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Checking the Arithmetic Mean Option Formula’s Numerical Example

Article Quant Q&A · Author: Henry P
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Summary

This exchange examines a numerical discrepancy in a paper’s purported exact formula for pricing an arithmetic-average Asian option. The questioner reports difficulty reproducing a stated value and asks whether the paper or their Monte Carlo implementation is wrong.

The response says the paper’s formula is correct but identifies an arithmetic error in the cited example: evaluating the displayed logarithmic expression gives approximately 0.500417, rather than the reported 0.498. That discrepancy could explain why the example does not match an implementation. The exchange is narrowly focused on checking one calculation; it supplies no fuller derivation of the pricing formula, no Monte Carlo convergence analysis, and no broader validation of the paper’s assumptions or applicability.

Key ideas

  • A mismatch with a published example can result from an arithmetic error in the example itself.
  • The response confirms the formula and recalculates the disputed logarithmic expression.
  • The exchange does not analyze the Monte Carlo method or establish the formula’s broader limits.

Tags

OptionsDerivatives pricingStatistics
Full text
# Monte Carlo for Asian Pricing


# Monte Carlo for Asian Pricing












I'm trying to verify the accuracy of my Monte Carlo method for pricing mean options. I came across this paper that supposedly gives an 'exact' solution for the arithmetic mean option (asian). It's a relatively short paper, but is it reliable?

I can't seem to reproduce the results shown in the paper and it is a simple calculation (or so I thought).

Please look at this extract from the paper (pg. 3):

I tried to put this into MATLAB, but I don't get the 0.498 mention, instead I get 0.5004. Does this paper make a mistake? When I compare this solution to my Monte Carlo, the error doesn't seem to get smaller as it should (by LLN etc.).

Any pointers in the right direction? Here is my attempt to implement the formula given in the paper into MATLAB:

## Answer by Sanjay (score 2)

https://quant.stackexchange.com/a/44731

The paper is reliable and the formula is correct. However as you mention yourself there is an error. $$ \frac{\log \left(\frac{e^{0.01}-1}{0.01}\right)}{0.01} = 0.500417 \neq 0.498 $$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.