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1

The first statement is kind of clear. If all investors are risk-neutral, they simply do not care about risk and do not pay more or less regardless how risky an asset is. As a consequence, the return of all assets is the risk-free rate. Regarding the second statement. What does risk-neutral pricing really do? We change the probabilities from the real world, $... 0 "Cannot figure out where this reasoning fails. The question is: am I misunderstanding the real meaning of arbitrage-free pricing or the only reason for which this it seems to be strange to me is that I miss some economical / "real markets world related" point of view?" In opposite to the first comment, I think you do miss something a bit subtle here. First ... 2 You can use such an approximation but there are known analytical prices. You have a special case in which the stock price is normally distributed. See Bachelier Model. Set$\mu=r-q$(if you have dividends, or simply$\mu=r$if there are no dividends). So if you change from the real worl probability measure$\mathbb{P}$to the risk-neutral measure$\mathbb{Q}...

2

I'm not sure I'd call it 'subjective' or 'pre-conditioned'. Traditionally, ensuring the absence of arbitrage is a guiding principle for pricing, while the risk-neutral measure is most frequently used for theoretical results. Assuming you'll be using your forecasting for trading activities, using AoA to determine price is the right way to go. Consider ...

2

The equation $f_T = \max \{ S_T - K, 0 \}$ is not an assumption, this is true by definition of what a call option is. It's an option which, at the time of maturity $T$, gives the value $\max\{ S_T - K, 0\}$ to the holder. And yes, options $f_t$ follow the diffusion $dF_t = \mu dt + \sigma dW_t$ because the underlying stock (or forward) also follows an Ito ...

6

The first equation expresses the option price as a discounted expected value of the payoff contingent on an asset price $S \geqslant 0$. Without loss of generality, we assume that the probability density function has support in $[0,\infty)$, and rewrite as \begin{align} P_{t,T}(K) &=e^{-r(T-t)} \int_{-\infty}^{\infty}\left(K-S\right)^+ q_T^S(S)\,dS \\...

3

What you say is perfectly true and there is no contradiction. Arbitrage means risk free profit , so your ‘statistical arbitrage’ is not arbitrage at all. It just says that if you take risk, your expected returns can be higher than the risk free rate. How much higher depends in the risk aversion of market participants.

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