Step 1: Concept Overview:
The problem involves determining the probability density function (PDF) of a transformed random variable \(Y\) given the PDF of \(X\) and the transformation \(Y=g(X)\). The change of variable formula is employed. \(X\) follows a Weibull distribution initially.
Step 2: Core Formula:
The change of variable formula is expressed as:
\[ f_Y(y) = f_X(g^{-1}(y)) \left| \frac{d}{dy} g^{-1}(y) \right| \]
where \(g^{-1}(y)\) represents the inverse transformation.
Step 3: Detailed Solution:
1. Identify Transformation and Inverse: The transformation is \( Y = X^\beta \). The inverse is uniquely defined since \(x>0\). Solving for \(X\) yields \( X = Y^{1/\beta} \), thus \( g^{-1}(y) = y^{1/\beta} \). The range of \(Y\) is \(y>0\), given \(x>0\) and \(\beta>0\).2. Compute the Jacobian: \[ \frac{dx}{dy} = \frac{d}{dy}(y^{1/\beta}) = \frac{1}{\beta} y^{(1/\beta) - 1} \] Given \(y>0\) and \(\beta>0\), the absolute value simplifies to: \( \left| \frac{dx}{dy} \right| = \frac{1}{\beta} y^{(1/\beta) - 1} \).3. Apply the Transformation Formula: Substitute \(x = y^{1/\beta}\) into \(f_X(x)\): \[ f_X(y^{1/\beta}) = \alpha \beta (y^{1/\beta})^{\beta-1} e^{-\alpha (y^{1/\beta})^\beta} \] \[ = \alpha \beta y^{(\beta-1)/\beta} e^{-\alpha y} \] 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 terms: \[ f_Y(y) = \alpha \beta \frac{1}{\beta} y^{(\beta-1)/\beta} y^{(1-\beta)/\beta} e^{-\alpha y} \] The \(\beta\) terms cancel. Sum the y exponents: \[ \frac{\beta-1}{\beta} + \frac{1-\beta}{\beta} = 0 \] Thus, \( y^0 = 1 \). The result is: \[ f_Y(y) = \alpha e^{-\alpha y} \]
Step 4: Solution:
The PDF of \(Y\) is \( f_Y(y) = \alpha e^{-\alpha y} \) for \(y>0\), representing an exponential distribution with rate \(\alpha\).