Why Implied Volatility Can Exceed 100%

Summary

The document addresses the misconception that volatility is bounded by the largest possible percentage decline in a stock price. In the Black–Scholes framework, volatility scales the standard deviation of the asset’s log return over the option’s life. That standard deviation has no mathematical upper bound, and quoting it as a percentage does not mean it is a cap on the stock’s possible price move. The discussion also notes that a positive price distribution can have a standard deviation much larger than its mean when it has a sufficiently heavy right tail.

A second explanation connects very high implied volatility to the model used to infer it. If traders price in occasional large jumps, especially downward jumps, options near expiration may be expensive relative to a smooth Black–Scholes model; fitting that model to those prices can produce a high implied volatility. The exchange offers intuition rather than a calibration example. Its jump explanation is a possibility, not a claim that every high implied volatility signals jumps or that the Black–Scholes interpretation fully describes real returns.

Key ideas

• Implied volatility is tied to the standard deviation of log returns in the Black–Scholes model, not a bound on price changes.
• A log-return standard deviation can exceed 100% because it has no mathematical upper limit.
• A positive price distribution can have a standard deviation larger than its mean when its right tail is sufficiently heavy.
• Prices that reflect possible jumps may translate into high implied volatility when interpreted through a smooth Black–Scholes model.

Tags

# Why an option has sometimes and implied volatility greater than 100%?


# Why an option has sometimes and implied volatility greater than 100%?












Sometimes, in an option chain, the implied volatility of an option is greater than 100% .

How is this possible? I mean, it is possible for 100$ stock to increase more than 100%, but not decrease more than 100%.

So, how to interpret this number when it is greater than 100%?

## Answer by AFK (score 16, accepted)

https://quant.stackexchange.com/a/16708

It seems that you are thinking of the volatility as some sort of standard deviation of your stock price. It is not.

In the BS model, $\sigma\sqrt{T}$ is the standard deviation of the log-return $\log(\frac{S_T}{S_0})$. There is no mathematical upper bound to its standard deviation. There is also no mathematical problem with returns being negative either. Quoting volatility as a percentage is common practice but does not necessarily make sense (in stochastic volatility models, vol of vol parameters can often be calibrated to $\sim 300\%$).

Note that even a positive random variable's standard deviation can be much larger than its mean if its right tail is fat enough. Consider the family of lognormal distributions for example, the standard deviation can be arbitrary large for a given mean.

## Answer by Richi Wa (score 3)

https://quant.stackexchange.com/a/16718

The answer of AFK is very good and accurate in a BS setting.

Thinking of jumps I would add the following: If we assume that stocks sometimes move in jumps (usually downwards) then it is clear that ATM or OTM options shortly before expiration with a price that accounts for the possibility of a jump - which is therfore quite high - only fit in the BS framework with large implied volatilities.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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