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In the book "Financial Modelling with jump processes" by Cont and Tankov there is a chapter that explains martingale pricing principles. It is not extremely formal, but gives the idea underlying the method. There he shows that any linear positive pricing rule can be associated with a probability measure (or expectation operator defined on a linear subspace of random variables representing the contingent claims). That's perfectly clear, and argument is rather simple.

The next step is to show that under some non-arbitrage argument, the discounted prices of traded securities must be martingales w.r.t. to the latter measure. The argument is as follows:

Consider now an asset $S^i$ traded at price $S^i_t$. This asset can be held until $T$, generating a terminal payoff $S^i_T$, or be sold for $S^i_i$: the resulting sum invested at the interest rate $r$ will then generate a terminal wealth of $\mathrm e^{r(T-t)}S^i_t$. These two buy-and-hold strategies are self-financing and have the same terminal payoff so they should have the same value at $t$.

The bold part in the quote is unclear to me: how do we know that terminal payoffs are the same? I even think, that in the simplest case of BS model they won't be: if $S^i_t$ is a GBM with drift $\mu$ and volatility $\sigma$, then $S^i_T \neq \mathrm e^{r(T-t)}S^i_t$ under the physical measure, and since we are looking for the equivalent measures, neither payoffs will be same (a.s.) under such measures.

Can somebody clarify, whether this is indeed a mistake in the argument, or am I missing something here?

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I am guessing that the argument is as follows.

They certainly have the same value at time t since they are both worth $S_t$ then. If they have the same value at $t$ they should have the same value at time $0.$

So if we are pricing by expectation our measure has to give the same discounted expectation price to both portfolios. So we must have

$$ e^{-rT} E( S_T) = e^{-rT} E(e^{r(T-t)}S_t) $$

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  • $\begingroup$ Does it mean, that no matter how we invest $S_t$ at time $t$, and which strategy $\varphi$ we choose until time $T$, we must have $E(S_T) = E(V_T(\varphi))$? I.e. even if we are trying to make as less money as possible (e.g. 1/2 $S_T$) we still get $E(S_T)$ on average? $\endgroup$ – Ulysses Nov 19 '14 at 7:55
  • $\begingroup$ yes that is correct $\endgroup$ – Mark Joshi Nov 20 '14 at 2:04
  • $\begingroup$ I do not really see how does that follow. That is, we have two strategies $\varphi$ and $\varphi'$ such that $V_t(\varphi) = V_{t'}(\varphi') = S_t$. From now on they do differ, and lead to two different payoffs. But that does not mean that there is necessarily an arbitrage opportunity: the latter requires that one payoff is a.s. bigger than another, which may not be the case in our setting. $\endgroup$ – Ulysses Nov 20 '14 at 11:47
  • $\begingroup$ i haven't read the section of Cont and Tankov. However, i think there are two questions 1) are prices given by expectations 2) if prices are given by expectations what properties must the measure have. I think it is 2) that is being addressed. A portfolio worth $X_t$ at time $t$ must then have $E(X_T) = X_t$ (taking r=0) no matter what by 1). Proving the existence of such a measure is a different matter entirely. Sorry if this is all off beam and actually you are asking something different. $\endgroup$ – Mark Joshi Nov 20 '14 at 22:45

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