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For simplicity assume zero interest rates in the following.

Given the price of a (European) put option with strike K and maturity T at time point t. $P_t(K, T)$ for a given underlying S with values $S_t$ at time point t.

Assuming you are currently at timepoint $t_0$. The return of S over the lifespan of the option is given by $r_T=\frac{S_T-S_0}{S_0}$.

Assuming that $r_t \sim f$. For some density $f$. Is the risk-neutral price of the option equal to the expected value of the option (if it exists, i.e. is finite)? So is this true? If not, why not?

$$P_{t_0}(K, T) = \mathbb{E}(P_T(K,T)=\int_{0}^{K} K-S_T \: dP(S_T) = \int_{-1}^{\frac{K-S_0}{S_0}} K-(1+r_T)S_0\: dP(r_T) \\ = \int_{-1}^{\frac{K-S_0}{S_0}} (K-(1+x)S_0) \: f(x)\: dx$$

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After reviewing further literature, I have come to the conclusion that indeed this method gives the correct answer. This thought process can be used to derive BS-formula, given the (risk-neutral) density of the "results" of the underlying stochastic process.

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  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Nov 27, 2023 at 16:38
  • $\begingroup$ Risk-neutral option pricing involves more than just taking an expectation. The core of it is to replicate a payoff by trading in assets. Under a certain numeraire these asset prices must be martingales so that we don't have arbitrage. Assuming $r_t\sim f$ for some density is far from meeting those requirements. $\endgroup$
    – Kurt G.
    Nov 27, 2023 at 17:02
  • $\begingroup$ @KurtG. I think I know what you mean, but I want to double check and learn. My hypothesis is: The value of an option is only equal to the expected value (ignoring interest rates) if you are risk-indifferent (may be a better word than risk-neutral). I don't really get the replication argument as this assumes that you can dynamically hedge options (doesn't work under certain conditions for density of $r_T$). I do not necessarily see the problem in assuming just a density, although I acknowledge that further "no-arbitrage" arguments have to be respected. Note: my model assumes only 1 timestep. $\endgroup$
    – MrLCh
    Nov 28, 2023 at 16:39
  • $\begingroup$ The title of your question starts with "Risk-neutral option pricing". Black, Scholes and Merton have shown that under the assumption of risk-neutrality of the probability measure the price of an option is the expected value of the payoff. For this they received a Nobel prize in economics. They used a replication argument to prove mathematically that this expectation must be the price. If you think that any expectation under a "risk-indifferent" measure is also the option price the bar for proving that hangs very very high. $\endgroup$
    – Kurt G.
    Nov 28, 2023 at 16:45
  • $\begingroup$ We cannot just mechanically calculate expectations despite the fact that the entire finance industry is doing just that. :) $\endgroup$
    – Kurt G.
    Nov 28, 2023 at 16:46

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