I recently discovered the LSMonteCarlo library in R which basically determines the price of American options via Longstaff Schwartz method.

I tried the AmerPutLSM which per description first simulates paths of a geometric Brownian Motion and then uses the Longstaff Schwartz method.

The first few applications did non convince so I tried to price an European Put Option with this function:

s0 <- 100
strike <- 100
sigma <- 0.03

AmerPutLSM(Spot = s0, sigma = sigma, n = 10000, 
            m = 12, Strike = strike, r = 0, mT = 1, dr = 0)

American Put Option
Price: 1.187689

So basically, I consider 10000 paths of a geometric Brownian motion with $\sigma = 0.03$ and an ATM put option (strike = $S_0$ = 100). The maturity of the option is set to 1 (mT = 1) and the number of time steps in the simulation is also set to 12 (m = 12). This means that the option can be exercised at the end of each month (Bermudan type). For simplicity I assumed no interest rate (r = 0) and zero dividends (dr = 0). This function tells me that the price of this option is about 1.188

But, if we compare this with the Black-Scholes Put Price of a European Put Option we get that

$$ V_{\text{European}} = \text{Strike} \cdot \Phi\Bigl(\frac \sigma 2 \Bigr) - S_0 \cdot \Phi\Bigl(-\frac \sigma 2 \Bigr) = 1.196782. $$

strike * pnorm(sigma/2) - s0 * pnorm(-sigma/2)
[1] 1.196782

This makes no sense since the value of the European Option should always be less then its American (or Bermudan) counterpart.

Does anyone have an explanation for this?

Thank you.

  • $\begingroup$ Well, you are assuming no dividends therefore the price of the American option should coincide with the European price. The difference between your 2 prices is 0.77% which for monthly steps seems already rather low, so it might be purely numerical imprecision. Have you tried augmenting the number of time steps? $\endgroup$ Jun 7, 2018 at 15:12
  • 1
    $\begingroup$ Thank you for your comment. I think only call prices coincide when there are no dividends. In general Put prices are not equal, even if dividends are zero. And yes I also tried different parameters. Results were always the same. $\endgroup$
    – Cettt
    Jun 7, 2018 at 15:17

1 Answer 1


It is due the number of timestamps in your case. Actually, as the ZC rate is zero, the price of European and american options should be the same.


You know that for american options (see proof in pages 4,5 HERE): $S_T-K\leqslant C-P \leqslant S_T-Ke^{-rT}$

When the risk free rate is zero, you get that the call put parity remains valid $S_T-K= C-P$ As the european call price is the same as american call price without dividends, you can conclude that the the put prices are also the same. IN ADDITION, when the strike and spot are equal, you have: $C_A=C_E=P_A=P_E$

On the other hand, more you increase the number of timestamps, more the american option price should be higher (more stopping times) and converges to BS put price when the number of timestamps goes to $+\infty$. It explains the reason behind the fact that LS montecarlo price is lower than european put price.

I've done the test using DMS software and I've a smaller value than BS price:

enter image description here

  • $\begingroup$ Hi, as far as I know the American Option (or Bermudan option) should be more expensive even if the ZC rate is zero. Also the number of time steps should not matter: if I only have 12 time steps than this would be the same as considering a Bermudan option instead of an American option. Also I tried the same calculation for weekly time step and the results were similar. I think there is some bug in the Longstaff-Schwarz implementation and this causes this error. $\endgroup$
    – Cettt
    Mar 3, 2020 at 15:48
  • $\begingroup$ no it not more expansive. With no discounting, there isn't much benefit to exercising early, which is the only way it can be worth more. $\endgroup$ Mar 3, 2020 at 15:53
  • $\begingroup$ Do you have a reference for that? But anyway, the American Option price should not be below the European price. $\endgroup$
    – Cettt
    Mar 3, 2020 at 16:05
  • $\begingroup$ i edited the answer with the proofs. $\endgroup$ Mar 3, 2020 at 16:28
  • $\begingroup$ thank you, I upvoted your answer, however I am not sure about the numerics part. $\endgroup$
    – Cettt
    Mar 3, 2020 at 17:03

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