Option price in a neutral risk world is the same as in the real world. I can not understand! [closed]

Question

Good evening. I know there are several posts on the subject but unfortunately I can not fully understand this concept and I hope you can help me.

To price the option the fundamental assumption needed is that arbitrage opportunities do not exist. The absence of arbitrage implies that the option price is equal in the risk-neutral world and in the real world. Why? For example if we consider the simple binomial model: In this case we can build a risk-free portfolio to use as a discount rate the risk-free rate. But I do not understand why the value I get is the same in the real world? In this case we get:

$$f=\frac{f_u(1-de^{-rT})+f_d(ue^{-rT}-1)}{u-d}$$ That we can rewrite in a risk-neutral world: $$f=e^{-rT}[pf_u+(1-p)f_d] \qquad p=\frac{e^{rT}-d}{u-d}$$ In the real world, instead, we would have a different yield rate $\mu$ $$f^*=e^{-\mu T}[p^*f_u+(1-p^*)f_d] \qquad p^*=\frac{e^{\mu T}-d}{u-d}$$ In a risk-neutral world the expected value of the action is: $$E(S_T)=p S_0 u+(1-p)S_0 d$$ In the real world the expected value of the action is: $$E^*(S_T)=p^* S_0 u+(1-p^*)S_0 d$$ These two values are different!

Answer 1

Of course $E_t[S_T]\neq E^*_t[S_T]$, but you forgot that you have to discount at different rates. In particular you can verify that $S_t =e^{-r(T-t)}E^*_t[S_T]=e^{-\mu(T-t)}E_t[S_T]$

EDIT: just to give you a little bit of intuition. When you move from physical to risk neutral probability you are simply saying that you can always rewrite the ratio $\frac{P(\omega)}{R(\omega)}$ in terms of risk neutral probabilities $\frac{Q(\omega)}{R_f}$ if $Q(\omega)\equiv\frac{P(\omega)R_f}{R(\omega)}$