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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 Out of the money calls are low as the probability that they will turn into an In the money calls is less.

As a result, volatility is lower than Black-Scholes-Merton formula assumption. So, their prices will go up. But BSM formula assumes constant volatility. So it underprice an Out of the money calls and In the money puts.

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 Out of the money calls are low as the probability that they will turn into an In the money calls is less.

As a result, volatility is lower than Black-Scholes-Merton formula assumption. So, their prices will go up. But BSM formula assumes constant volatility. So it underprice an Out of the money calls and In the money puts

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 Out of the money calls are low as the probability that they will turn into an In the money calls is less.

As a result, volatility is lower than Black-Scholes-Merton formula assumption. So, their prices will go up. But BSM formula assumes constant volatility. So it underprice an Out of the money calls and In the money puts.

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Source Link

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 OTMOut of the money calls are low as the probability that they will turn into an ITMIn the money calls is less.

As a result, volatility is lower than BSMBlack-Scholes-Merton formula assumption. So, their prices will go up. But BSM formula assumes constant volatility. So it underprice an OTMOut of the money calls and ITMIn the money puts

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

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 Out of the money calls are low as the probability that they will turn into an In the money calls is less.

As a result, volatility is lower than Black-Scholes-Merton formula assumption. So, their prices will go up. But BSM formula assumes constant volatility. So it underprice an Out of the money calls and In the money puts

Source Link

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