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I am not sure if this question is appropriate here, but I am just going to give a shot. It is related to an empirical observation about the price of a call option. Specifically, I was looking at a very out-of-the-money call on AMZN, namely AMZN Jan 15 '21 $3000 call. I noticed that it follows the price of the stock (AMZN) very closely.

I then plotted the correlation coefficient between the price of the call option and the price of the underlying stock and here is the result (according to trading platform I am using):

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According to this plot, the correlation coefficient between these two fluctuates, but often it is very high, i.e. $\ge 0.9$. Current stock price (at close) is about \$2408 and option price is roughly \$104 (taking the midpoint between bid and ask). Again according the trading platform, the $\delta$ for this option is $0.286 \approx 0.3$ (with an IV of 33.6%). Assuming they are using the Black-Scholes model to price the option and they are doing the calculation correctly, it seems that there is a big mismatch between the model and the actual data (i.e., between 0.3 and 0.9)?

Is there an explanation for this? Am I missing something? I am not an expert in quantitative finance, but do people in the field care about these discrepancies between theory and empirical evidence? Are there any good papers on it to explore more?

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6 Answers 6

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Correlation of 1 means the option price moves in the same direction of the underlying with perfect association, it says nothing about how much the option price moves. In general, correlation coefficients measure the strength and direction of a linear statistical relationship, not the magnitude of that relationship

In your example, a correlation coefficient of .9 implies a strong, positive linear relationship between the price of the AMZN call and the price of AMZN equity. Delta quantifies the magnitude of that linear relationship. A Delta of .3 implies that the price of the AMZN call will rise by 30% relative to the rise in price of AMZN equity.

Correlation = Strength and direction of linear relationship

Delta = Option price sensitivity (in dollar terms) of that linear relationship

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  • $\begingroup$ Thanks. Based on your answer, I realized my mistake. I will add an answer explaining it. In statistical terms, delta would be more related to the covariance between the two rather than correlation. Thanks for the references. $\endgroup$
    – passerby51
    Apr 17, 2020 at 5:58
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The delta is not a measure of correlation. You have misinterpreted the two. Options are called derivatives because they are precisely highly correlated, to the underlying stock so in statistical terms estimating the stock/option correlation is a non sequiter, of no purpose. The delta measures two things and this is what you want to concentrate on, firstly it measures how much in or out of the money the option is, .5 being at the money above in the money below out of the money. And secondly it is a proxy for probability, it roughly estimates the likelihood of the option expiring in or out of the money, it by extension also measures the probability of profit. Though it slightly exaggerates this because it adds the interest rate whereas the true probability of profit formula minuses interest rate, but with 1% interest rates the error is marginal but if interest was 9% it could significantly overstate probability of profit. Options are derived from stocks so have to be highly correlated. The real issue to option trading is delta and theta and gamma, and volatility, theta measures change in volatility and its effect on the option price. gamma measures the change in delta when delta changes. Volatility is key I suggest working through Introduction to quantitative finance by Paul Willmot. For day to day trading McMillans Options as a Strategic Investment. And also the classic on options and volatility by Sheldon Natenberg Option Volatility and Pricing. Successful Trading!!!!!

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  • $\begingroup$ OK, thanks. I realized my mistake. I will add an answer explaining it. Thanks for the references. (I understand Black-Scholes formula to some extent and the role of volatility and all that. Option's delta would be the derivative of the option price w.r.t. the stock prices, so there is a natural relation with regression, correlation coeff., etc.) $\endgroup$
    – passerby51
    Apr 17, 2020 at 5:57
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Based on the answers given, I realized my mistake (which is a bit embarrassing in the hindsight.) So here is roughly what is going on form a statistician's perspective:

Let us say the price of the call option and the stock are $C_t$ and $S_t$ and the relation is $C_t = f(S_t;t)$. Let us say that the interval we consider is small enough that the dependence of $f(\cdot,t)$ on $t$ can be ignored, i.e., we assume $C_t = f(S_t)$. By a first-order Taylor expansion around $S_{t_0}$ (again rough approx.), we would have $$ C_t \approx C_{t_0} + \delta (S_t - S_{t_0}) $$ where $\delta = f'(S_{t_0}) = \partial f / \partial S\mid_{S=S_{t_0}}$. Letting $\alpha= C_{t_0} - \delta S_{t_0}$, we can model the price as $$ C_t = \alpha + \delta S_t + \epsilon_t $$ in a small interval around $t_0$. Fitting the regression by least squares will give $$\hat \delta = \frac{\rho_{CS}}{\rho_{SS}}$$ where $\rho_{CS} = \frac1{|I|} \sum_{t \in I} (C_t - \bar C)(S_t - \bar S)$ is the empirical covariance between $C_t$ and $S_t$ and $\rho_{SS} = \frac1{|I|} \sum_{t \in I} (S_t - \bar S)^2$ is the empirical variance of $S_t$. On the other hand the (empirical) correlation coefficient between the two would be $$ \hat r = \frac{\rho_{CS}}{\sqrt{\rho_{CC} \rho_{SS}}}. $$ The correlation coefficient would be the sensitivity of the normalized price of the call option to the normalized price of the stock when both are standardized by their standard deviations.

