I read that the Fundamental Theorem of Asset Pricing states, that a market is arbitrage-free if and only if there exists an equivalent martingale measure Q, under which the discounted asset price process becomes a martingale.

Why does the existence of Q matter for arbitragefree property, if we have the physical measure P under which there potentially actually is an arbitrage opportunity?

From my understanding, if the market is not arbitragefree under physical probability measure P, why would it be so just if there exists such theoretical measure Q?

  • 1
    $\begingroup$ There is no physical measure P. The measure Q is not the probability of anything. The FTAP is a geometric result, not a probabilistic. See kalx.net/fms/fms.html for the details. $\endgroup$ Commented Dec 3, 2014 at 13:05

1 Answer 1


In the derivatives context, "arbitrage free" means almost surely for the probability measure under consideration. This is in opposition with statistical arbitrage used at high frequencies for example.

More precisely the assumption is that there is no $T\geq 0$ and self-financed portfolio $V$ such that $V_0 = 0$, $P(V_T < 0) = 0$ and $P(V_T > 0) > 0$. This property remains true if we replace $P$ by any equivalent measure $Q$. Simply apply $P(A) = 0 \Leftrightarrow Q(A) = 0$ to $A = \{V_T < 0\}$ and $A = \{V_T > 0\}$.

If you have an equivalent martingale measure $Q$, then you can't have arbitrage. Let's prove this. Let $V$ be a self-financed portfolio such that $V_T \geq 0$ $Q$-almost surely and $Q(V_T > 0) > 0$. Write $\widetilde{V}_t = e^{-\int_0^t r_sds} V_t$ then we also have $\widetilde{V}_T \geq 0$ $Q$-almost surely and $Q(\widetilde{V}_T > 0) > 0$ (these properties are invariant under change of numeraire). Since the discounted value $\widetilde{V}$ is a $Q$-martingale, $V_0 = \widetilde{V}_0 = E^Q[\widetilde{V}_T] > 0$. This proves that you can't have an arbitrage strategy under $Q$ so you can't have one under $P$ either. This proves the easy half of the fundamental theorem of asset pricing.

Now you might and should wonder why bother with finding a martingale measure $Q$? The answer is that it is very useful because by definition discounted assets (and all self-financed portfolios) are martingales. So computations of derivatives prices are much easier since the expectation of a martingale at $T$ is equal to its initial value (observable at least in theory on the market). The difference with the real world (the $P$-world) is that you don't need to estimate the drift (return) of the underlying assets (predicting if a stock is going to go up or down on a given day is a very difficult thing to do for deep statistical reasons). If you set jumps aside, the value of an option will only depend on the volatility of the underlying. The difference between the real returns are already priced in the market risk premium $\lambda$ used to define the change from the historial to the risk neutral measure (cf. Girsanov's theorem: $\frac{dQ}{dP}|_{\mathcal{F}_t} = \exp( -\int_0^t \lambda_s dW_s - \frac{1}{2}\int_0^t |\lambda_s|^2 ds)$).

  • $\begingroup$ Where did that last formula come from? $\endgroup$
    – Trajan
    Commented May 15, 2021 at 20:55

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.