No-Arbitrage Proofs of the Law of One Price

# The Law of One Price in a discrete model


# The Law of One Price in a discrete model












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.

- 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?

- 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?

## Answer by AFK (score 2, accepted)

https://quant.stackexchange.com/a/17562

(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.

## Answer by torbonde (score 2)

https://quant.stackexchange.com/a/17560

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.

## Answer by eltigrechino (score 2)

https://quant.stackexchange.com/a/17561

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.