> [!tldr] > An American option is the right to buy/sell the underlying product at an agreed-upon strike price $K$ at any time $t \le T$, where $T$ is the option's expiry. Different from [[The Black-Scholes Formula of European Options|European options in, say, Black-Scholes world]], no-arb guarantees that *American options always have positive time value* (regardless of the SDE model of underlying prices, not just BSM) since American options can be exercised immediately. That is, for an American call of fair price $C_{t}$, for example, $\forall t \le T,~ C_{t} \ge (S_{t}-K)_{+}.$ *American options (without underlying paying dividends)* is never exercised early: - Partly because selling the option always gives more PnL at present compared to exercising, and if the holder wants to hold the underlying, they can just use the option money to buy it. - A no-arb argument involves the portfolio $C_{t}+Ke^{-r\tau}-S_{t},$which is guaranteed to worth more than $0$ at expiry. If exercised early, it is worth $-K(1-e^{-r\tau})$, and when carried to expiry, the accumulating debt makes it worth less than $0$. ## American Options Under Black-Scholes Under [[Geometric Brownian Motion#Application in Black-Scholes|the Black-Scholes model of underlying movements]], American options satisfy an relationship similar to the [[Black-Scholes Equation|Black-Scholes equation of European options]], but an inequality: $rV\ge\frac{ \partial V }{ \partial t } + r S \frac{ \partial V }{ \partial S } +\frac{\sigma^{2}S^{2}}{2}\frac{ \partial ^{2}V }{ \partial S^{2} }.$ To see the inequality, consider the same hedged portfolio $V_{t}-\Delta_{t}S_{t}$, - By the same no-arb argument as for the European options, longing the portfolio cannot grow faster than the risk-free rate (otherwise borrowing to long the portfolio is risk-free money). - The other inequality does not hold, when shorting the option can be exercised against, making it not perfectly hedged; therefore, *the equality holds only for prices $S_{t}$ where it is not optimal to exercise*. ## Perpetual American Options Perpetual American options basically have $T=\infty$, i.e. no expiry. This means that *the fair price of perpetual options is independent of time*. - The independence is because an regular American option's value does not depend on $t$ if $\tau$ (time-till-expiry) is the same; now letting $\tau \to \infty$ gives the value of the perpetual option. One important consequence is that *its price can be solved in closed form using Black-Scholes*: - ${ \partial V } / { \partial t }=0$ in the Black-Scholes (in)equality, making it an ODE in $S$ only, which is solvable for the region where exercising is suboptimal and the equality holds. - The option has an exercise boundary $S^{\ast}$ constant in $t$, so we know exactly in which region the equality holds -- outside it, the price is just the intrinsic value because the option will be exercised immediately. Now Black-Scholes reduces to the ODE $rV=rS\frac{ \partial V }{ \partial S }+ \frac{\sigma^{2}S^{2}}{2} \frac{ \partial V }{ \partial S^{2} },$for $S$ outside the exercise region (lt; S^{\ast}$ for calls, gt; S^{\ast}$ for puts). Its boundary conditions are: - **Extreme stock prices**: for example a put's value $\to 0$ when $S \to \infty$. Usually, applying this condition forces one of $A, B$ to be $0$. - **Value matching**: $V(S)=P(S)$ if $S$ is in the exercise region. ### Solving the BSM ODE and Exercise Boundary The time-independent ODE has base-solution $V=S^\beta$, for which the ODE further reduces to the characteristic equation, $(r-D)\beta+\frac{\sigma^{2}}{2}\beta(\beta-1)-r=0, $which has roots (big quadratic formula!) $\begin{align*} \beta_{\pm}&~~= -\frac{r-\sigma^{2}}{\sigma^{2}}\pm \frac{\sqrt{ \left( r-D-\frac{\sigma^{2}}{2} \right)^{2} +2r\sigma^{2}}}{\sigma^{2}}.\\[0.8em] &\left(=1,-\frac{2r}{\sigma^{2}} ~~~~\text{~if }D=0\right) \end{align*}$ Note that *$\beta_{+}$ is always positive and $\beta_{-}$ negative* (because the parabola opens upwards, and it is $-r<0$ at $\beta=0$). The general form of option value is then $AS^{\beta_{+}}+BS^{\beta_{-}}$ -- we shall use the boundary conditions to determine $A,B$. > [!exposition] Solving for $A,B$. > Consider a perpetual American call. > > The **limiting behavior** can rule out one of the constants: > - When $S \to 0$, a call is worthless, so its $B=0$ (otherwise $S^{\beta_{-}}$ blows up). > - Similarly, a put must have $A=0$ (consider $S \to \infty$). > > The remaining constant can be written as a function of the decision boundary $S^{\ast}$: for example the call has $S^{\ast}-K=C(S^{\ast})=AS^{\ast \beta_{+}} \Longrightarrow A=\frac{S^{\ast}-K}{S^{\ast \beta_{+}}}.$Its value will follow once we find the optimal $S^{\ast}$. > [!exposition] Solving for $S^{\ast}$ > Because only the holder (not the writer) gets to decide the American option's decision boundary, naturally they will maximize the option value with it: $S^{\ast}_{\mathrm{optimal}}=\underset{S^{\ast}}{\arg\max}~V(S;S^{\ast}).$Therefore, *one way of solving $S^{\ast}_{\mathrm{optimal}}$ is to simply differentiate $V$*: find $S^{\ast}: \frac{ \partial V }{ \partial S^{\ast} }=0. $ For example, a call will need $\begin{align*} S^{\ast}: \frac{ \partial C }{ \partial S^{\ast} }&= \frac{ \partial A }{ \partial S^{\ast} }S^{\beta_{+}}\propto \frac{-\beta_{+}(S^{\ast}-K)}{S^{\ast\beta_{+}+1} }+\frac{1}{S^{\ast \beta_{+}}}\equiv{0}\\[1em] &\Longrightarrow S^{\ast}_{\mathrm{optimal}}=\frac{\beta_{+}}{\beta_{+}-1}K. \end{align*}$ > > --- > > The other method is to apply the **smooth pasting condition**: $\frac{ \partial V }{ \partial S }|_{S=S^{\ast}}=\frac{ \partial P }{ \partial S }|_{S=S^{\ast}}.$This is because if the equality does not hold the holder can adjust the exercise boundary $S^{\ast}$ to increase its value: > > ![[PerpetualAmericanexerciseBoundary.png#invert|w80|center]] > > However, *not all options satisfy the smooth pasting assumption*: for example, it is impossible to smoothly paste a digital/binary option (a derivative of $0$ everywhere), and its exercise boundary is just the strike.