Tag Info

New answers tagged

4

It can also be proved by Jenson's inequality. It can only be optimal to exercise the American option if the option is below its intrinsic value; but since the "max" function is convex, the European price satisfies the following inequality: $$c(S_t, t)=e^{-rT}\mathbb{E}[(S_T-K)^+]>=e^{-rT}\left(\mathbb{E}[S_T]-\mathbb{E}[K]\right)^+=S_t-Ke^{-rT} $$ The ...


5

it's a model-free result. The conditions are $d\leq 0, r\geq 0.$ The proof is that for a european $$ C_t > S_t - Ke^{-r(T-t)} \geq S_T - K $$ and the American is worth at least as much so you never early exercise. So it's worth the same as European. To prove the inequality, observe if $B_T =1,$ take $K$ units of $B_t$ and one of $C_t$ to get something ...


0

at the time of exercise, you don't know what the final expiry stock value is. Consider the portfolio consisting of the option and $E$ zero coupon bonds worth $B_t \leq 1.$ At expiry its value is $$ \max(S,E) > S. $$ since can you exercise and get the stock if $S>E$ and have $E$ otherwise. So at all times previously, $$ C_t + K B_t > S_t $$ ...


1

The theoretical answer is: If you feel you should exercise at $t<T$ then you can get $S_t-K$ (obviously $S_t>K$). But price of the option at any$t<T$ is $C_t>S_t-K$ always. So rather than exercising just sell the option and let someone else hold it. But then practically there are transaction costs etc.



Top 50 recent answers are included