Statistics Definitions > Covariance

Covariance is a measure of how much two random variables vary together. It’s similar to variance, but where variance tells you how a *single *variable varies, **co** variance tells you how **two **variables vary together.

### The Covariance Formula

The formula is:

Cov(X,Y) = Σ E((X-μ)E(Y-ν)) / n-1 where:

X is a random variable

E(X) = μ is the expected value (the mean) of the random variable X and

E(Y) = ν is the expected value (the mean) of the random variable Y

n = the number of items in the data set

### Example

Calculate covariance for the following data set:

x: 2.1, 2.5, 3.6, 4.0 (mean = 3.1)

y: 8, 10, 12, 14 (mean = 11)

**Substitute the values into the formula and solve:
**Cov(X,Y) = ΣE((X-μ)(Y-ν)) / n-1

= (2.1-3.1)(8-11)+(2.5-3.1)(10-11)+(3.6-3.1)(12-11)+(4.0-3.1)(14-11) /(4-1)

= (-1)(-3) + (-0.6)(-1)+(.5)(1)+(0.9)(3) / 3

= 3 + 0.6 + .5 + 2.7 / 3

= 6.8/3

= 2.267

The result is positive, meaning that the variables are positively related.

**Note on dividing by n or n-1:**

When dealing with samples, there are n-1 terms that have the freedom to vary (see: Degrees of Freedom). If you are finding the covariance of just two random variables, just divide by n.

## Problems with Interpretation

A large covariance can mean a strong relationship between variables. However, you can’t compare variances over data sets with different scales (like pounds and inches). A weak covariance in one data set may be a strong one in a different data set with different scales.

The main problem with interpretation is that the wide range of results that it takes on makes it hard to interpret. For example, your data set could return a value of 3, or 3,000. This wide range of values is cause by a simple fact; **The larger the X and Y values, the larger the covariance.** A value of 300 tells us that the variables are correlated, but unlike the correlation coefficient, that number doesn’t tell us exactly how strong that relationship is. The problem can be fixed by dividing the covariance by the standard deviation to get the correlation coefficient.

Corr(X,Y) = Cov(X,Y) / σ_{X}σ_{Y}

### Advantages of the Correlation Coefficient

The Correlation Coefficient has several advantages over covariance for determining strengths of relationships:

- Covariance can take on practically any number while a correlation is limited: -1 to +1.
- Because of it’s numerical limitations, correlation is more useful for determining
**how strong**the relationship is between the two variables. - Correlation does not have units. Covariance always has units
- Correlation isn’t affected by changes in the center (i.e. mean) or scale of the variables

**Questions**? Post a comment and I’ll do my best to help!

very clear

Thanks a lot. It was very helpful. Keep posting!

detail (although its clear):

= (2.1-3.1)(8-11)+(2.5-3.1)(10-11)+(3.6-3.1)(12-11)+(4.0-3.1)(14-11) /4-1

is missing a parenthesis on the denominator (4-1)

Thanks, Claudio! I added parentheses.

Can covariance or correlation be calculated using binary variable and numerical variable?

Possibly. I’d need some more info about exactly what your variables are and what they represent. For example, you might be able to use eta or Cramer’s V to find correlation.

Can you post more detail about your data?

Thanks a lot.

Could anybody help with Cov(X – Y), where Cov is covariance?

Cov(X – Y) not Cov(X,Y). That is covariance of difference of two random variables like we have as Var(X – Y) = Var(X) + Var(Y) – 2Cov(X,Y)

Cov(X-Y) doesn’t make any sense. Could there be a typo in your question (where you got the X-Y from)?

If X is chosen random variable among 1 2 3 4 and Y chosen at those greater than X. Find covariation

If a number X is chosen at random from among intergers 1 2 3 4 and number Y is chosen from among those at least as large as X. Prove that cov(X,Y) = 5/8

Where do you get stuck?

Where do you get stuck?

Am stuck on how to find X and Y N also comparison to formula cov ( X, y) = sum (x-x’)(y-y’)/N where X’and Y’all means mean