Here is the question:

Let $(Ω,\mathscr F,\mathbb P,\mathbb F= (\mathscr F_k)_{k=0,...,T})$ be a filtered probability space and $S=(S_k)_{k=0,...,T}$ a discounted price process. Show that the following are equivalent:

a) $S$ satisfies Non-arbitrage.

b) For each $k = 0, . . . , T − 1$, the one-period market $(S_k, S_{k+1})$ on $(Ω, \mathscr F_{k+1}, \mathbb P, (\mathscr F_k, \mathscr F_{k+1}))$ satisfies Non-arbitrage.

I know how to prove from a) to b). But proving from b) to a) seems quite difficult. So if anyone can help, please share your idea here. Thanks!

  • 1
    $\begingroup$ If you explain what non-arbitrage is, maybe some mathematicians that are not economists will be able to help. $\endgroup$ – Erel Segal-Halevi Oct 25 '13 at 11:32
  • $\begingroup$ prices also discrete ? you could string together you one period martingale measures in the simplest possible way, one which makes the stock price markov. I would also not be afraid to try to prove that overall arbitrage implies one period losses somewhere. After all, if you must take a loss at each step (with some prob) how can you avoid a loss overall ? $\endgroup$ – mike Oct 25 '13 at 13:15

No-arbitrage is equivalent to existence of state prices or stochastic discount factor. The proof usually is based on separating hyperplane theorem (see Duffie's Dynamic Asset Pricing Theory for example) but the conclusion is that prices do not admit arbitrage from date $t_1$ to date $t_2$ if and only if there exist state prices $\pi_{t_1,t_2}$ such that $S_{t_1}=E^{t_1}[\pi_{t_1,t_2}S_{t_2}]$.

Part b) implies existence of $\pi_{k,k+1}$ such that $S_{k}=E^{k}[\pi_{k,k+1}S_{k+1}]$ for $k=0,...,T-1$. For any $t_1<t_2$, using law of iterated expectations, $S_{t_1}=E^{t_1}[\prod_{k=t_1}^{t_2-1}{\pi_{k,k+1}}S_{t_2}]$ for $k=0,...,T-1$ so $\prod_{k=t_1}^{t_2-1}{\pi_{k,k+1}}$ represents state prices or discount factor between dates $t_1$ and $t_2$. This means there is no-arbitrage between dates $t_1$ and $t_2$. You can choose $t_1=0$ and $t_2=T-1$.


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.