3

Example option chain quote

I'm trying to understand implied volatility (IV) better. Recently, I was looking at $YUMC's option chain (date: 07/10/2017) and in particular the 27.5 strike price with an IV of 0%.

I'm trying my best to understand why the 27.5 strike price has an IV of 0%, but the strike price directly above it and below it have an IV of 56.63% and 97.75% respectively.

As I understand it, implied volatility represents the expected gyrations of an options contract over it's lifetime.

So it bewilders me how the IV can be 0% at 27.5, but greater than 0% for strike prices just above and below, it makes no sense and I have no idea how to interpret it.

I even tried staring at the math behind the options pricing model to see if that could make more sense for me but that didn't help.

BS-Model

If anyone could help me better understand and interpret this I would really appreciate it. Thanks!

  • Are you doing a school exercise or something ? Do you have a text book from which you are reading ? Any good text book will explain this. If you do want somebody else to answer this try posting it in the Quantitative Finance forum. – DumbCoder Jul 10 '17 at 9:00
  • This isn't for a class I study this stuff on my own, If you have a book to recommend that would be useful. – HappyTurtle Jul 10 '17 at 16:18
4

As I understand it, Implied Volatility represents the expected gyrations of an options contract over it's lifetime.

No, it represents that expected movement of the underlying stock, not the option itself. Yes, the value of the option will move roughly in the same direction the value of the stock, but that's not what IV is measuring.

I even tried staring at the math behind the Options pricing model to see if that could make more sense for me but that didn't help.

That formula is correct for the Black-Scholes model - and it is not possible (or at least no one has done it yet) to solve for s to create a closed-form equation for implied volatility. What most systems do to calculate implied volatility is plug in different values of s (standard deviation) until a value for the option is found that matches the quoted market value ($12.00 in this example). That's why it's called "implied" volatility - the value is implied from market prices, not calculated directly.

The thing that sticks out to me is that the "last" quoted price of $12 is outside of the bid-ask spread of $9.20 to $10.40, which tells me that the underlying stock has dropped significantly since the last actual trade.

If the Implied Vol is calculated based on the last executed trade, then whatever algorithm they used to solve for a volatility that match that price couldn't find a solution, which then choose to show as a 0% volatility.

In reality, the volatility is somewhere between the two neighbors of 56% and 97%, but with such a short time until expiry, there should be very little chance of the stock dropping below $27.50, and the value of the option should be somewhere around its intrinsic value (strike - stock price) of $9.18.

  • Hmm Interesting. So whenever I see an Implied Volatility of 0% should I be interpreting it as something with '0% expected movement of the underlying stock' or should I just be interpreting it as 'No Solution/Error'? – HappyTurtle Jul 10 '17 at 17:46
  • @HappyTurtle No solution/Error - There would be no point in buying an option on something with no volatility (since the underlying value would never move). – D Stanley Jul 10 '17 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.