Question

Let \( X \) and \( Y \) be two independent exponential random variables with \( E(X^2) = \frac{1}{2} \) and \( E(Y^2) = \frac{2}{9} \). Then \( P(X<2Y) \) (rounded off to two decimal places) is equal to ________

Hint:

When dealing with independent exponential random variables, use their respective CDFs and PDFs to compute probabilities involving linear combinations.

Answer 1

Correct Answer: 0.55

Let \( X \sim \text{Exp}(\lambda_1) \) and \( Y \sim \text{Exp}(\lambda_2) \) be independent exponential random variables. The mean and variance of an exponential random variable are related to the rate parameter as follows: \[ E(X) = \frac{1}{\lambda_1}, \quad E(X^2) = \frac{2}{\lambda_1^2}, \] \[ E(Y) = \frac{1}{\lambda_2}, \quad E(Y^2) = \frac{2}{\lambda_2^2}. \] We are given \( E(X^2) = \frac{1}{2} \) and \( E(Y^2) = \frac{2}{9} \), which gives: \[ \frac{2}{\lambda_1^2} = \frac{1}{2} \quad \Rightarrow \quad \lambda_1 = 2, \] \[ \frac{2}{\lambda_2^2} = \frac{2}{9} \quad \Rightarrow \quad \lambda_2 = 3. \] Now, we need to compute \( P(X<2Y) \). Since \( X \) and \( Y \) are independent, the probability can be computed as: \[ P(X<2Y) = \int_0^\infty P(X<2y) f_Y(y) \, dy. \] For \( X \sim \text{Exp}(2) \) and \( Y \sim \text{Exp}(3) \), the cumulative distribution function (CDF) of \( X \) is \( F_X(x) = 1 - e^{-2x} \), and the PDF of \( Y \) is \( f_Y(y) = 3e^{-3y} \). Thus, the integral becomes: \[ P(X<2Y) = \int_0^\infty (1 - e^{-4y}) 3e^{-3y} \, dy. \] After solving the integral, we find that \( P(X<2Y) \approx 0.55 \). Thus, \( P(X<2Y) \) is approximately \( \boxed{0.59} \).