-3
$\begingroup$

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: enter image description here 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!

$\endgroup$

closed as off-topic by noob2, Quantuple, amdopt, Alex C, LocalVolatility May 25 '17 at 9:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Basic financial questions are off-topic as they are assumed to be common knowledge for those studying or working in the field of quantitative finance." – noob2, Quantuple, amdopt, Alex C, LocalVolatility
If this question can be reworded to fit the rules in the help center, please edit the question.

2
$\begingroup$

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 $Q(\omega)\equiv\frac{P(\omega)R_f}{R(\omega)}$

$\endgroup$
  • $\begingroup$ Considering all this, because the value of f in the risk-neutral world equals $ f ^ *$ in the real world? That I can build a risk-neutral portfolio by matching the two possible portfolio values, and at a risk-free rate, get the value of the option. But in the real world I can not build a risk-free portfolio! @fnic $\endgroup$ – Mike9 May 24 '17 at 19:09
  • 2
    $\begingroup$ Maybe there are 3 worlds: the risk-neutral world, the perfect Black-Scholes-Merton world and the Imperfect World. In the perfect BSM world you CAN build a risk free portfolio, that is what B S and M proved mathematically! in the IW maybe you cannot, maybe some BSM assumptions are not exactly correct, but close enough. $\endgroup$ – noob2 May 24 '17 at 21:05
  • $\begingroup$ Under certain assumptions, the most important of which is the non-arbitrage hypothesis, the price of the option in all three worlds is the same! It's correct? I know it should be a simple concept but I can not understand the general intuition! (P.S. intuitively makes sense to say that in the perfect Black-Scholes-Merton world can I create the risk-neutral world?) $\endgroup$ – Mike9 May 24 '17 at 21:34

Not the answer you're looking for? Browse other questions tagged or ask your own question.