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*} ?


1 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)...

  • 1
    $\begingroup$ Very nice and concise answer! Since we average (integrate) expected (log) asset levels over time, do you second the idea that Asian options are only ‚mildly‘ exotic as we can use observed European option prices to calibrate a pricing model? $\endgroup$ Commented Aug 19, 2021 at 3:10
  • 1
    $\begingroup$ Kevin, thank you very much for your answer. $\endgroup$
    – Count
    Commented Aug 19, 2021 at 7:28
  • $\begingroup$ @Kermittfrog Great comment! You’re saying that barrier/American/Asian options are all mildly exotic, because I can observe European vanilla prices for one maturity and use those market prices to infer the parameters of a model and then price the (mildly) exotic options in accordance with these parameters.. As I understand it, this is what one commonly does with exotic options. It’s our only possibility (as only vanilla prices are quoted) but I feel it’s not perfect (see next comment) $\endgroup$
    – Kevin
    Commented Aug 19, 2021 at 10:45
  • 1
    $\begingroup$ @Kevin yes, you are right. As we are not interested in the expectation of $\sum_t S_t$ (or its continuous time equivalent), but some $(\cdot)^+$ of it, we introduce the dependence on the joint distribution, hence we require more than ‘just’ European options for calibration. Thank you very much for the discussion and the example. $\endgroup$ Commented Aug 19, 2021 at 20:40
  • 1
    $\begingroup$ @KapesMate It's a very good question but way too much for a comment here. You could ask your own question. There are several ways of how you can hedge exotics using vanilla options. I guess that is what people may do in practice: price it using a calibrated model and hedge all the Greeks using (tradable) vanillas $\endgroup$
    – Kevin
    Commented Dec 17, 2023 at 18:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.