Textbook
For jointly defined random variables $X,Y$ with finite second moments, covariance is $\operatorname{Cov}(X,Y)=\mathbb E[(X-\mathbb E[X])(Y-\mathbb E[Y])]$. Equivalently, $\operatorname{Cov}(X,Y)=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]$.
Deviations in the same direction produce a positive product; opposite directions produce a negative one. Averaging shows which contributions prevail. Covariance depends on scale and uses the product of the variablesβ units; magnitudes cannot be compared without considering those units.
If $Y=2X$ and $X$ is equally likely to be 2, 4, or 6, then $\operatorname{Cov}(X,Y)=2\operatorname{Var}(X)=16/3$. Independence implies zero covariance, but zero covariance does not imply independence.
In a counterexample, $U$ takes $-1,0,1$ with equal probability and $V=U^2$. Then $\mathbb E[U]=\mathbb E[UV]=0$, giving zero covariance. Yet $U=0$ guarantees $V=0$, although $P(V=0)=1/3$. Dependence exists; positive and negative contributions merely cancel.
Expanding the centered product gives $XY-X\mathbb E[Y]-Y\mathbb E[X]+\mathbb E[X]\mathbb E[Y]$. Linearity of expectation and the constant means reduce these four terms to the shorter formula above. Adding a constant to one variable preserves its deviations and therefore its covariance. Multiplying one variable by a number $a$ multiplies covariance by $a$, so a negative rescaling reverses its sign.
Definitions and notation: Mathematics for Machine Learning.
Related concepts: Variance: the mean squared deviation, Statistical independence.
In plain language
Covariance compares paired measurements with their respective means. Deviations in the same direction produce a positive product, while opposite directions produce a negative one. The result combines that balance with probability weights, but zero can reflect cancellation rather than an absence of dependence. This bookkeeping neither explains causes nor detects every possible relationship; a zero balance can conceal plenty of activity.
Example
The constructed counterexample $V=U^2$ shows how zero covariance coexists with a deterministic relationship.
Covariance: deviations together
Linked values
Each of the three pairs has probability 1/3.
Step 1 of 4: Linked values
Read all steps
Linked values
Each of the three pairs has probability 1/3.
- U
- β101
- V
- 101
Centering
Means are 0 and 2/3. Each is subtracted from the corresponding row.
- Deviations of U
- β101
- Deviations of V
- 1/3currentβ2/3current1/3current
Products of deviations
The outer products are opposite; the middle product is zero.
- Products
- β1/3excluded01/3included
Dependence check
Knowing U = 0 leaves only V = 0, making its probability 1.
- Remaining pair
- U = 0includedV = 0included
- Unconditional; conditional
- 1/31current
At the outer outcomes, the deviation of $V$ is the same $1/3$, while the deviations of $U$ are opposite: β1 and 1. Their products cancel. The middle pair contributes zero because $U=0$, although its $V$ deviation is $-2/3$. Dividing by three is justified by the equal pair probabilities. Other weights would require weighted averaging, and that cancellation could no longer be assumed.
Covariance measures a balance of joint deviations. Establishing independence requires a stronger check.