Path-dependent options valuation

Question

Assume that we have an arbitrage-free and complete market. The well known formula for the arbitrage-free price of an attainable derivative $X$ at time $0 \leq t \leq T$ is given by: \begin{align*} V(t)=e^{-r(T-t)}E_Q(X \vert {\cal F}_t) \end{align*} Where $r$ is the risk-free interest rate and $E_Q$ is the expected value under the risk-neutral measure.

From my understanding, given a path-independent derivative with payoff $\Psi(S(T))$ for some function $\Psi$, we can calculate the arbitrage-free price by evaluating $E_Q(\Psi(S(T))\vert {\cal F}_t)$ right ? For instance, for the European call we have $\Psi(S(T))=\max(S(T)-K;0)$, where $S(T)$ is the stock price at time $T$ and $K$ is the strike price.

Now I wonder if this formula holds for path-dependent options too. For instance, if I want to calculate the arbitrage-free price of a fixed strike Asian call with payoff $\max(A(S)-K;0)$, where $A(S)$ is some sort of average, can I calculate \begin{align*} V(t)=e^{-r(T-t)}E_Q(\max(A(S)-K;0)\vert {\cal F}_t) \end{align*} ?

Answer 1

Risk-neutral pricing

A time-$T$ payoff is an integrable, $\mathcal{F}_T$-measurable random variable $\xi$. The value process of the discounted payoff is then a $\mathbb{Q}$-martingale, i.e., \begin{align*} V_t=\mathbb{E}^\mathbb{Q}_t\left[\frac{B_t}{B_T}\xi\right], \end{align*} where $B_t$ is a locally risk-free bank account ($\text{d}B_t=r_tB_t\text{d}t$).

• The above result essentially follows from the definition of $\mathbb{Q}$ and the fact that $M_t=\mathbb{E}_t[X]$ is a martingale if $X$ is integrable (due to the tower law).

• If $r_t\equiv r$ is constant, we have $B_t=e^{rt}$ and $V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\xi\right]$.

Does it work? Yes!

The only requirement is that $\xi$ is known (observable, measurable) at maturity $T$. There is no requirement that $\xi$ needs to be path-independent. Thus, $\xi$ can indeed be an average and standard risk-neutral pricing applies to (European-style) Asian options, i.e., $\xi=\max\{A-K,0\}$ is allowed! It makes no difference whether you consider arithmetic or geometric averages here, or whether you use averages as strike prices. Risk-neutral pricing also applies to other path-dependent exotic options such as (European-style) barrier options.

Indeed, semi-closed-form numerical methods for Asian options rely on explicitly on this risk-neutral pricing framework.

We also get some simple results: The identity $\max\{x-K,0\}-\max\{K-x,0\}=x-K$ gives a put-call parity for Asian options.

Where's the problem?

The only problem is that computing the first moment of the option payoff is darn difficult. Most often, we're interested in arithmetic Asian options but we tend to model stock prices in an exponential form. That makes closed-form solutions very rare. Essentially, the distribution of the average $\int_t^T S_u\text{d}u$ is not really known for sensible stock price models. For geometric averages, the situation is a bit better.

American options

The risk-neutral pricing formula does not apply to early exercise features (e.g., American put options). Their prices relate to the Snell envelope, which is a supermartingale, see this answer. Their prices can thus be decomposed into a European option (a martingale) and an early exercise correction term (Riesz decomposition or Doob-Meyer decomposition). The maths for these early exercise features is more difficult. Obviously, pricing American-style Asian options is a really difficult task (I'd opt for MC simulations)...