American-Options

Pricing of American Deriviatives

  • December 9, 2015

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) ?

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,

$$ \begin{align*} \max\left(X_{T-1}, , \frac{1}{1+r}\mathbb{E}^Q\left(X_T \mid \mathscr{F}_{T-1} \right) \right). \end{align*} $$

引用自:https://quant.stackexchange.com/questions/22220