Covariance and Correlation Coefficient

The covariance $\text{Cov}(X, Y)$ measures the direction of the linear relationship between two random variables — positive if they tend to move together, negative if oppositely. The correlation coefficient $\rho_{X,Y}$ standardizes it into the range $[-1, 1]$ .

$\text{Cov}(X, Y) = E[XY] - E[X]\,E[Y]$

$\rho_{X,Y} = \frac{\text{Cov}(X, Y)}{\sigma_X \, \sigma_Y}$

Correlation is unit-free and comparable across scales, whereas covariance carries the product of the two variables' units.
$\rho_{X,Y} = 0$ means no linear relationship, but not necessarily independence.
Covariance is bilinear: $\text{Cov}(aX + bY,\, Z) = a\,\text{Cov}(X, Z) + b\,\text{Cov}(Y, Z)$ , adding a constant to either variable leaves it unchanged, and $\text{Cov}(X, X) = \text{Var}(X)$ .
See also Correlation and Covariance.

Covariance of Study Hours and Exam Score A joint PMF is $p(1,4)=0.2$ , $p(1,8)=0.1$ , $p(3,4)=0.1$ , $p(3,8)=0.6$ , where $X$ = hours studied and $Y$ = exam score. Compute $\text{Cov}(X, Y)$ .
Answer First the marginal means:
 $\begin{align*} E[X] &= 1(0.3) + 3(0.7) = 2.4 \\ E[Y] &= 4(0.3) + 8(0.7) = 6.8 \end{align*}$ 
Then $E[XY]$ : $E[XY] = 1{\cdot}4(0.2) + 1{\cdot}8(0.1) + 3{\cdot}4(0.1) + 3{\cdot}8(0.6) = 17.2$ Therefore: $\text{Cov}(X, Y) = 17.2 - (2.4)(6.8) = 17.2 - 16.32 = 0.88$ The positive covariance confirms that more study hours is associated with higher scores.