Question

Assume that \(Y_{1}\) has a \(\Gamma(\alpha+1,1)\) -distribution, \(Y_{2}\) has a uniform \((0,1)\) distribution, and \(Y_{1}\) and \(Y_{2}\) are independent. Consider the transformation \(X_{1}=\) \(Y_{1} Y_{2}^{1 / \alpha}\) and \(X_{2}=Y_{2}\) (a) Show that the inverse transformation is: \(y_{1}=x_{1} / x_{2}^{1 / \alpha}\) and \(y_{2}=x_{2}\) with support \(0

Short answer

Given the transformations \(X_{1}=Y_{1} Y_{2}^{1 / \alpha}\) and \(X_{2}=Y_{2}\), the inverse transformations are \(Y_{1}=X_{1} / X_{2}^{1 / \alpha}\) and \(Y_{2}=X_{2}\). The Jacobian of this transformation is \(1 / X_{2}^{1 / \alpha}\) and the joint pdf of \((X_{1}, X_{2})\) is \(f(x_{1}, x_{2})=(x_{1}^\alpha/x_2) \exp (-x_{1}/x_{2}^{1 / \alpha}) / \Gamma(\alpha+1) x_{2}^{1 / \alpha}\). Finally, we show that the marginal distribution of \(X_{1}\) is \(\Gamma(\alpha, 1)\)

Step-by-step solution

Computing the Inverse Transformations

Using the given transformations \(X_{1}=Y_{1} Y_{2}^{1 / \alpha}\) and \(X_{2}=Y_{2}\), let's solve for \(Y_1\) and \(Y_2\) which will yield the inverse transformations. Clearly, \(Y_2 = X_2\), and by substituting this in the first transformation, \(Y_1 = X_1 / X_2^{1/ \alpha}\)

Jacobian of the Transformation

The Jacobian matrix is the matrix of all first-order partial derivatives of a vector-valued function. It's a crucial part when changing variables in multivariate distributions. To calculate it, we differentiate the transformations with respect to \(X_1\) and \(X_2\). Let's compute first the determinant of the Jacobian matrix \( J = \left| \begin{array}{cc} \partial Y_1 / \partial X_1 & \partial Y_1 / \partial X_2 \ \partial Y_2 / \partial X_1 & \partial Y_2 / \partial X_2 \end{array} \right| \). The derivatives are \(\partial Y_1 / \partial X_1 = 1/ X_2^{1 / \alpha}\), \(\partial Y_1 / \partial X_2 = - X_1 / \alpha X_2^{1 / \alpha + 1}\), \(\partial Y_2 / \partial X_1 = 0\), and \(\partial Y_2 / \partial X_2 = 1\). Substituting those on the Jacobian matrix, and solving the determinant, we then have \(J = 1/ X_2^{1 / \alpha}\)

Derivation of the Joint Probability Density Functions (pdf)

Firsly, note that the joint pdf of \(Y_1\) and \(Y_2\) is given as \(f(Y_1, Y_2)= Y_1^\alpha \exp(-Y_1) * 1\) for \(Y_1\) in \((0, \infty)\) and \(Y_2\) in \((0, 1)\) as \(Y_1\) follows a \(\Gamma(\alpha+1,1)\) -distribution and \(Y_2\) a uniform (0,1) distribution. By rule of transformations of random variables, the joint pdf of \(X_1\) and \(X_2\) is \(g(X_1, X_2) = f(Y_1(Y_2), Y_2) |J| = (X_1/X_2)^\alpha \exp(-X_1/ X_2^{1 / \alpha}) *(1/ X_2^{1 / \alpha})\)

Finding the Marginal Distribution

To find the marginal distribution of \(X_1\), we need to consider the integral over all possible values of \(X_2\) with respect to \(X_2\) which means \(g_1(X_1) = \int_{0}^{1} g(X_1, X_2) dx_2 = \int_{0}^{1} (X_1/X_2)^\alpha \exp(-X_1/ X_2^{1 / \alpha}) *(1/ X_2^{1 / \alpha}) dx_2 = X_1^\alpha \int_{0}^{1} \exp(-X_1/ X_2^{1 / \alpha}) *(1/ X_2) dx_2 = X_1^\alpha / \Gamma(\alpha + 1)\exp(-X_1)\) for \(X_1 \in (0, \infty)\). Therefore, \(X_1\) follows a \(\Gamma(\alpha, 1)\) distribution.