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If the distribution is skewed to the right,Black-Scholes overprices out-of-the-money puts and in-the-money calls. It underprices in-the-money puts and out-of-the-money calls.

How?

Stock price log-returns distribution is skewed to the right means it is a log-normal distribution.

I know the moments of log-normal distribution and how it relates to normal distribution.

But how does Black-Scholes-Merton formula overprice out-of-the money puts and in-the-money calls and underprice in-the-money puts and out-of-the calls?

It is just because of volatility of option prices at different strike prices or any other reason?

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    $\begingroup$ If (in Excel) I discount the 30 cash flows at an annual rate 10.185% (periodic rate 0.050925) I get a price of 833.82, which is too high. If I discount at 10.942% I get a price 785.52 which is quite close to the desired price. The exact answer seems to be 10.9424%. $\endgroup$ – Alex C Sep 3 at 17:33
  • $\begingroup$ How did you calculate your result? I get 10.9424 as well. $\endgroup$ – D Stanley Sep 3 at 18:40
  • $\begingroup$ @Alex, Yes, You are right. I wrongly computed total number of payments. Would you answer my second question? $\endgroup$ – Dhamnekar Winod Sep 3 at 18:53
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    $\begingroup$ @DhamnekarWinod I would just remove your first question altogether since you found your mistake, and posts should be limited to one topic. $\endgroup$ – D Stanley Sep 3 at 19:21
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    $\begingroup$ The distribution they are talking about is the distribution of log-returns (not of prices). It is normal (no skewness) in the BSM case. The alternative case would be a distribution of $\ln r$ with positive skewness (skewed to the right), but with the same variance as the previous case. These are the alternatives being compared. $\endgroup$ – Alex C Sep 4 at 4:18
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Suppose the distribution of the stock price $S_t$ is positively skewed and thus, assigns more weight (higher probability) to outcomes with high stock prices. The exercise price of an OTM call lies to the right of the current price and thus, the exercise probability increases noticeably which results in a higher option price.

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I am giving answer to my question.

If the stock price log returns distribution is skewed to the right, then $mode<median<mean$ in most of the cases.

The strike price of an OTM calls lies to the right of the current price. So the demand for an OTM calls are low as the probability that they will turn into an ITM calls is less.

As a result, volatility is lower than BSM formula assumption. So, their prices will go up. But BSM formula assumes constant volatility. So it underprice an OTM calls and ITM puts

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