The joint distribution of X ∈ {0, 1} and Y ∈ {0, 1, 2} is P(0, 0) = 1/12, P(0, 1) = 1/6, P(0, 2) = 1/12, P(1, 0) = 1/6, P(1, 1) = 1/6 and P(1, 2) = 1/3. What is Cov(X, Y)?

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Covariance from a Table

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The joint distribution of $X \in {0, 1}$ and $Y \in {0, 1, 2}$ is $P (0, 0) = \frac{1}{12}$ , $P (0, 1) = \frac{1}{6}$ , $P (0, 2) = \frac{1}{12}$ , $P (1, 0) = \frac{1}{6}$ , $P (1, 1) = \frac{1}{6}$ and $P (1, 2) = \frac{1}{3}$ . What is $Cov (X, Y)$ ?

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Hint

Find the marginal means first, then the mean of the product over the cells where both are nonzero.

Solution

Summing the row for $X = 1$ gives $P (X = 1) = \frac{2}{3}$ , so $E [X] = \frac{2}{3}$ .
The marginals of $Y$ are $\frac{1}{4}$ , $\frac{1}{3}$ and $\frac{5}{12}$ , so $E [Y] = \frac{1}{3} + \frac{5}{6} = \frac{7}{6}$ .
Only $X = 1$ contributes to the product, so $E [X Y] = 1 \times \frac{1}{6} + 2 \times \frac{1}{3} = \frac{5}{6}$ .
Then $Cov (X, Y) = \frac{5}{6} - (\frac{2}{3}) (\frac{7}{6}) = \frac{15}{18} - \frac{14}{18}$ , which equals $\frac{1}{18}$ .

Answer: $\frac{1}{18}$

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