What does that mean? I have seen it a lot at top of chains, but sometimes even in the middle of the chain, with "normal" values above and below it. Please explain, if you would.
Potaire
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IV is 0.00%
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Where did you see that, do you have an example of a specific chain? IV means implied volitility...
an IV of zero might happen for a stock that has already been boughten out a specific price and isn't expected to move at all. Other than that, I can't think of why you'd have an IV of 0.
I see it all the time, all over the place. It is almost always the first few strikes on a given chain. Example: CELG
First 6 strikes for July have 0% implied vol, and August has the first 3 at 0. How could an option have zero vol at one strike, but there IS vol at another strike??? In fact, I don't see how there could be zero vol for ANY strike, unless the company closed down.
Potaire
Perhaps there is a zero in the numerator and/or denominator in the equation for computing implied volitility?? If so, what are we being told, and how should we react??
Potaire
Here is what I think is happening. All the options with 0 IV are extremely illiquid. In fact most have 0 open interest. There is probably some moving average to compute IV. Let,s say IV is a 30 day moving average. If it hasn't traded in 30 days we get an undefined IV. All the options i saw with 0 IV also hac 0 volume. So I think they haven't tdaded recently enough for the IV to be measured. This doesn't occur until CELG is into the leaps. Also notice when this happens the infomation is given in red.
runningpair, I thought you might be on to something, but not. Look at the CELG chain for July:
IV Delta Open Int Strike
0.00% 1.00 370 35
107.61% 1.00 1267 40
0.00% 1.00 1848 45
0.00% 1.00 4607 50
3 outa 4 of the first strikes for July have 0% for IV. But, none of them have 0 for open interest, nor are any of them in red. I've seen the red you're talking about--that means it is some kind of special option, and you must contact your broker for more info. Soooo, we're back to where we started--what does IV=0 mean/imply??
Potaire
I believe what you are seeing is the last two times it traded it traded at the same price. Hence Zero change. Not sure but I think that is what it means.
Hmmmmm, not feeling strong about that one, snowman. By the same token that would mean the 40 strike more than doubled in price from one trade to the next (107%)?? Perhaps one of you staffers can answer this?? Neal or Nicole??
I am not a TK staffer but let me take a stub and you guys correct me where I am wrong. Basically option prices (American options, which can be excercised at any time before expiration) consist of intristic value and time value. Intrinstic value is how much the option is in the money. Time value is determined by many factors including time to expiration, interest rates and implied volatility. If an option becomes deep in the money, the time value disappears and IV drops to zero. Basically the market thinks that the option will alwys expire in the money and the value moves with the underlying, thus delta = 1. Because of rounding models may produce different results for some strikes
I think there is an article on investopedia describing the concept. Google "implied volatility" and "time value"
If IV=0 means the feeling is that the option will NOT ever go out-of-the-money, than that would make sense EXCEPT-----------
notice the strikes above--there is the 107% basically in the middle of all the 0%. So, how could the 45 & 50 strikes have 0% chance of leaving the money, but the 40 strike, which is FARTHER in the money than the 45 & 50, has a 107% chance of leaving the money? The 45 & 50 would be out-of-the-money long before the 40 would be. Your theory sounded good up until that point.
Potaire
Potaire, good catch. My take though is the same - computer rounding error based on current market conditions. If I am right the 107% IV at the middle strike will disappear in a while (maybe even a few minutes) and something else may appear at a different strike. IV is derived of current option prices and it requires reversing a differential equation which may not have exact solutions. IV is meaningless when Delta =1, options move as the underlying, however the computer does not know this. In addition market makers will adjust their bid/ask spreads which will introduce a different variable for some time until it converges back to IV = 0
I still think that it must be in the equation itself. As you said, there are several different components thrown into the IV formula. Something and another something must add up to zero, or they subtract out to zero. Either way, a zero in the numerator or in the denominator will result in the entire equation equalling a big, fat zero (actually, a zero in the denominator is technically considered undefined--you can not divide by a zero in any circumstance. But, only us math freaks would be aware of that). If we could find a place that explains the Black-Scholes model in ENGLISH (instead of math symbols without definitions), we could probably see that it is possible for, say, the number of days left minus the interest equals zero on one particular day, or something along those lines. The Black-Scholes models I have seen leave me more confused than I was BEFORE I looked at their formula! lol And I have a lot of math from my college days!
