I tried to pricing the American Call option using "Longstaff-Schwartz" least squares method. However, I found the American call option is always lower than the Monte Carlo European call option (they should be equal to each other).

  1. I collected one stock's daily returns over past 10 years.
  2. Plot the frequency distribution of all daily returns.
  3. Found the "t Location-Scale Distribution" is the best fitted distribution, where $\mu=1.0118\times10^{-4}$, $\sigma = 0.0076$ and $\nu=2.5977$, the probability density function is given by \begin{equation} p(x)=\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sigma\sqrt{\nu\pi}\Gamma\left(\frac{\nu}{2}\right)}\left[\frac{\nu+\left(\frac{x-\mu}{\sigma}\right)^2}{\nu}\right]^{-\frac{\nu+1}{2}}. \end{equation}
  4. The set of Monte Carlo daily returns is given by $x_{MC}=\{-20\% : 0.001\%:20\%\}$.
  5. Select 2-million numbers randomly from set $x_{MC}$ according to the "t Location-Scale Distribution".
  6. Initial stock price $S_0 = 16.86 $, generate Mont-Carlo price paths for $60$ days.
  7. The strike price $K=S_0=16.86$, risk-free rate $r=3.95\%$.

One of results is: American Call option $V_0^{A_c} = 0.7895$ and European Call option $V_0^{E_c} = 0.7907$.

  • $\begingroup$ For "Longstaff-Schwartz" least squares method, I also tried different orders of "basis functions", but it doesn't change the situation. $\endgroup$
    – Stephen Ge
    Mar 7, 2019 at 2:35
  • $\begingroup$ One suggestion: Check to see how many paths are such that early exercise is occurring. $\endgroup$
    – dm63
    Mar 7, 2019 at 3:40
  • $\begingroup$ Regarding point 3, what distribution is this $p(x)$ fitting to? $\endgroup$
    – Hans
    Mar 7, 2019 at 8:21
  • $\begingroup$ @Hans p(x) is the pdf of t-location scale distribution $\endgroup$
    – Stephen Ge
    Mar 7, 2019 at 8:39
  • $\begingroup$ You are fitting this t-distribution to something (a distribution of some sort). What is this target distribution you are fitting to? $\endgroup$
    – Hans
    Mar 7, 2019 at 9:35

1 Answer 1


The LS algo only approximates the continuation value no matter which and how many basis functions you use (unless it's infinite). Therefore it will always undervalue any option and sure enough also an American Call with no dividends will be under-priced too. I also remember wondering about this when I first tried it, but like you I found it always gives a price lower than the European for an American call w/o dividends. That is if you do everything else right: use a lot of time steps, so that it's practically "continuous exercise" and not Bermudan (that would also take care of the time discretization error by the way), and don't introduce something else that would give it an upward bias (like say using the same random numbers for both the fitting and the pricing part).

  • $\begingroup$ Thanks, about "using the same random numbers for both the fitting and the pricing part", do you mean that I should generate 2-million paths for estimating parameters in basis function, and then re-generate 2-million paths for option pricing? $\endgroup$
    – Stephen Ge
    Mar 12, 2019 at 1:14
  • $\begingroup$ I think LS method is a widely used method for option pricing, do you know how to deal this problem in practice? $\endgroup$
    – Stephen Ge
    Mar 12, 2019 at 2:07

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