Equal Terminal Payoffs Imply Equal Prices Under Martingale Pricing

Summary

The question asks whether two assets with identical payoffs at a future time must have identical prices at earlier times in continuous time. The answer uses a money-market account as the numeraire and assumes that each asset price discounted by this account is a martingale under the pricing measure.

Conditional expectations of the discounted terminal values then give the discounted prices at any earlier time. Since the terminal payoffs agree almost surely, their conditional expectations agree, so the discounted prices—and therefore the asset prices—are equal. This provides a direct martingale argument for the law of one price in the setting described. The conclusion depends on the stated pricing assumptions, including the existence of the relevant martingales and common numeraire; the document does not provide a counterexample or explore cases where those assumptions fail.

Key ideas

• Discounting by the money-market account turns each asset price into a martingale under the assumed pricing measure.
• A martingale price equals the conditional expectation of its discounted terminal value.
• Equal terminal payoffs yield equal conditional expectations at earlier times.
• The argument relies on its martingale and numeraire assumptions.

Tags

# Law of one price in continuous time


# Law of one price in continuous time












The law of one price (i.e. for assets $S^{(i)}$ and $S^{(j)}$, $S^{(i)}_T = S^{(j)}_T $ almost surely implies that $S^{(i)}_t = S^{(j)}_t $ almost surely for all $ 0 \leq t \leq T$) is known to hold in discrete time when there is no arbitrage.

However, my lecturer claims that this statement might not hold in continuous time. Can anyone give me an example for that?

## Answer by Gordon (score 1)

https://quant.stackexchange.com/a/30874

Let $B_t=e^{\int_0^t r_s} ds$ be the money-market account value at time $t$, where $r_t$ is the short interest rate rate. Then both $\{\frac{S_t^i}{B_t}, \, t \ge 0\}$ and $\{\frac{S_t^j}{B_t}, \, t \ge 0\}$ are martingales. Therefore, for $0 \le t \le T$, \begin{align*} \frac{S_t^i}{B_t} &= E\left(\frac{S_T^i}{B_T} \mid \mathcal{F}_t \right)\\ &= E\left(\frac{S_T^j}{B_T} \mid \mathcal{F}_t \right)\\ &= \frac{S_t^j}{B_t}. \end{align*} That is, $S_t^i = S_t^j$.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.