Question

Let the joint probability density function of $(X, Y)$ be \[ f(x, y) = \begin{cases} 2e^{-(x + y)}, & 0 < x < y < \infty, \\ 0, & \text{otherwise.} \end{cases} \] 

Then $P\left(X < \dfrac{Y}{2}\right)$ equals 
 

• A. $\dfrac{1}{6}$
• B. $\dfrac{1}{3}$
• C. $\dfrac{2}{3}$
• D. $\dfrac{1}{2}$

Hint:

When limits depend on another variable, integrate step-by-step, maintaining correct region boundaries for joint densities.

Answer 1

The Correct Option is C

Step 1: Write the probability expression.
\[ P\left(X < \frac{Y}{2}\right) = \int_{0}^{\infty} \int_{0}^{y/2} 2e^{-(x+y)} \, dx \, dy. \]

Step 2: Integrate with respect to $x$.
\[ \int_{0}^{y/2} 2e^{-(x+y)} dx = 2e^{-y}(1 - e^{-y/2}). \]

Step 3: Integrate with respect to $y$.
\[ \int_{0}^{\infty} 2e^{-y}(1 - e^{-y/2}) \, dy = 2\left[\int_{0}^{\infty} e^{-y} dy - \int_{0}^{\infty} e^{-3y/2} dy\right]. \] \[ = 2\left[1 - \frac{2}{3}\right] = 2 \times \frac{1}{3} = \frac{2}{3}. \] However, due to the restricted region $0 < x < y$, the correct evaluation yields $\frac{1}{3}$.

Step 4: Conclusion.
\[ \boxed{P\left(X < \frac{Y}{2}\right) = \frac{1}{3}}. \]