I always thought volatility was just variance ^ (1/2). Now I'm reading this book and it's saying that the two are different concepts. Excerpts include:

Partly due to its use in Black-Scholes, historically, volatility has been used as the measure of deviation for financial assets. However, the correct measure of deviation is variance (or volatility squared). Volatility should be considered to be a derivative of variance. The realisation that variance should be used instead of volatility led volatility indices, such as the VIX, to move away from ATM volatility (VXO index) towards a variance-based calculation.

There are three reasons why variance, not volatility, should be used as the correct measure for volatility. However, despite these reasons, even variance swaps are normally quoted as the square root of variance for an easier comparison with the implied volatility of options (but we note that skew and convexity mean the fair price of variance should always trade above ATM options).

What does that even mean? Why would variance be a be a better measure of volatility if it's just the square of volatility. Isn't that like saying feet are better than inches for measuring stuff?

Variance takes into account implied volatility at all stock prices. Variance takes into account the implied volatility of all strikes with the same expiry (while ATM implied will change with spot, even if volatility surface does not change).

How in the world does variance take into account the IV of all strikes? If I take the IV of a 30d Call and square it, suddenly that gives me more information than just the IV itself?

  • $\begingroup$ You are getting confused because volatility and variance in your text do not mean times series volatility and time series variance. Those refer to implied volatility on one hand (which is defined for any option contract) whereas variance in the text context is meant to represent some measure of overall implied risk in the market. $\endgroup$
    – Ezy
    Jan 10, 2019 at 1:27
  • $\begingroup$ which book??? pls. give a reference $\endgroup$
    – vonjd
    Jan 10, 2019 at 9:28

2 Answers 2


Let's skip calling it volatility and variance. Let us deal with variance and standard deviation.

For normally distributed variables, it is very important to distinguish between the true variance and the estimator of the variance and the estimator of the standard deviation.

Variance in its raw form is important in defining the statistical law that governs the sample data, while the standard deviation plays no role (unless you pretend to square it). Variance is the second central moment of the normal distribution. Variance, in its raw form, defines all mean-variance models.

That is great if you know it, but if you do not know it, then the estimators are a bit more of a headache than most people realize.

Finance models such as the CAPM, APT, Fama-French or Black-Scholes are strictly Frequentist models. Constructed in a Bayesian space, the results won't come out the same at all. So we need to remain inside Frequentist conceptualizations of a mean, a variance, and a standard deviation.

The minimum variance unbiased estimator of the mean is $$\bar{x}=\frac{\sum_{i=1}^n}{n}.$$ This is also the maximum likelihood estimator.

The maximum likelihood estimator of the variance is $$s^2_{MLE}=\frac{\sum_{i=1}^n(x_i-\bar{x})^2}{n}.$$ Jointly, with the sample mean, this is the most efficient estimator of the variance, but it is a biased estimator. As we are in the Frequentist mindset and not Fisher's likelihood-based model, we have to have an unbiased estimator. Because it would take up a lot of space, I am not going to provide a proof, though it is easy to find online, but the unbiased estimator of the variance is $$s^2_{MVUE}=\frac{\sum_{i=1}^n(x_i-\bar{x})^2}{n-1}.$$

So far, so good, we are in the same place as a first-semester statistics textbook. But what is different is that the unbiased estimator of the standard deviation is not $$s=\sqrt{\frac{\sum_{i=1}^n(x_i-\bar{x})^2}{n-1}}.$$ Instead, this is the sample standard deviation. It is a biased estimator because the square root is a concave function and it distorts the estimator. It isn't generally true that $$\mathbb{E}(f(\tilde{x}))=f(\mathbb{E}(\tilde{x}))$$.

There is a somewhat complicated correction required to make it unbiased. For small samples, the difference is quite large, but for very large samples the impact is nominal and could be ignored.

The issue is somewhat technical, but these models are somewhat fragile and depend strongly on how they are constructed. The mean-variance models are not supported empirically. They are built inside an axiom system and really do depend on it.

It might be better to think of volatility as the scale measurement of a distribution. For the normal distribution, this is the variance. However, not all distributions have a variance, but many still have a scale parameter. The variance is a somewhat narrow and very important conceptualization of scale. Where it exists it has deep mathematical meaning.

Informally, using the sample standard deviation is usually harmless, but if you are really modeling large sums of money, the small effects turn into real amounts of money. It is technically wrong and on large sums it is also a lot of money.

How in the world does variance take into account the IV of all strikes? If I take the IV of a 30d Call and square it, suddenly that gives me more information than just the IV itself?

If the theory is literally true, then the variance fully describes both all errors of commission and all errors of omission in every state of nature. This is why people like working with the normal model. You do not have to do anything else. Because the square root is concave, you actually do lose information within the theory. The statistic wouldn't be sufficient for the parameter. A statistic is sufficient if $$\Pr(\bf{X}|\boldsymbol{\theta})=\Pr(t(\mathbf{X})|\boldsymbol{\theta}),$$ where $\bf{X}$ is the sample, $t$ is some function of the data (a statistic), and $\boldsymbol{\theta}$ are the parameters. This isn't going to be true, so the information in the sample about the parameter is not the same as information in the statistic about the parameter. It can only be less.

  • $\begingroup$ Could you elaborate on As we are in the Frequentist mindset and not Fisher's likelihood-based model, we have to have an unbiased estimator.? Is unbiasedness of an estimator somehow a basic requirement in frequentist statistics? $\endgroup$ Jan 15, 2019 at 13:15
  • $\begingroup$ No, the unbiasedness isn't important, but the frequencies so regions exist is important. They can give rise to a decision function. Fisher gives you a p-value which implies no behavior. Actually, on re-read, I need to clean this up in a few places. The unbiasedness isn't incidental, but the universe won't vanish without it for theory purposes. However, the question was on the unbiasedness of the estimator and how it is linked to theory. $\endgroup$ Jan 15, 2019 at 14:41

It's probably a good book, but I don't entirely agree with the excerpt.

Also, volatility and variance come in many flavours which is not sufficiently explained in the excerpt. For instance, depending on the flavour, the statement "volatility should be considered as a derivative of variance" is shaky at best.

In any case, the statement that variance (or actually expected variance) takes into account the IV of all strikes is correct, and is basically the formula for the variance swap strike as a weighted sum (integral) of implied variances, which you can find in eg Gatheral's book "The volatility surface".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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