# 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$stock to increase more than 100%, but not decrease more than 100%. So, how to interpret this number when it is greater than 100%? ## 2 Answers It seems that you are thinking of the volatility as some sort of standard deviation of your stock price. It is not. In the BS model,$\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\%$). Note that even a positive random variable's standard deviation can be much larger than its mean if its right tail is fat enough. Consider the family of lognormal distributions for example, the standard deviation can be arbitrary large for a given mean. • Very good answer! Log-return can be less than$-100\%\$ and the price is still positive. And the standard deviation can be whatever. – Ric Feb 24 '15 at 16:30
• Thanks. So it is not right to say: if IV = 110%, it means that the stock has a 33%(length of 1 SD) chance of increasing 110%, and 33% chance of decreasing 110%? – Victor123 Feb 24 '15 at 18:21
• What I am really getting at is: if IV =110%, what guess can i make about the possible size of the stock move? – Victor123 Feb 24 '15 at 18:22

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.