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Covariance on a triangle

ProbabilityHard

$(X, Y)$ is uniform on the triangle $0 < y < x < 1$ . What is $Cov (X, Y)$ ? Give the answer as a fraction in lowest terms.

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Hint

Given $X = x$ , the variable $Y$ is uniform on $(0, x)$ . Work out every moment through $X$ .

Solution

The triangle has area $\frac{1}{2}$ , so the density is $2$ and the marginal is $f_{X} (x) = 2 x$ . That gives $E [X] = \frac{2}{3}$ and $E [X^{2}] = \frac{1}{2}$ .
Given $X = x$ , the mean of $Y$ is $\frac{x}{2}$ , so $E [Y] = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$ .
In the same way $E [X Y] = \frac{E [ X ^{2} ]}{2} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ .
So $Cov (X, Y) = \frac{1}{4} - \frac{2}{3} \times \frac{1}{3} = \frac{1}{36}$ .

Answer: $\frac{1}{36}$