Why does implied volatility show an inverse relation with strike price when examining option chains? - Quantitative Finance Stack Exchange most recent 30 from quant.stackexchange.com 2019-06-25T13:45:02Z https://quant.stackexchange.com/feeds/question/27 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://quant.stackexchange.com/q/27 18 Why does implied volatility show an inverse relation with strike price when examining option chains? Arjen Kruithof https://quant.stackexchange.com/users/53 2011-01-31T21:57:05Z 2015-12-18T14:11:18Z <p>When looking at option chains, I often notice that the (<strong>broker calculated) implied volatility has an inverse relation to the strike price</strong>. This seems true both for calls and puts. </p> <p>As a current example, I could point at the SPY calls for MAR31'11 : the 117 strike has 19.62% implied volatility, which decreaseses quite steadily until the 139 strike that has just 11.96%. (SPY is now at 128.65)</p> <p>My intuition would be that volatility is a property of the underlying, and should therefore be roughly the same regardless of strike price.</p> <p>Is this inverse relation expected behaviour? What forces would cause it, and what does it mean? </p> <p>Having no idea how my broker calculates implied volatility, could it be the result of them using alternative (wrong?) inputs for calculation parameters like interest rate or dividends?</p> https://quant.stackexchange.com/questions/27/-/56#56 5 Answer by barrycarter for Why does implied volatility show an inverse relation with strike price when examining option chains? barrycarter https://quant.stackexchange.com/users/59 2011-02-01T03:32:58Z 2011-02-01T03:32:58Z <p>"My intuition would be that volatility is a property of the underlying, and should therefore be roughly the same regardless of strike price". </p> <p>I agree, but the market doesn't. People who buy out-of-money calls tend to be more optimistic than those who buy at-the-money calls, so out-of-money calls are "overpriced" and thus have a higher volatility. </p> <p>Oddly, people who buy deep-in-the-money calls ALSO tend to be more optimistic, so these calls ALSO have a higher implied volatility. Why?<br> You can buy a deep-in-the-money call for much less than the stock price. When the stock goes up 1 point, the deep-in-the-money call goes up almost 1 point too, so you get the same gain for less investment (ie, leverage). </p> <p>You probably also noticed implied volatility varies with expiration date too. </p> <p>Ultimately, the market determines how much an option is worth, and thus the volatility. Black-Scholes' belief that volatility was a fundamental characteristic of an instrument isn't really accurate. </p> https://quant.stackexchange.com/questions/27/-/58#58 11 Answer by Joel Spolsky for Why does implied volatility show an inverse relation with strike price when examining option chains? Joel Spolsky https://quant.stackexchange.com/users/83 2011-02-01T03:43:35Z 2011-02-01T03:43:35Z <p>The skew is almost always bid for puts on the stock market. When stocks go down, people tend to panic and volatility goes up as a result. Since the puts get more vega when the market goes down, they trade at higher vols. Read up on stochastic volatility for a more in-depth explanation.</p> https://quant.stackexchange.com/questions/27/-/95#95 8 Answer by nicolas for Why does implied volatility show an inverse relation with strike price when examining option chains? nicolas https://quant.stackexchange.com/users/172 2011-02-01T14:44:06Z 2011-02-01T14:44:06Z <p>It can be shown using a combination of calendar and butterfly that one can lock now the future variance <em>conditionally</em> to the spot being around some specific level (local vol). So if you bought it and it gets realized higher <em>and</em> the spot is there, you get money. if the spot is not there, you are neutral. Another way to look at the dependency of spot level and vol level is just using a regular delta hedge strategy whose PL is path dependent on what spot level is regarding strike when volatility gets realized.</p> <p>These dependency combined with the market sentiment that volatility is higher when spot goes down leads to higher vol price for options with lower strike.</p> https://quant.stackexchange.com/questions/27/-/237#237 5 Answer by glyphard for Why does implied volatility show an inverse relation with strike price when examining option chains? glyphard https://quant.stackexchange.com/users/214 2011-02-07T22:56:54Z 2011-02-07T22:56:54Z <p>Short answer: volatility skew. Longer answer: investors are willing to pay more for out of the money puts (disaster hedge).<br> This buying bids up the price of puts, which makes the volatility implied by those prices go up.<br> calls and puts at the same strike must trade roughly at the same implied volatility otherwise there is arbitrage, this is why you see the same phenomenon for lower strike calls. (investors are less willing to do this when buying out of the money calls(higher strikes), and so those options typically trade at lower bids, and lower implied volatility.