3
$\begingroup$

I am reading the chapter 15 of Options, futures, and other derivatives by John Hull.

Specifically, 15.12 Dividends-American Call Options.

I am stuck while proving the fact that exercising an American options with dividend stock just before the last dividend date is optimal, rather than holding it til the maturity.

The investor will get $S(t_n)-K$ when he/she exercise before the ex-dividend date. After the ex-dividend date, the stock price would down to $S(t_n) - D_n$. And the lower bound of this option is known to be $S(t_n)-D_n-Ke^{-r(T-t_n)}$.

So, using proof by contradiction, the following is derived. $$S(t_n)-D_n-Ke^{-r(T-t_n)}\geqslant S(t_n)-K$$ $$-D_n-Ke^{-r(T-t_n)}\geqslant - K$$ $$D_n + Ke^{-r(T-t_n)}\leqslant K$$ $$D_n \leqslant K\left[1-e^{-r(T-t_n)}\right]$$

From the last equation, the book concluded that it cannot be optimal to exercise at time $t_n$(the ex-dividend date).

However, I don't understand how come we can conclude like this and the implication of right hand side of the last equation.

Can anyone help me to understand this proof?

$\endgroup$
2
  • $\begingroup$ Do you mean “cannot be optimal” or “is optimal”? $\endgroup$ Commented Apr 26, 2021 at 22:38
  • $\begingroup$ The book says the last equation implies it cannot be optimal to exercise an American option with dividend stock, which contradicts what we prove $\endgroup$
    – Turtle203
    Commented Apr 27, 2021 at 2:36

1 Answer 1

1
$\begingroup$

$$S(t_n)-D_n-Ke^{-r(T-t_n)}\geqslant S(t_n)-K$$ means that if LHS (lower bound of option price) $\geqslant$ RHS (what you get if you exercise early), it cannot be optimal to exercise. This is an assumption, not a claim that it is true or must hold under any circumstances.

The rest is just reformulation to have $D$ on one side. Hence, iff $$D_n \leqslant RHS$$ it is not optimal to exercise at time $t_n$ (simply because the first equation says it is not and nothing changed). That said, it's also possible that $D_n > RHS$. Particularly if $T-t_n$ is small and / or the dividend is large. Starting with first equation, you can also argue that there exists a $S(t_n)$ that justifies early exercise.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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