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Under a risk-neutral measure $\mathbb{Q}$, the discounted stock price is a $\mathbb{Q}$-martingale. Does it mean that under the actual probability measure $\mathbb{P}$ the discounted stock price is NOT a $\mathbb{P}$-martingale? Assuming that we are working with a Black-Scholes market model where the stock price $S(t)$ is a geometric Brownian motion and the bond dynamic is given by

$dB(t) = r(t)B(t)dt$

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    $\begingroup$ In the real world measure, the whole of the pensions industry is based on the fact that over time, the value of equities will rise faster than the risk free rate! In 6 months I can't guarantee that, but over 30 years it's nearly certain (or at least, this is what history seems to tell us) $\endgroup$
    – StackG
    Oct 5, 2020 at 8:48
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    $\begingroup$ The martingale property of a process $X_t$ depends on a given filtration and a probability measure. If you change one of them, $X_t$ may no longer be a martingale. Recall that $\mathbb Q$ is constructed such that $S_t/B_t$ is a martingale. That’s the entire point of bothering with $\mathbb Q$. So no, there is no reason why the discounted stock price should be a martingale under $\mathbb P$. $\endgroup$
    – Kevin
    Oct 5, 2020 at 9:14
  • $\begingroup$ Thanks @KeSchn for your answer. $\endgroup$
    – Matteo Campagnoli
    Oct 6, 2020 at 12:23

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