91Ó°ÊÓ

When transforming variables, such as in the exercise with \(U_1 = \frac{Y_1}{Y_2}\) and \(U_2 = Y_2\), the Jacobian determinant plays a critical role. It helps us understand how the density changes after the transformation.
To calculate the Jacobian, determine the partial derivatives of the transformation functions. For example, the transformation is given by \(g(y_1, y_2) = (\frac{y_1}{y_2}, y_2) = (u_1, u_2)\).
The Jacobian matrix is computed by:

• \( \frac{\partial y_1}{\partial u_1} = u_2 \)
• \( \frac{\partial y_1}{\partial u_2} = u_1 \)
• \( \frac{\partial y_2}{\partial u_1} = 0 \)
• \( \frac{\partial y_2}{\partial u_2} = 1 \)

The Jacobian determinant thus becomes the determinant of this 2x2 matrix, resulting in \(\left| J \right| = u_2\).
This transformation scales the density function by \(|u_2|\), ensuring it remains a valid probability distribution.

The marginal density function allows you to isolate one variable by integrating the joint density over the range of other variables.
In our exercise, we find the marginal density of \(U_1\) by integrating over \(U_2\).
This means:

• Start with the joint density function: \(f_{U_1, U_2}(u_1, u_2) = f_{Y_1, Y_2}(u_1 u_2, u_2) \cdot |u_2|\).
• Integrate over all possible values of \(u_2\): \[f_{U_1}(u_1) = \int_{-\infty}^{\infty} f_{Y_1, Y_2}(u_1 u_2, u_2) \cdot |u_2| \, du_2\]

By performing this integration, we essentially "sum out" the effects of \(U_2\) and focus solely on \(U_1\). This process allows us to understand the distribution of \(U_1\) alone, simplifying the overall problem.

Independence is a fundamental concept in probability theory. When two random variables are independent, knowledge about one does not provide information about the other.
In the exercise, if \(Y_1\) and \(Y_2\) are independent, their joint density is the product of their individual densities: \(f_{Y_1, Y_2}(y_1, y_2) = f_{Y_1}(y_1) \cdot f_{Y_2}(y_2)\).
For part (c) of the problem, independence simplifies the marginal density function for \(U_1\).
By assuming independence, the function becomes:

• The integral focuses on \(f_{Y_1}(u_1 u_2)\), removing dependency on \(f_{Y_2}(u_2)\) due to integration.
• The result is: \[f_{U_1}(u_1) = \int_{-\infty}^{\infty} f_{Y_1}(u_1 u_2) \cdot |u_2| \, du_2\]

This simplification is powerful because it often allows us to solve complex problems with more straightforward calculations. Understanding independence enables clearer insights into how distributions interact, significantly aiding in probability-related challenges.