</p> https://quant.stackexchange.com/questions/27/-/411#411 11 Answer by Brian B for Why does implied volatility show an inverse relation with strike price when examining option chains? Brian B https://quant.stackexchange.com/users/254 2011-02-10T19:50:21Z 2015-12-18T14:11:18Z <p>That implied volatility you are observing was calculated using the standard Black-Scholes model (BSM). As we all know, no model is a perfect representation of reality. The variation (or <em>skew</em>) you observe is a consequence of the model being wrong.</p> <p>Let's think about the implications of the BSM <strong>not</strong> being exactly correct and everybody knowing that fact. Market prices cannot come solely from the model in this case. In particular, an important result is that (since the model is incorrect) even if you were to plug in the "right" value for every parameter, you would not get the market option prices. </p> <p>Any model, including the BSM, can be run "backwards", by which we mean here that it can start with an option price and derive an <em>implied</em> parameter. If the model has $M$ parameters $p_1, p_2, \dots, p_M$ that are normally used to find a price $V$, then we can also choose any one of the parameters, call it $p_n$, to derive from an observed price $W$ (normally by root-finding techniques).</p> <p>That is to say,</p> <p>$$V = f(p_1, p_2, \dots, p_M)$$</p> <p>gets inverted to $g=f^{-1}$ in parameter $n$ to form</p> <p>$$p_n^{\text{impl}} = g(W, p_1,\dots,p_{n-1},p_{n+1},\dots,p_M).$$</p> <p>It so happens that for the BSM most of the parameters are reasonably easy to observe (strike, interest rate, etc.) while volatility is a rather more mysterious quantity, especially because the BSM needs <em>future</em> volatility rather than past volatility. Therefore, the market practitioners tend to pick on that parameter and talk about <em>implied volatility</em> even though in principle we could do everything in terms of, say, <em>implied dividend yield</em>.</p> <p>In any case, since the model is wrong, we don't expect to get the exact right option prices when we run the model forward, and therefore don't expect to get one "right" parameter when we run it backward. That's why you see variation in volatility by option strike.</p> <p>Now, as to the exact shape of that variation (decreasing implied volatility with strike), there are quite a few explanations and they are not mutually exclusive. For example, a somewhat more credible model than the plain old BSM is Black-Scholes With Jumps (BSJ), where the underlying price can take a sudden dive. You need extra parameters to describe the jumps of course, but the result is a model whose implied volatility skew is "flatter". Because those jumps are to the downside, they show up as higher prices(=higher BSM implied volatility) for the low-strike options.</p> <p>Other explanations involve transaction costs, discrete stock price processes, bankruptcy, stochastic volatility, market psychology, etc.</p> https://quant.stackexchange.com/questions/27/-/720#720 4 Answer by user586 for Why does implied volatility show an inverse relation with strike price when examining option chains? user586 https://quant.stackexchange.com/users/0 2011-03-14T07:37:28Z 2011-03-14T07:37:28Z <p>It's well know by everybody that when the market prices goes up, the <a href="http://www.livevol.com/historical_files.html" rel="nofollow">implied volatility</a> goes down and vice versa. So the strike prices is having the reverse relationship when examinig the option chains.</p> https://quant.stackexchange.com/questions/27/-/1603#1603 5 Answer by John Channing for Why does implied volatility show an inverse relation with strike price when examining option chains? John Channing https://quant.stackexchange.com/users/1161 2011-08-06T14:03:18Z 2011-08-06T14:03:18Z <p>The Black-Scholes model is based on a set of assumptions about the distribution of asset returns which are incorrect for real markets. For mathematical simplicity, returns are assumed to be normally distributed, but in reality the distributed is asymetrical (skew) and has fat tails (kurtosis).</p> <p>If you back out implied volatility from option prices across a range of strikes at the same expiry, you will observe that it is not a constant, but a function of strike that tilts downwards (skew) and curves upwards (smile).</p> <p>You can think of the value of an option in a number of ways, including:</p> <ul> <li><p>the <a href="http://en.wikipedia.org/wiki/Expected_value" rel="noreferrer">Expected Value</a> e.g. the sum of the size of the pay out multiplied by the probability of getting it</p></li> <li><p>The cost of hedging the option</p></li> </ul> <p>Where implied volatility demonstrates skew it is because the Black-Scholes model is an approximation and to get the right price for the option you need to adjust its value.</p> https://quant.stackexchange.com/questions/27/-/10262#10262 0 Answer by prespbj for Why does implied volatility show an inverse relation with strike price when examining option chains? prespbj https://quant.stackexchange.com/users/7136 2014-02-15T13:57:50Z 2014-02-15T13:57:50Z <p>you get a volatility skew by imposing a neumann-like barrier</p> <p>if market makers think a stock won't surpass a certain threshold, a skew is inevitable if one were to match the pricing under a barrier with the BS formula</p> <p><a href="https://en.wikipedia.org/wiki/User:Barrieroption/sandbox" rel="nofollow">https://en.wikipedia.org/wiki/User:Barrieroption/sandbox</a></p>