Covariance and Correlation Coefficient for Discrete Random Variables

If equity and bond returns move in opposite directions, portfolio variance shrinks, diversification works. The covariance and correlation coefficient quantify the direction and strength of linear dependence between two random variables.

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

HIGH-FREQUENCY: The computational formula $E[XY] - E[X]E[Y]$ is the standard approach. Compute $E[XY]$ from the joint PMF; compute $E[X], E[Y]$ from the marginals.

For discrete random variables:

$$E[XY] = \sum_x \sum_y xy \cdot p_{X,Y}(x,y)$$

• $\text{Cov}(X,X) = \text{Var}(X)$
• $\text{Cov}(X,Y) = \text{Cov}(Y,X)$, symmetric
• $\text{Cov}(aX+b, cY+d) = ac \cdot \text{Cov}(X,Y)$, constants vanish
• $\text{Cov}(X+Y, Z) = \text{Cov}(X,Z) + \text{Cov}(Y,Z)$, additive
• If independent, $\text{Cov}(X,Y) = 0$

TRAP: The converse of property 5 is FALSE. $\text{Cov} = 0$ does NOT imply independence. Covariance only captures linear dependence. Two variables can be maximally dependent with zero covariance if the dependence is nonlinear.

Classic counterexample: Let $X$ be uniform on $\{-1, 0, 1\}$ and $Y = X^2$. Then $E[X] = 0$, $E[XY] = E[X^3] = 0$, so $\text{Cov} = 0$. But $Y$ is completely determined by $X$.

Common mistakes

• Forgetting to subtract $E[X]E[Y]$. Reporting $E[XY] = 41/9 \approx 4.56$ as the covariance. The correct covariance is $8/81 \approx 0.099$. Trap: 41/9.
• Using $\text{Cov} = 0$ to conclude independence. Zero covariance means no linear relationship, but the variables may still be dependent through nonlinear association.
• Omitting the $2\text{Cov}$ term in $\text{Var}(X+Y)$. Under independence, $\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y)$. With dependence, the covariance term is essential. Dropping it gives 88/81 instead of 104/81. Trap: 88/81.

Bottom line

• Covariance: $\text{Cov}(X,Y) = E[XY] - E[X]E[Y]$; a positive value means they move together
• Correlation: $\rho = \text{Cov}(X,Y)/(\text{SD}(X) \cdot \text{SD}(Y))$, always bounded $[-1,1]$
• $\text{Cov} = 0$ does NOT imply independence: it captures only linear association
• Variance of sum: $\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y) + 2\text{Cov}(X,Y)$

Exam shortcut

Compute five quantities in order: $E[X]$, $E[Y]$, $E[X^2]$, $E[Y^2]$, $E[XY]$. For $E[XY]$ with a table, sum only cells where both $x$ and $y$ are nonzero. "Covariance = Cross minus Product of means." Pattern matches variance: $\text{Var}(X) = E[X^2] - (E[X])^2$. "Variance of a sum = sum of variances plus twice the covariance." The 2 is there because $(X+Y)^2$ has a cross-term $2XY$.

The full lesson (about 2,319 words, 15 min read) adds 3 worked examples, all 6 common mistakes, a self-check, free in the app.

Learning objectives

• 3e

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