2
$\begingroup$

Reading the book by Andrea Pascucci "PDE and Martingale Method in Option Pricing" I am struggling with a very simple issue. Suppose we want to find the price of an American derivative $X$ in an arbitrage-free and complete market. Let $\mathbb{Q}$ be then the (unique) equivalent martingale measure with numeraire $B$ (the deterministic bond) and let $X_t$ be the value of the American derivative a time $t$ ($X_t$ is thus, in the most general formulation, a $\mathcal{F}_t$-adapted stochastic process). Let $H$ be the no-arbitrage price of $X$. Clearly it must be $H_T=X_T$. At time $t=T-1$ the price is determined as

$$ H_{T-1} = \max\left(X_{T-1},\frac{1}{1+r}\,\mathbb{E}^{\mathbb{Q}}\left[ X_T\mid\mathcal{F}_{T-1}\right]\right). \quad(1) $$

I clearly understand that $\frac{1}{1+r}\,\mathbb{E}^{\mathbb{Q}}\left[ X_T\mid\mathcal{F}_{T-1}\right]$ is the no-arbitrage price at time $T-1$ of an European derivative with maturity $T$ and payoff $X_T$, but which is the no-arbitrage argument behind equation (1) ?

$\endgroup$
2
$\begingroup$

For an American option, you have the right to exercise at any intermediate time. Then, at time $T-1$, if you exercise your option, you obtain the payoff $X_{T-1}$. However, if you wait to exercise at the maturity $T$, your value is $\frac{1}{1+r}\mathbb{E}^Q\left(X_T \mid \mathscr{F}_{T-1} \right)$. Your option value at time $T-1$ is the maximum of these two values, that is, \begin{align*} \max\left(X_{T-1}, \, \frac{1}{1+r}\mathbb{E}^Q\left(X_T \mid \mathscr{F}_{T-1} \right) \right). \end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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