This is pretty cool, though---lots of opinions and theories from several different people. We are all fairly smart or we wouldn't be involved with the stock market to begin with (although this last week could prove that the smart people are NOT involved with the stock market!! lol ). Certainly one of us will will stumble across the correct path eventually. Let us stick with this and we will find out soon enough.
Thanks--keep it going
Potaire
http://www.erieri.com/scripts23/blackscholes/EuropeanCall.htm This website does a decent job of explaining the Model.
Potaire
I agree with you Fart as far as the Delta--obviously Delta being at, or VERY CLOSE (.997), to 1.0 has something to do with this. All of the zero IVs I've seen have a Delta equal to .997 or higher--usually 1.0. Hmmmmm.
Potaire SM (Saint Mary's) also has a zero IV for one strike
I hope you don't mind if I jump in here. This is a very candid and mostly accurate conversation about Implied Volatility (IV). First to address the skewing of implied volatilities in the chain between in-the-money and out-of-the money options, here is a blog post on the topic from 2/04/08 that explains this concept in detail.
Next, is the legit zero implied volatility on options with very little time premium in the option. Time premium is also referred to as extrinsic value. This part of an option's price has a time component and a volatility component. The more time to expiration adds value and the larger the IV also adds value to this portion of the option's price. You're correct to state that the IV is usually zero when the delta is close to one. Now this is because the option is so deep in the money that there is a very high probability that the option will be exercised. So it will probably become a stock position eventually. So what happens to the option? The option starts to look like a stock position and all the time premium goes away. Without extrinsic value there can not be an implied volatility number for that option. When the extrinsic value for way out of the money options goes away the exact opposite thing will happen to IV, the implied volatility number will sky rocket. Why? Because it really is a worthless option that has an ask price of .05. So the IV has to be extremely large to justify the 5 cent price, the market marker is not going to let you buy it from them for nothing so they leave a offer out there of 5 cents. Bottom line here is that the model is breaking down in each case because the deep in the money option is becoming a stock position and the way out of the money option is becoming a nothing position. So to be frank - ignore it.
To address the next situation is when the IV is zero and it is not a legit situation. Some examples mentioned here are just situations where the chains are picking receiving bad data from the quote vender. So if you think there should be an IV and it is shows zero go to a different source. Maybe try the options calculator under the tools menu.
Moral to all of this, unless you're trying to do some type of arbitrage play the implied volatility of deep in-the-money and far out-of-the-money options is not very helpful to us retail traders. It is important to understand the concept of skew and once again it is explained in this blog post, but besides the reasons mentioned in that post it does not provide much use to retail option traders.
The main implied volatility number to worry about is the at-the-money strike volatility. And a different sent of blog posts explains this concept. Here is the link to the first post in the 5 part series titled "why do we care about volatility?". Hope this helps...
Regards,
Brian (Og)
Ahhhh, excellent Brian!!! You are who I was hoping would help us to the answer! Many thanx!
Sooo, much of OldFart's theory is pretty close to the mark---good job Fart. The thing is so far in the money that IV is really a non-issue at that point. The example of CELG must have been a non-legit case of the zeros, cuz the 45 and 50 were 0 but not the 40, which is even more ITM. I DID run it on TK's Option Tool, and did indeed come up with different IVs than zero. Musta been bad info on that one--I shoulda checked another source. Also sounds like I might have been in-the-neighborhood with the math angle---at extreme edges of a chain the math would yield infinitely large, or infinitely small results for IV--either of which would be completely useless for our purposes. Hopefully I understand Brian's explanation correctly.
Many Thanx, Brian, and all who took a stab at this,
Potaire
Thank you Potaire for the summary and kind words, everybody added solid fodder to the forum. The reason the option calculator will sometimes provide IV information when zeros are on the chains is that the calculator uses a different data provider. The option calculator data is provided by iVolatility.com, usually one of the two sources will have the data needed.
Regards,
Brian (Og)
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