Crash O Phobia and the reason for higher pricing of out of the money puts

Question

In my stochastic finance course we are currently talking about Implied Volatility and Crash O'Phobia.

According to Rubinsteins Crash O'Phobia, put-sellers attach a higher probability to the left tail (instead of the log normal distribution of BMS) and therefore price their put option higher. This makes sense, but i don't understand why with a lower strike price there is a higher implied volatility, for example as in this picture:

Shouldn't it be the opposite: If the market crashes, then the put buyer can excercise the put option I sold him and the higher the strike price of that put option, the higher my loss will be / the higher will be the profit of the put option buyer?

Answer 1

Option prices encode the market's consensus on the probabilities of future moves in the underlying price. (Thus, if you speculate in options, considering some to be undervalued or overvalued, you are betting on your own probabilistic forecast that differs from the consensus.) The Black-Scholes model is based on an assumed lognormal distribution of the underlying. That is, the log return over the next time interval T is normally distributed with standard deviation IV * sqrt(T), and this implies a pricing formula for all options on that underlying in terms of a constant implied volatility (IV) parameter. Note that the option price is always an increasing function of IV (all else equal, a higher IV gives a higher option price).

Since the 1987 crash, as you note, the consensus probabilities for stock returns typically include heavier-than-lognormal tails, especially on the downside. This is reflected in higher prices for out-of-the-money puts than the Black-Scholes model predicts, because the chance of those puts becoming in the money is higher. (For example, 10 times standard deviation or "10-sigma" moves are not unheard of in the stock market, even though they would be vanishingly rare according to the normal distribution.) Thus, if we still use the Black-Scholes formula to define a per-option IV from the market prices, we find that the pricing at those downside strikes corresponds to a higher IV.

See this answer and also this question.

Answer 2

If you plot implied volatility (IV) against strike prices, several curves occur:

• Volatility Smile is a U-shaped curve

• Reverse Skew (aka Volatility Smirk) is where lower strikes have higher IV than higher strikes (ITM calls and OTM puts are more expensive than OTM calls and ITM puts). This is the pattern depicted in your link. The popular explanation for this is that investors are more concerned about market crashes and therefore they buy protection puts. Another explanation is ITM calls are a good alternative to stock ownership, offering leverage and higher ROI. Both of these scenarios would lead to greater demand for ITM calls and OTM puts.

• Forward Skew is the mirror image of of Reverse Skew (OTM calls and ITM puts are in greater demand).

There are a number of Questions (and Answers) on Stack about this. Perhaps some of them might better address your question..

Answer 3

With the disclaimer that I acknowledge that Bob (the member whom I consider our resident expert and author of the current answer) is a few levels above me in options knowledge, I'll offer a layman's answer -

A stock has a volatility.

BS (The options pricing equation) offers a 'fair value'.

Since one can use BS to reverse-engineer the equation, an option has an 'implied volatility', i.e. the number that makes the equation fit.

Out of the money options tend to have a price that is exaggerated (i.e. people wiling to pay more than the model says it's worth), and therefore, the IV shows as higher than it otherwise would.