# Why do we calculate variance if standard deviation serves the ends well?

I don't understand why we even care for square units. How does it make sense that if you take the squared difference of each data set and mean, then divided it by $$n-1$$ gives us a measure of spread? Variance is not as intuitive to me as standard deviation, which makes absolute sense. Can anyone help me understand the importance of variance and how its formula makes sense?

• Is your problem with the scale of variance, or the fact that its formula (and that of standard deviation) has a squared term? Jan 27 at 10:20
• One advantage of variance is that for sums of independent or uncorrelated random variables, you can sum the variances. Jan 27 at 10:21
• How do you calculate standard deviation? I know a couple of ways to do it, but the last step is always taking a square root to get the final answer. Just before you take the square root you have a number which is called the variance. So if I take your headline literally, the answer to "why do we calculate variance" is that if you want standard deviation, you're going to get it by calculating the variance first. Jan 27 at 14:11
• Have you ever in your entire life calculated the standard deviation of a set of data points? Have you ever in your entire life seen someone calculate a standard deviation? Have you ever seen the formula for standard deviation? If the answer to any of those questions is "yes," I am astounded that you could make the comment you just did. It seems you need to do some studying; perhaps here: khanacademy.org/math/statistics-probability/… Jan 28 at 5:27

My opinion :

A naive approach for measuring the spread around the mean would be $$\mathbb{E}[|X-\mathbb{E}[X]|]$$, but absolute value have bad properties for different reasons and I guess thats why we prefer to use the square instead of absolute value. The square function is a smoother function and therefore have better properties for optimization etc..., which lead to $$\mathbb{E}[(X-\mathbb{E}[X])^2]$$. However due to square it will change the magnitude of the values involved and to rescale it we put a square root, leading to standard deviation.

Standard deviation is the square root of variance, so they're measuring the same thing. The standard deviation approximates a typical element's distance from the mean (as you pointed out), so it's useful for visually understanding a distribution, whereas the variance has more convenient algebraic properties and tends to show up directly in probability theorems more often. Luckily, it's easy to convert between the two.

A couple of reasons involving statistical practice.

(1) For a random sample of size $$n$$ from a normal population where $$\mu$$ and $$\sigma^2$$ are both unknown, one has the relationship $$\frac{(n-1)S^2}{\sigma^2}\sim\mathsf{Chisq}(\nu= n-1),$$ which can be used to make a CI for $$\sigma^2$$ and to test hypothesis involving $$\sigma^2.$$ [To find a CI for $$\sigma,$$ take square roots of endpoints of a CI for $$\sigma^2.]$$

Using R, we get the CI $$(40.04,\, 70.12)$$ for $$\sigma^2$$ from a sample of size $$n = 100$$ from a normal population known to have $$\sigma^2 = 49.$$

set.seed(131)
x = rnorm(100, 50, 7)   # norm samp size 100. var 49
CI = 99*var(x)/qchisq(c(.975,.025), 99)
CI
[1] 40.05443 70.11714   # 95% CI for pop variance
sqrt(CI)
[1] 6.328857 8.373598   # 95% CI for pop SD


(2) $$E(S^2) = \sigma^2,$$ but for normal data $$E(S) < \sigma.$$ The bias of $$S$$ as an estimate of $$\sigma$$ is especially noticeable for small $$n.$$

The following simulation in R, using a million samples of size $$n=4,$$ illustrates the bias of the sample standard deviation $$S.$$ (Any one normal sample of size 4 can have an unusually large or small standard deviation. However, by looking at a million samples, it becomes clear that the bias is toward values of $$S$$ that underestimate $$\sigma.)$$

 set.seed(2022)
n = 4;  sg = 15
s = replicate(10^6, sd(rnorm(n, 100, sg)))
mean(s)
[1] 13.81432  # noticeably smaller than 15
mean(s^2)
[1] 224.8197  # approx = pop variance 225