Question

If, \(f(x; \alpha, \beta) = \begin{cases} \alpha \beta x^{\beta-1} e^{-\alpha x^\beta} & ; x>0 \text{ and } \alpha, \beta>0 \\ 0 & ; \text{otherwise} \end{cases}\), then the probability density function of \(Y=x^\beta\) is

• A. \( \begin{cases} \alpha \beta e^{-\alpha \beta y} & ; y>0\\ 0 & ; \text{otherwise} \end{cases} \)
• B. \( \begin{cases} \alpha e^{-\alpha y} & ; y>0 \\ 0 & ; \text{otherwise} \end{cases} \)
• C. \( \begin{cases} \beta e^{-\beta y} & ; y>0 \\ 0 & ; \text{otherwise} \end{cases} \)
• D. \( \begin{cases} \frac{1}{\beta} e^{-y/\beta} & ; y>0 0 & ; \text{otherwise} \end{cases} \)

Hint:

This is a standard result: if X follows a Weibull distribution with parameters \( \alpha \) (scale) and \( \beta \) (shape), i.e., \( f(x) = \alpha\beta x^{\beta-1} e^{-\alpha x^\beta} \), then the random variable \( Y = X^\beta \) follows an Exponential distribution with rate parameter \( \alpha \). Recognizing this can lead to the answer immediately.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem requires finding the probability density function (PDF) of a transformed random variable. We are given the PDF of X and a transformation \(Y=g(X)\). We can use the change of variable formula to find the PDF of Y. The initial distribution of X is a Weibull distribution.

Step 2: Key Formula or Approach:
The change of variable formula for a transformation \(Y=g(X)\) is: \[ f_Y(y) = f_X(g^{-1}(y)) \left| \frac{d}{dy} g^{-1}(y) \right| \] Where \(g^{-1}(y)\) is the inverse transformation.

Step 3: Detailed Explanation:
1. Identify the transformation and its inverse: The transformation is \( Y = X^\beta \). Since \(x>0\), we can find a unique inverse. Solving for X, we get the inverse transformation: \( X = Y^{1/\beta} \). So, \( g^{-1}(y) = y^{1/\beta} \). The range of Y is also \(y>0\) since \(x>0\) and \(\beta>0\). 2. Calculate the Jacobian (the derivative of the inverse): \[ \frac{dx}{dy} = \frac{d}{dy}(y^{1/\beta}) = \frac{1}{\beta} y^{(1/\beta) - 1} \] Since \(y>0\) and \(\beta>0\), the absolute value is just the expression itself: \( \left| \frac{dx}{dy} \right| = \frac{1}{\beta} y^{(1/\beta) - 1} \). 3. Apply the change of variable formula: Substitute \(x = y^{1/\beta}\) into the PDF of X, \(f_X(x)\): \[ f_X(y^{1/\beta}) = \alpha \beta (y^{1/\beta})^{\beta-1} e^{-\alpha (y^{1/\beta})^\beta} \] \[ = \alpha \beta y^{(\beta-1)/\beta} e^{-\alpha y} \] Now, multiply by the Jacobian: \[ f_Y(y) = f_X(g^{-1}(y)) \left| \frac{dx}{dy} \right| = \left( \alpha \beta y^{(\beta-1)/\beta} e^{-\alpha y} \right) . \left( \frac{1}{\beta} y^{(1/\beta) - 1} \right) \] Combine the terms: \[ f_Y(y) = \alpha \beta \frac{1}{\beta} . y^{(\beta-1)/\beta} y^{(1-\beta)/\beta} . e^{-\alpha y} \] The \(\beta\) terms cancel out. For the y exponents, we add them: \[ \frac{\beta-1}{\beta} + \frac{1-\beta}{\beta} = 0 \] So, \( y^0 = 1 \). The expression simplifies to: \[ f_Y(y) = \alpha e^{-\alpha y} \]
Step 4: Final Answer:
The PDF of Y is \( f_Y(y) = \alpha e^{-\alpha y} \) for \(y>0\). This is the PDF of an exponential distribution with rate parameter \(\alpha\).