The Fundamental Theorem of Asset Pricing (FTAP) is invoked when we say the time $0$ price of a European option with payoff $g$ is $e^{-rT}E_Q(g(S_T))$, with the hypothesis that $e^{-rt}S_t$ is a $Q$-martingale. This martingale condition is satisfied if we assume $S_t$ follows a geometric Brownian motion with drift $r$, so we can use Monte Carlo to estimate the price as $$ e^{-rT}\frac{1}{N}\sum_{i=1}^N g(S_T^i) $$ where $\{S_T^i\}$ are lognormal.

Fair enough, but what permits one to do this same Monte Carlo estimation assuming $S_t$ follows the dynamics of the Heston model? In particular, under what conditions is $e^{-rt}S_t$ a martingale for some measure? We don't know the distribution of $S_t$ under this model, so it's not as straightforward as Black-Scholes. Certainly we can simulate the paths of this model and compute the sum above, which is what it seems people do, but what justifies this theoretically?

  • $\begingroup$ Why distribution $S_t$ is important for you? $\endgroup$ – user16651 Jul 21 '16 at 12:15
  • $\begingroup$ @BehrouzMaleki I actually just need to be able to compute the conditional expectations for the martingale property. But knowing the distribution makes everything easier, so I'm just saying it's not as straightforward as the lognormal case. $\endgroup$ – bcf Jul 21 '16 at 12:19
  • $\begingroup$ @bcf You don't need to know the distribution to check the martingale condition. Ito's lemma and martingale representation theorem is enough. Basically as long as, under Q, your diffusion model writes: $dS_t/S_t = rdt + \dits dW_t^Q$, $S_t/B_t$ is a Q-martingale. $\endgroup$ – Quantuple Jul 21 '16 at 17:45

You don't need any assumption about the distributional properties of $S_t$. What matters for the FTAP is the drift only.

By definition, the risk neutral measure $Q$ is the measure, equivalent to the natural measure $P$ (*), under which the local rate of return (i.e. the instanteneous drift of the SDE of $S_t$ per unit of $S_t$) of "any" traded asset $S_t$ (but also the price of the derivative $g_t$) is $r$, the market risk-free rate.

Equivalently, this consists in requiring that the price of any traded asset measured in unit of the fundamental reference unit of the value of the bank account (aka numeraire) $B_t=e^{rt }$ ($B_0=1$) is a martingale: $$ E^Q_0 \left[\frac{g_T}{B_T}\right] = \frac{g_0}{B_0} $$ that is the price is, as you mentioned $$ g_0=e^{-rT} E^Q_0 \left[g_T\right] $$ and this holds for any traded asset, disregarding its nature (stock, commodity, index, derivative...).

This truly is an equivalence. Indeed, being $\mu$ the local rate of return of $g_t$ under $Q$: $$ dg_t= \mu g_t dt + (\cdots) dW $$ you have for $m_t=g_t/B_t$, by Ito's lemma: $$ d m_t = [-r m_t + (\mu g_t) (1/B_t)] dt + (\cdots) dW $$ which is a martingale (no drift condition) provided that $$\mu = r$$.

As you can see, the diffusion term plays no role in this context. Being the asset dynamics normal, lognormal or with stochastic volatility doesn't matter (**)

(*) the one in which you "live", that is the one under which you require a risk premium, on the top of the risk free return $r$, for taking the risk of holding the risky asset $S_t$.

(**) for your future reference: the volatility doesn't change under change of measure (take a look at Chap 2 of Brigo-Mercurio on Interest rates derivatives). It's an advanced concept, maybe better start from the basis: Bjork "Arbitrage theory in continuous time" Chap 10.

  • $\begingroup$ +1 for I agree with the gist. However, there's a little caveat. It is not true that any traded asset, when expressed in the numéraire $B_t$ should emerge as a Q-martingale. Rather, any self-financing portfolio should. This is important since investing in a dividend paying stock is not a self-financing strategy....thus the drift under Q of such a dividend paying stock will not be $r$. $\endgroup$ – Quantuple Jul 22 '16 at 16:18

You can use this article

Let $$\begin{align} & d{{S}_{t}}=r{{S}_{t}}dt+\sqrt{{{\nu }_{t}}}\left( \rho dW_{1}^{Q}(t)+\sqrt{1-{{\rho }^{2}}}dW_{2}^{Q}(t) \right) \\ & d{{v}_{t}}=\kappa (\theta -{{v}_{t}}){{d}{t}}+{{\sigma }_{v}}\sqrt{{{\nu }_{t}}}dW_{1}^{Q}(t) \\ \end{align}\ $$

we can show

$${{S}_{T}}={{S}_{t}}\exp(X_t)\tag 1$$ where $$X_t= \left( r\tau-\frac{1}{2}\int_{t}^{T}{{{v}_{s}}}ds+\rho \int_{t}^{T}{\sqrt{{{v}_{s}}}}dW_{1}^{Q}(s)+\sqrt{1-{{\rho }^{2}}}\int_{t}^{T}{\sqrt{{{v}_{s}}}}dW_{2}^{Q}(s) \right)\tag 2$$ and $$v_t=v_s+\kappa\theta(t-s)-\kappa\int_{s}^{t}{{{v}_{u}}}du+\sigma_v\int_{s}^{t}\sqrt{v_u}dW_1^{Q}(u)$$ As you know, the conditional on a realized value of $v_s$, the random variable $2c_t v_t$ follows a non-central chi-square distribution, where $$c_t=\frac{2\kappa}{\sigma_v^2(1-e^{-\kappa(t-s)})}\tag 3$$ therefore you can do this procedure :

  • Generate a sample from the distribution of $v_t$ given $v_s$.
  • Generate a sample from the distribution of $\int_{s}^{t}{{{v}_{u}}}du$ given $v_t$ and $v_s$.
  • Recover $\int_{t}^{T}{\sqrt{{{v}_{s}}}}dW_{1}^{Q}(s)$ from $(1),(2)$ given $v_t$, $v_s$ and $\int_{t}^{T}{{{v}_{s}}}ds$.
  • Generate a sample from the distribution of $S_t$ given $\int_{t}^{T}{\sqrt{{{v}_{s}}}}dW_{1}^{Q}(s)$ and $\int_{t}^{T}{{{v}_{s}}}ds$.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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