X and Y have joint density f(x, y) = c(x + 2y) on the unit square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and zero elsewhere. Find P(X ≤ 1/2). Give the answer as a fraction in lowest terms.

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$X$ and $Y$ have joint density $f (x, y) = c (x + 2 y)$ on the unit square $0 \leq x \leq 1$ , $0 \leq y \leq 1$ , and zero elsewhere. Find $P (X \leq \frac{1}{2})$ . Give the answer as a fraction in lowest terms.

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Hint

Find $c$ from the total, integrate $y$ out, then integrate the marginal up to $\frac{1}{2}$ .

Solution

The total without $c$ is $\int_{0}^{1} \int_{0}^{1} (x + 2 y) d y d x = \frac{1}{2} + 1 = \frac{3}{2}$ , so $c = \frac{1}{\frac{3}{2}} = \frac{2}{3}$ .
The marginal is $f_{X} (x) = \int_{0}^{1} c (x + 2 y) d y = c (x + 1)$ .
Then $\int_{0}^{\frac{1}{2}} (x + 1) d x = \frac{1}{8} + \frac{1}{2} = \frac{5}{8}$ .
So $P (X \leq \frac{1}{2}) = \frac{2}{3} \times \frac{5}{8} = \frac{5}{12}$ .

Answer: $\frac{5}{12}$

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