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.


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?


1 Answer 1


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.


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.