Question

Let \( X_1, X_2, X_3 \) be independent random variables with the common probability density function 

Let \( Y = \min \{ X_1, X_2, X_3 \}, \, E(Y) = \mu_Y \, \text{and} \, \text{Var}(Y) = \sigma_Y^2. \) Then \( P(Y>\mu_Y + \sigma_Y) \) equals} 
 

Hint:

When dealing with the minimum of independent random variables with exponential distributions, use the CDF and the properties of the exponential distribution to find probabilities.

Answer 1

Correct Answer: 0.13

Step 1: Understanding the distribution.
We are given that \( X_1, X_2, X_3 \) are independent random variables with the same exponential probability density function. The minimum of these variables, \( Y \), will also follow an exponential distribution, but with a different rate parameter.
Step 2: Finding the rate parameter for \( Y \).
The cumulative distribution function (CDF) of \( Y \) is given by: \[ F_Y(y) = 1 - P(Y>y) = 1 - \prod_{i=1}^3 P(X_i>y). \] For each \( X_i \), \( P(X_i>y) = e^{-2y} \), so: \[ F_Y(y) = 1 - e^{-6y}. \] Thus, \( Y \) has an exponential distribution with rate \( 6 \).
Step 3: Finding \( P(Y>\mu_Y + \sigma_Y) \).
The mean and variance of an exponential distribution with rate \( \lambda \) are \( \mu_Y = \frac{1}{\lambda} \) and \( \sigma_Y^2 = \frac{1}{\lambda^2} \). Here, \( \lambda = 6 \), so: \[ \mu_Y = \frac{1}{6}, \quad \sigma_Y = \frac{1}{6}. \] We need to find \( P(Y>\mu_Y + \sigma_Y) = P(Y>\frac{2}{6}) \). Using the CDF, we find that: \[ P(Y>\frac{2}{6}) = e^{-6 \times \frac{2}{6}} = e^{-2}. \] Thus, \( P(Y>\mu_Y + \sigma_Y) \) is approximately between 0.13 and 0.14.

Step 4: Conclusion.
Thus, the value of \( P(Y>\mu_Y + \sigma_Y) \) is approximately \( 0.13 \) to \( 0.14 \).