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It seems to be a foolish question but I can't take my mind off from , Is it true that there is no analytic formula for the value of an American put option on a non-dividend-paying stock (or a divident paying stock either) has been produced ?

Thanks for your helping in advanced.

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There is no closed formula for American put option. However, there is an analytic solution for perpetual American put option. The only difference is that the maturity of the perpetual American option is infinite.

Why that makes such a difference? That's because we can determine the optimal stopping time (and therefore optimal exercise price) if we don't have to worry about maturity. Unfortunately, there is no such solution if the optimal exercise price is a function of maturity. Consider:

enter image description here

L is the optimal exercise price. It's a convex function of maturity. Far away from maturity, the price is significantly lower than K because we'd expect a deep in-the-money intrinsic value to compensate giving up the option right early. The price approaches to the strike price as shown in the plot. This function is clearly non-linear.

You might need to do approximation. For example, the Barone-Adesi-Whaley quadratic approximation. Google for the paper if you're interested.

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it all comes down to how you define analytic. If you push the definition far enough there are some.

An exact and explicit solution for the valuation of American put options DOI:10.1080/14697680600699811 Song-Ping Zhu pages 229-242

However, it's an infinite sum of recursively defined double integrals.

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    $\begingroup$ Interesting. What's the real benefit of it practically? The computational complexity seems too much.... $\endgroup$ – SmallChess Jul 15 '15 at 10:12
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Yes, there is none. Quoting Higham (2004):

"The mathematical problem defined by (...) is much more difficult than the Black–Scholes PDE that arose without the early exercise facility. In general, there is no closed form expression for $P^{Am}(S, t$) and we must use numerical methods to obtain approximate values."

Where (...) refers to the American Option PDE.

Please check chapter 18 of this book.

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In the Black-Scholes Model or Heston Model, the American option satisfies the same PDE, but with different boundaries.For an American call option $C_A(S,\tau )$, we can therefore write \begin{align} \frac{\partial {{C}_{A}}}{\partial \tau }=+\frac{1}{2}{{\sigma }^{2}}{{S}^{2}}\frac{{{\partial }^{2}}{{C}_{A}}}{\partial {{S}^{2}}}+(r-q)S\frac{\partial {{C}_{A}}}{\partial S}-r{{C}_{A}} \end{align}

or(Heston) $C_A(S,v,\tau )$ satisfy \begin{align} \frac{\partial {{C}_{A}}}{\partial \tau }=\,+\frac{1}{2}v{{S}^{2}}\frac{{{\partial }^{2}}{{C}_{A}}}{\partial {{S}^{2}}}+\rho \sigma \,vS\frac{{{\partial }^{2}}{{C}_{A}}}{\partial S \partial v}+\frac{1}{2}{{\sigma }^{2}}v\frac{{{\partial }^{2}}{{C}_{A}}}{\partial {{v}^{2}}}-rC_A+(r-q)S\frac{\partial {{C}_{A}}}{\partial S}+\kappa (\theta -v)\,\frac{\partial {{C}_{A}}}{\partial v} \end{align} where $\tau$ is the time until maturity. The PDE holds for $0 ≤ \tau < T$, where $T$ is the maturity calendar time, and for $0 < S ≤ b(v, \tau)$, where $b(v, τ )$ is the early exercise boundary. Essentially, this means that as long as the stock price is within the early exercise boundary, the American call option behaves like its European counterpart and the PDE holds. Building on the work of Chiarella and Ziogas(2006), Tzavalis, and Wang(2003). approximate the early exercise boundary $b(v,\tau )$ with the log-linear function.

\begin{align} b(v,\tau )=exp(b_0(\tau)+b_1(\tau)v) \end{align}

They show the American call is obtained by adding the early exercise premium to the price of the European call

\begin{align} C_A=C_E+V \end{align} where $V$ is the early exercise premium on an American call with strike $K$ and maturity is $\tau$

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  • $\begingroup$ This is also used by Barone-Adesi-Whaley. They approximate the early exercise premium with a quadratic function. $\endgroup$ – SmallChess Jul 15 '15 at 12:39
  • $\begingroup$ Yes, but even for the calculation of European option we need to approximate it. $\endgroup$ – user16891 Jul 15 '15 at 19:09

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