_Q[\kappa_t|S_t]=E_Q[\kappa_t]?$ $\endgroup$ – Daneel Olivaw Oct 23 '18 at 22:02
  • $\begingroup$ @DaneelOlivaw: It may not be, given that it is adaptable but not necessary independent. $\endgroup$ – Gordon Oct 23 '18 at 22:12

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.

  • $\begingroup$ Thank you for your post. Couldn't we, assuming that the option value function is a bijection (i.e. invertible), write $S(t) = V^{-1}(V,t)$ and substitute this into the SDE, so that where we had functions of $S(t)$, like in $\mu(S,t)$, we now have functions of the inverse expressed in terms of $V(t)$? $\endgroup$ – Confounded Oct 23 '18 at 17:30
  • $\begingroup$ Theoretically speaking, that might be a possibility, but I don't think any of the standard option pricing models yield an evaluation formula which is easily invertible in the stock price (e.g. Black-Scholes formula is not invertible in volatility). Technically, you might be able to do that but at the price of 1) tractability and 2) precision. $\endgroup$ – Daneel Olivaw Oct 23 '18 at 17:51
  • $\begingroup$ Based on @Gordon’s answer it does seem like you could have a tractable formula. $\endgroup$ – Daneel Olivaw Oct 23 '18 at 22:05

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