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No I think your math might be not completely correct. options are non linear so can not be compared to linear regression derived statistics, it is derived from ito's lemma which states a taylor expansion of a terms around a Brownian motion function only go to the second degree hence the mean and variance describe the function, and the resultant equation of the black scholes model is a Partial differential equation, non-linear. There is no beta, no correlation co-efficient etc, it is a whole different ball game.

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  • $\begingroup$ You have to think a little bit outside the box to see where I coming from. Any nonlinear function can be approximated by a linear function. The question would be whether that approximation is useful. (It generally is, hence the usefulness of Taylor expansion). Forget about Black-Scholes for a moment. You are given two time series and you want to see their relation. The simplest thing I would do is to regress one onto the other and see what happens. Does it capature the relation necessarily perfectly? No. But it is a first step (and often quite powerful at that.) $\endgroup$
    – passerby51
    Apr 22, 2020 at 2:14
  • $\begingroup$ In other words, suppose you want to test whether Black-Scholes formula for the price matches the data. One thing that you can do is to see if the derivate of the option price w.r.t. the stock price matches what is predicted by the theory. How to do estimate the derivative of a function from the data? We fit a rolling linear regression! There are other approaches as well... you could fit more complex models, say a higher-order polynomial smoother and analytically compute the derivaties of those. $\endgroup$
    – passerby51
    Apr 22, 2020 at 2:29
  • $\begingroup$ Taylor expansion goes to the second degree polynomial and can not be linearised for option pricing, not meaninfully. Learn to walk before run and try and re-invent the wheel what you are saying has no meaning or purpose. The black scholes price never matches the market price, you have derive the implied volatility to get the market price, with hist vol, you get an idealised riskless perfect market price, which is only good as a guidepost as to how far from the mean value the price in the market has moved. $\endgroup$
    – Con Fluentsy
    Apr 22, 2020 at 22:37
  • $\begingroup$ Linear regression is absolutely pointless, I am a fully trained econometrician, my stock in trade is linear multivariate regression and its extensions, and it has never ever been even slightly useful in option trading. $\endgroup$
    – Con Fluentsy
    Apr 22, 2020 at 22:40
  • $\begingroup$ The higher order derivatives of the BS model have already been derived, the delta, theta, gamma, rho, vega, and their is a whole range of second order derivatives. These are universally called "THE GREEKS" for obvious reasons. read Wilmott please, stop wasting time. $\endgroup$
    – Con Fluentsy
    Apr 22, 2020 at 22:42
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I am not sure if you are aware volatility of stock price is meaningless , what volatility actually is, is the measure of the standard deviation of the daily change in stock price over a year. You are comparing apples and peaches when you say "The correlation coefficient would be the sensitivity of the normalized price of the call option to the normalized price of the stock when both are standardized by their standard deviations."

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  • $\begingroup$ I don't think I mentioned volatility of stock prices, but volatility of stock prices is not meaningless, at least according to me! I consider volatility as just related to the spread of a distribution. Any random variable has volatility (i.e., non-zero spread or variance, etc.) being it the price or the returns or log returns or whatever. Stock prices are definitely volatile, i.e., not determistic. Whether their variance is useful is a different story. I think we have very different perspectives and it would be hard to settle here. $\endgroup$
    – passerby51
    Apr 22, 2020 at 2:11
  • $\begingroup$ n the context described volatily is meaningless, all I do in Option trading is based on Volatility, I just do volatility arbitrage, to me volatility is the key, It is far more than the range or variation around a stock and it is the second moment of a distribution not a distribution in itself. Your knowledge is shallow, read a good quant finance book such as willmot. $\endgroup$
    – Con Fluentsy
    Apr 22, 2020 at 22:31
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No you are off on a completely irrelevant tangent, you can not say forget black scholes, unless you price with the Cox Ross or other Binomial or Trinomial models, these are the absolute essence of options trading, there is no parrallel linearisation of this, and as I pointed out in BS model Taylor expansion goes to the second degree polynomial, please refer to ito's lemma. The black scholes price never ever matches the real price, first you have to derive the market implied volatility and then feed that back to get the real price. Using one of the many other volatility methods GARCH, EWMA, Standard Deviation, etc, you can come up with an idealised perfect riskless market price, but it bears no resemblance to the market price, and is only helpful because it shows the possibility of regression to the mean. Options trading is literally based on rocket science as it is, why are you so insistent on unnecessarily irrelevant complicated mathematics, learn before you try and reinvent the wheel, you are not sure of what you are doing at all.

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