Why an Equivalent Martingale Measure Rules Out Arbitrage
Summary
The document explains one direction of the fundamental no-arbitrage result: if discounted asset prices admit an equivalent martingale measure, an arbitrage strategy cannot exist. Under that measure, the discounted value of a self-financing strategy is a martingale, so its expected terminal value equals its initial value. A zero-cost strategy therefore has zero expected discounted payoff.
An arbitrage payoff is nonnegative in every state and positive with positive probability. Equivalence ensures those positive-probability states remain possible under the martingale measure. Such a payoff must have strictly positive expectation, contradicting the zero expectation. The accompanying answer illustrates this with a nonnegative random variable: a zero mean requires it to be zero almost surely. The document presents only this implication, not the reverse direction of the theorem, and its density-based explanation is informal for distributions that do not have ordinary probability densities.
Key ideas
- Under an equivalent martingale measure, discounted values of self-financing strategies are martingales.
- A zero-cost strategy must have zero expected discounted terminal value under that measure.
- An arbitrage payoff that is nonnegative and sometimes positive has strictly positive expectation.
- The contradiction rules out arbitrage when an equivalent martingale measure exists.
- A zero-mean nonnegative random variable must equal zero almost surely.
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Full text
# No-arbitrage theorem: a proof
# No-arbitrage theorem: a proof
The market is arbitrage free iff there exists an equivalent martingale measure for the discounted price process of the stock.
My course only provides me part of the entire proof that shows that whenever such martingale exists, it implies the market must be arbitrage free.
Proof:
Assume $P^*$ is an equivalent martingale for the discounted stock price process $S'$. For any self-financing strategy $\phi$, we have that $V'^\phi$ is a $P^*$ martingale. ($V'$ here is the discounted value of such strategy). So: $$ E_{P^*}[V'^\phi_T] = V^\phi_0$$
Suppose now that $\phi$ is an arbitrage oppurtunity. Then $P(V^\phi_0=0) = 0$ (the initial cost of the strategy is zero), so also $P^*(V^\phi_0=0) = 0$ and thus $ E_{P^*}[V'^\phi_T] = B^{-1}_TE_{P^*}[V^\phi_T] = B^{-1}_TV^\phi_0$.
We must have $$ P^*(V^\phi_T\geq0) = 1 , P^*(V^\phi_T\gt0) > 0 $$
Together with the fact that for each possible outcome $\omega$, we have $P^*({\omega}) \gt 0$, this leads to a contradiction.
Now as explanation for the contradiction, it says that we can not have a non negative random variable with a zero mean and positive mass on the positive real numbers. Which I don't understand. Could someone shed some light on this explanation?
## Answer by LocalVolatility (score 4, accepted)
https://quant.stackexchange.com/a/32017
Consider a random variable $X$ that has a probability density function (PDF) $f(x)$. $X$ being non-negative means that $f(x) = 0$ for $x < 0$. The expectation of $X$ is thus
\begin{equation} \int_{-\infty}^\infty x f(x) \mathrm{d}x = \int_0^\infty x f(x) \mathrm{d}x. \end{equation}
Since $f(x) \geq 0$ for it to be a valid PDF, it follows that
\begin{equation} \int_0^\infty x f(x) \mathrm{d}x = 0 \qquad \Leftrightarrow \qquad f(x) = \delta(x), \end{equation}
where $\delta(x)$ is the Dirac delta function. I.e. for $X$ to have a zero-mean, its PDF needs to be a point mass as zero.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
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