1
$\begingroup$

The following question assumes familiarity with the discrete model described in chapter 5 of Steven Roman's "Introduction to the Mathematics of Finance", 2nd edition, Springer 2012. I will not describe the model or the associated notation in this post.

  1. The Law of One Price (p. 132) states that, in the absence of arbitrage opportunity in the market, $$ \mathcal{V}_T(\Phi_1) = \mathcal{V}_T(\Phi_2) \implies \mathcal{V}_k(\Phi_1) = \mathcal{V}_k(\Phi_2) $$ for all times $0 \leq k \leq T$ and for all self-financing trading strategies $\Phi_1$ and $\Phi_2$.

    Unfortunately, no proof is provided in the text (in fact, this law is stated as a definition rather than a theorem). Why does this law hold?

  2. Additionally, it is implied by the text following the statement of the Law of One Price, that, if the market has no arbitrage opportunity, then, given an attainable alternative $X$, if a new asset $a^*$ is introduced into the market and is priced in such a way that its payoff at time $t_T$ is $X$ and its pricing is consistent with the Law of One Price, i.e. for every $k \in \{0, 1, \dots, T\}$, $S_{a^*, k} := \mathcal{V}_k(\Phi)$, where $\Phi$ is any self-financing trading strategy such that $\mathcal{V}_T(\Phi) = X$, then the resulting, extended market will still have no arbitrage opportunity.

    Why is this so?

$\endgroup$
2
$\begingroup$

(1) To get an arbitrage, buy low and sell high.

Consider the following strategy: at $k$, in the event $V_k(\Phi_1) < V_k(\Phi_2)$, buy $\Phi_1$ and sell $\Phi_2$ invest the difference at the risk free rate. At maturity, your portfolio is worth what the you put in the bank plus interest.

Formally, if $V_t(\Phi_\alpha) = \sum_{i=0}^d \Phi^i_{\alpha,t} S^i_t$ where $S^0_t = (1+r)^t$ is the risk free asset, then the strategy corresponds to $$ V_t = \sum \delta^i_t S^i_t $$ where $\delta^i_t = 0$ for all i and all $t<k$, then for $t\geq k$, and $i\neq 0$,
$$ \delta^i_t = 1_{V_k(\Phi_1)<V_k(\Phi_1)}(\Phi^i_{1,t} - \Phi^i_{2,t}) $$ The amount invested in the risk free asset is exactly what is left so that the strategy is self-financing. You can check that $$ \delta^0_t = 1_{V_k(\Phi_1)<V_k(\Phi_2)}\Big( (V_k(\Phi_2)-V_k(\Phi_1))\frac{S^0_t}{S^0_k} + (\Phi^0_{1,t} - \Phi^0_{2,t}) \Big) $$

Note that following this strategy, your portfolio satisfies

  • $V_0 = 0$,
  • $P(V_T \geq 0) = 1$
  • $P(V_T > 0) \ge P(V_k(\Phi_1) < V_k(\Phi_2))$ since every time the portfolio $1$ is worth strictly less than portfolio $2$ you end up with cash in the bank.

Since there is no arbitrage, this last probability has to be zero: $V_k(\Phi_1) \ge V_k(\Phi_2)$ almost surely. By symmetry the reverse inequality is also true so $V_k(\Phi_1) = V_k(\Phi_2)$ almost surely.

(2) The market is still without arbitrage because the asset you added it a linear combination of the previous ones. If you have a portfolio $$ V_t = \sum_i \delta^i_t S^i_t + \delta_{a^*,t}S_{a^*,t} $$ creating an arbitrage in the extended market, you can decompose $S_{a^*,t}$ into the other assets, rewriting $$ V_t = \sum_i (\delta^i_t + \delta_{a^*,t}\Phi^i_t) S^i_t $$ and this would give you an arbitrage in the original market.

$\endgroup$
  • $\begingroup$ Thank you. Regarding (1): What does it mean, in terms of portfolios, to "buy $\Phi_1$ and sell $\Phi_2$"? And what actions should I take (i.e. what should my portfolios be) before time $k$? $\endgroup$ – Evan Aad Apr 28 '15 at 7:11
  • $\begingroup$ Dear AFK. I don't want to nag, but I would really appreciate it if you could help me to understand how your answer to question (1) translates to a formal proof, i.e. what portfolio should I use at each of the times $1, 2, \dots, T$? If I understand correctly, at times $1, 2, \dots, k-1$ you suggest that I do nothing, in other words that I use the zero portfolio (i.e. the portfolio that is identically zero). Then what? $\endgroup$ – Evan Aad Apr 28 '15 at 21:12
  • $\begingroup$ I edited the answer to add the explicit solution. $\endgroup$ – AFK Apr 28 '15 at 22:10
  • $\begingroup$ Thank you very much! One more question, please. Could you please help me understand the mechanics of the action taken at time $k$? Up to time $k$ no assets are held which is to say we have no debt no credit no money. Then at time $k$ we wish to buy the cheaper portfolio and sell the more expensive one. But we have no money to carry out these transactions with. So we borrow money from the bank in the form of risk free assets at a value that amounts to the combined value of the two portfolios. Then we sell the more expensive portfolio and use the sale money to repay some of our debt to the bank. $\endgroup$ – Evan Aad Apr 29 '15 at 5:53
  • $\begingroup$ So now we have a short position on the more expensive portfolio, a long position on the cheaper portfolio and a debt to the bank that amounts to the value of the cheaper portfolio. Am I correct so far? However, if I'm correct then the total value of this position is negative, because the long position and the debt to the bank cancel each other out, and I am left with a short position, which translates to a negative value. However, the strategy is supposed to be self-financing, and before time $k$ we had no debt and no credit. So how can this be? $\endgroup$ – Evan Aad Apr 29 '15 at 5:53
2
$\begingroup$

Unfortunately, I do not know the model you talk about. However, the law of one price is a direct implication of the no-arbitrage assumption, which is assumed in many models (if not all).

I do agree that the law of one price should be stated as a theorem rather than a definition. Anyway. Consider the case in which two portfolios A and B have the same value at time $T$, $V^{(A)}_T = V^{(B)}_T$, and assume that their values are different at some time $0\leq t<T$. Assume without loss of generality that $V_t^{(A)} < V_t^{(B)}$. Then buy portfolio A and short portfolio B, which will give you a profit at time $t$ of $V_t^{(B)} - V_t^{(A)} > 0$ and no risk of losing money in the future. Thus we found an arbitrage, which is a contradiction.

$\endgroup$
2
$\begingroup$

Regarding (1).

Assume for some time $k$, $\mathcal{V}_k(\Phi_1) > \mathcal{V}_k(\Phi_2)$ (w.l.o.g.) with full knowledge that these strategies have equal value at $T$ ,($\mathcal{V}_T(\Phi_1)=\mathcal{V}_T(\Phi_2)$). I claim that this situation admits arbitrage.

I can sell $\mathcal{V}_k(\Phi_1)$ and buy $\mathcal{V}_k(\Phi_2)$ and pocket the difference as profit.

At time T deliver $\mathcal{V}_T(\Phi_2)$ (since it was self financing) to settle my short position in $\mathcal{V}_T(\Phi_1)$. The positions offset and I made a riskless profit.

$\endgroup$
  • $\begingroup$ Thank you. Could you please help me see how this strategy can be described in formal terms? What would my portfolios $\theta_i$ be at times $i = 1 \dots T$? $\endgroup$ – Evan Aad Apr 28 '15 at 7:07

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.