Question

The joint density function of X and Y is $f_{XY}(x,y)=\begin{cases}x+y & \text{for } 0<x<1, 0<y<1 \\ 0 & \text{elsewhere}\end{cases}$ then $P(X<2Y)$ is

• A. $4/3$
• B. $19/24$
• C. $1/4$
• D. $3/4$

Hint:

If the region of integration is complex, check if calculating the complement $P(X \ge 2Y)$ is simpler. In this case, both are relatively straightforward, but always ensure your limits correctly reflect the intersection of the inequality and the density's support.

Answer 1

The Correct Option is B

We need to calculate the volume under the joint density surface over the region where $X < 2Y$ within the unit square.

Step 1: \color{redDefine the Integration Limits
The region is $0 < x < 1, 0 < y < 1$ and $x < 2y$ (or $y > x/2$).
It is easier to integrate with respect to $y$ first:
For a fixed $x$, $y$ goes from $x/2$ to $1$.

Step 2: \color{redSet up the Double Integral
$P(X < 2Y) = \int_{0}^{1} \int_{x/2}^{1} (x+y) dy dx$.

Step 3: \color{redEvaluate the Inner Integral
$\int_{x/2}^{1} (x+y) dy = [xy + \frac{y^2}{2}]_{x/2}^{1}$
$= (x \cdot 1 + \frac{1}{2}) - (x \cdot \frac{x}{2} + \frac{(x/2)^2}{2})$
$= x + \frac{1}{2} - \frac{x^2}{2} - \frac{x^2}{8} = x + \frac{1}{2} - \frac{5x^2}{8}$.

Step 4: \color{redEvaluate the Outer Integral
$\int_{0}^{1} (x + \frac{1}{2} - \frac{5x^2}{8}) dx = [\frac{x^2}{2} + \frac{x}{2} - \frac{5x^3}{24}]_{0}^{1}$
$= \frac{1}{2} + \frac{1}{2} - \frac{5}{24}$
$= 1 - \frac{5}{24} = \frac{19}{24}$.
The probability is $19/24$.