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I am trying to connect risk-neutral and physical measure expectations, to understand the difference between a no-arbitrage price and an expected terminal value.

Imagine I have a European derivative with payoff $f(S_t)$, where $S_t$ is defined on the physical measure as usual: $$\text dS_t = \mu S_t \text dt + \sigma S_t \text d W^P_t\quad 0 \leq t \leq T.$$ Given a risk-free rate $r$, I define a risk-neutral measure with associated Brownian motion $W^Q_t$ and: $$\text dS_t = r S_t \text dt + \sigma S_t \text d W^Q_t\quad 0 \leq t \leq T.$$

The no-arbitrage price of $f$ at time $t$ is $\mathbb E^Q\left[ e^{-r(T-t)}f(S_T)\right]$. Changing measure from the risk-neutral to the physical one I get that the no-arbitrage price is $\mathbb E^P\left[ e^{-\mu(T-t)}f(S_T)\right]$, that is the "de-trending" idea. But then the expected value of the payoff at time $t$, that is what I expect to gain, is $\mathbb E^P\left[ f(S_T)\right]$.

I am aware of the idea that no-arbitrage price and expected value should be two different things, because the no-arbitrage price is designed to let you break-even always if you know how to, while the expected value is what you get if you buy it thousands of times.

How correct is to say: "Take the payoff and de-trend it, clearly you take away the expected gain because the price in expectation will just stay $S_0$, therefore you get a price you should always be able to break even with" ?

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  • $\begingroup$ Intuitively I would say that $\mathbb E^{\color{red}{P}}[e^{-\color{red}{r}(T-t)}f(S_T)]$ is the real world expectation of today's value (discounted) of the payoff $f(S_T)$ and that is what you should break even with when you buy it thousands of times for cash borrowed at the risk-free rate $r$. I don't understand this de-trending idea and what $\mathbb E^{\color{red}{P}}[f(S_T)]$ is telling us other than this will be the expected payoff at $T$. $\endgroup$
    – Kurt G.
    Apr 21 at 9:01

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