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I read that the Fundamental Theorem of Asset Pricing states, that a market is arbitrage-free if there exists a riskneutral equivalent martingale measure Q~P, 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?

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In a complete market the absence of arbitrage implies the existence of an equivalent probability measure under which asset prices are martingales. The fair value of a derivative security that can be replicated with a dynamic hedging strategy can then be determined as an expected value under this measure.

Proving this in a continuous-time stochastic framework is complicated.

To get a sense of how this works, consider a simple discrete model to illustrate how a risk-neutral probability measure arises. Assume there are only two possible states of the market in a single time period -- with "real-world" probabilities $p$ and $1-p$, respectively. We consider the pricing of a call option on an asset with random price $S$.

At time $t=0$, the asset price is $S_0$ and the call option price is $C_0$, to be determined. At time $t=1$, there are two possible asset prices $S_{1u} = S_0(1+u)$ with probability $p$ and $S_{1d} = S_0(1+d)$ with probability $1-p$. The payoff of the option at expiration is

$$C_1 = \max(S_1-K,0), $$

where the random variable $S_1$ is $S_{1u}$ or $S_{1d}$ depending on the future state of the market:

$$C_{1u} = \max(S_{1u}-K,0)\\\ C_{1d} = \max(S_{1d}-K,0)$$

The value of the option at time $t=0$ is the discounted expected value of the payoff. However this expected value is not calculated with real probability $p$ -- but rather the risk-neutral probability $\hat{p}$ that preclude arbitrage opportunities.

We can determine the no-arbitrage risk-neutral probability $\hat{p}$ by showing it is possible to construct a hedged portfolio of the option and the asset that is risk free -- it has the same value in both future states. Hence the value of the portfolio grows in time at the risk-free rate of interest $i$.

Suppose the portfolio is long $1$ call option and short $\Delta$ shares of the asset. The value at time $t=1$ is

$$V_t= C_t-\Delta S_t.$$ We can solve for the hedge ratio $\Delta$ so that the value of the portfolio at time $t=1$ is independent of the state of the market:

$$C_{1u}-\Delta S_{1u}=C_{1d}-\Delta S_{1d},$$

or

$$C_1-\Delta S_1=C_{1u}-\Delta S_0(1+u)=C_{1d}-\Delta S_0(1+d).$$

This value of the hedge ratio is independent of the probabilities for the value of the asset at time $t=1$:

$$\Delta = \frac{C_{1u}-C_{1d}}{S_0(u-d)}$$

Consequently, in the absence of arbitrage, the portfolio grows at the risk-free rate:

$$C_1-\Delta S_1 = (C_0 - \Delta S_0)(1+i)$$

and we can solve for the value of the call option at time $t=0:$

$$C_0 -\Delta S_0= \frac{C_{1u}-\Delta S_0 (1+u)}{1+i},$$

$$C_0 = \frac{\Delta S_0 (1+i) +C_{1u}-\Delta S_0 (1+u)}{1+i},$$

$$C_0 = \frac{-\Delta S_0 (u-i) + C_{1u}}{1+i}.$$

Substituting for $\Delta$, we obtain

$$C_0 = \frac{\hat{p}C_{1u}+(1-\hat{p})C_{1d}}{1+i}=\frac{1}{1+i}\{\hat{p}\max[S_0(1+u)-K,0)]+(1-\hat{p})\max[S_0(1+d)-K,0)]\},$$

where

$$\hat{p} = \frac{i-d}{u-d}.$$

This has the form of an expected value with a different probability -- the risk-neutral probability. The fair value of the call option is the discounted expected value under the risk-neutral probability measure.

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  • $\begingroup$ Thanks for your answer. I understand now, that Q is to be interpreted as condition for the existence of a riskfree replicating portfolio. $\endgroup$
    – emcor
    Commented Jun 25, 2014 at 0:08
  • $\begingroup$ You're welcome. $\endgroup$
    – RRL
    Commented Jun 25, 2014 at 0:10

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