# Confusion about optimal choices with exotic options

With exotic options, holders usually face choices at certain times. In my understanding, the price of the option is determined by assuming the optimal choice is taken and computing the discounted expectation of the payoff under the risk neutral measure with backward induction. My question is if I hold such options, how does taking the optimal choice benefit me? Such optimal choices are optimal under the risk neutral measure, so how does taking such choices imply/guarantee (even probabilistically) anything under the actual measure? If I take the optimal choice as determined by the pricing model, that would maximize the arbitrage-free price of the option that I am holding, but does that bear any optimality in the actual world? Does that tend to lead to a higher P/L? Say the option has intrinsic value $$x$$, extrinsic value $$y$$, and I just happen to hold this option, not for hedging purposes and without any particular value on the underlying. How can I benefit from the optionality/extrinsic value? I understand that I would lose some of the extrinsic value if I don't follow the optimal choice, but that's just a theoretical construct, where would I see the manifestation of this "loss of value"?

Consider a vanilla european option. The optimal strategy in the risk neutral measure is to exercise if $$S(T)>K$$. (because in the risk neutral world, I value my payoff at its expectation, which at that point is the payoff itself. At time T, if $$S(T)-K$$ is positive, I will take it over not exercising, which is a payoff of 0.

This is of course also optimal in the real world - more money is better than less money. So my 'optimal exercise strategy' matches.

The manifestation of loss in value is a PnL leak. Consider the same example as above. Say $$S(T)>K$$ but you don't exercise your option at all: so you've just paid something for the option, but sub-optimal exercise strategy means that you've got nothing. This is an extreme example of course but you see the idea.

The same idea extends to exotics. Consider a Bermudan with 2 exercise dates. Denote by E the immediate exercise value at the first date, and by C the continuation value. Denote by $$w$$ the state of the world, and $$T$$ denotes the 1st exercise date, and $$T_ex$$ denote the date I exercise my option on (which is random).

Let $$A={w:E(w)>C(w)}$$. Then $$Pr[(T_ex=T)|A]=1$$ in the risk neutral measure. Since real world and risk neutral measure are equivalent (they agree on what's possible and not), we get $$Pr[(T_ex=T)|A]=1$$ in the real world. This tells you that you will exercise in the real world exactly when you exercise in the 'risk neutral measure'. Which tells you that the 'optimal strategy' is exactly the same.

Appendix:

The proof of conditional probabilities being the same in equivalent measures is better thought of by considering (assume $$Pr(A)>0$$ and $$Pr(B)>0$$) say $$Pr(B|A)=0$$ in the R.N. measure. Then $$Pr(B,A)=0 => Pr(B,A)=0$$ in the real world measure. As $$Pr(A)>0$$, we must have $$Pr(B|A)=0$$ in the real world measure too.