Let the random variables $X$ and $Y$ have joint pdf $f(x,y)=2$ for $0<x<1$, $0<y<x$, and $f(x,y)=0$ otherwise. Find $P(Y<0.5,X>0.5)$ (round off to third decimal place).

• A. $0.375$
• B. $0.250$
• C. $0.125$
• D. $0.500$

Answer: (A)

Verified step by step guidance

1. First, understand the joint probability density function (pdf) given: \(f(x,y) = 2\) for \$0 < y < x < 1\(, and \)0\( otherwise. This means the joint distribution is uniform over the triangular region where \)y\( is between 0 and \)x\(, and \)x$ is between 0 and 1.
2. Identify the event for which we want to find the probability: \(P(Y < 0.5, X > 0.5)\). This corresponds to the region where \(y\) is less than 0.5 and \(x\) is greater than 0.5, but still within the support of the joint pdf.
3. Set up the double integral for the probability over the appropriate region. Since \(y\) must be less than 0.5 and also less than \(x\), and \(x\) must be greater than 0.5 but less than 1, the limits are: \(x\) from 0.5 to 1, and for each fixed \(x\), \(y\) from 0 to the minimum of \$0.5\( and \)x\(. Because \)x > 0.5\(, the upper limit for \)y$ is 0.5. So the integral is: \[ P = \int_{0.5}^{1} \int_{0}^{0.5} 2 \, dy \, dx \]
4. Perform the inner integral with respect to \(y\): \[ \int_{0}^{0.5} 2 \, dy = 2 \times (0.5 - 0) = 1 \]
5. Then perform the outer integral with respect to \(x\): \[ \int_{0.5}^{1} 1 \, dx = 1 \times (1 - 0.5) = 0.5 \] This gives the probability \(P(Y < 0.5, X > 0.5)\). You can now calculate the numerical value and round it to three decimal places.