# Option pricing under distribution assumption

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?

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

• 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. Nov 27, 2023 at 17:02
• @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. Nov 28, 2023 at 16:39