Given the classic Black-Scholes model, e.g.

$dS(t)/S(t)=rdt+\sigma dW^{\mathbb{Q}}(t)$ with $S(0)=S_0$ and $dB(t)=rB(t)dt$ with $B(0)=1$,

whereby $r$ and $\sigma$ are constants and $\mathbb{Q}$ denotes the risk-neutral measure. Given now an option with payout $\Phi_T$ only at $T$, e.g. an European call option $\Phi_T(S_{T})=\max(S_T-K,0)$.

  1. Do we need to assume that the option itself is tradable in order to determine the price of such an option as $e^{-rT}\mathbb{E}^{\mathbb{Q}}[\Phi_T(S_T)]$? If so, why? If not, why not?

  2. What if the option still can be only exercised at maturity $T$, but the payout depends on previous prices of the asset, e.g. $S_1, S_2,\ldots,S_T$. That is, do we need that the option itself is tradable in order for $e^{-rT}\mathbb{E}^{\mathbb{Q}}[\Phi_T(S_1,S_2,\ldots,S_T)]$?

  • 1
    $\begingroup$ The risk neutral pricing is based on an argument in complete markets for the hedging portfolio delivering the proper amount at maturity as long as one begins with the risk neutral option price and trades dynamically in the stock and money market. So the option doesn't need to be actively traded for the seller to price it using risk neutral pricing, they would just need to be liable for the payout at maturity. $\endgroup$ – Slade Sep 27 '19 at 12:30
  • $\begingroup$ OK, I somehow guesses this. If we find a hedging portfolio strategy that perfectly replicates the payoff of the option, their price need to be same due non-arbitrage. However, I was worried whether one runs into trouble with this argument as one can trade the portfolio over the duration of the option, yet the option is not tradable. $\endgroup$ – Hari Sep 28 '19 at 6:40
  • $\begingroup$ The hedging portfolio doesn't consist of the option itself so there's no need for the option to be traded for risk neutral pricing $\endgroup$ – Slade Sep 28 '19 at 13:53

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