Need help with joint distribution involving cauchy distributions
- Thread starter iiiiaann
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Since \( X_1, X_2 \) are i.i.d. standard Cauchy, you know the joint pdf of \( (X_1, X_2) \)
Now, \( (Y_1, Y_2) \) is a bijective linear transformation of \( (X_1, X_2) \). So you can express \( (X_1, X_2) \) as a function of \( (Y_1, Y_2) \), and also find out the Jacobian (determinant) of such transform. Then you can derive the joint density for \( (Y_1, Y_2) \)
Now, \( (Y_1, Y_2) \) is a bijective linear transformation of \( (X_1, X_2) \). So you can express \( (X_1, X_2) \) as a function of \( (Y_1, Y_2) \), and also find out the Jacobian (determinant) of such transform. Then you can derive the joint density for \( (Y_1, Y_2) \)
In general the Jacobian transformation of random variables involve several steps. Using your question as an example,
1. Write down the joint density \( f_{X_1, X_2}(x_1, x_2) \)
2. In the question you are given that \( Y_1 = \frac {X_1 + X_2} {2}, Y_2 = \frac {X_2} {2} \). Since this transformation is bijective, you just need to find out the inverse transform - i.e. find out the function \( g, h \) such that \( X_1 = g(Y_1, Y_2) \) and \( X_2 = h(Y_1, Y_2) \). As this is just a simple linear transformation, the inverse is very easy to find out.
3. Compute the absolute value of the Jacobian determinant
\( |J| = \left|\det\begin{bmatrix} {\displaystyle \frac {\partial g} {\partial y_1} } && {\displaystyle \frac {\partial g} {\partial y_2} }\\ {\displaystyle \frac {\partial h} {\partial y_1}} && {\displaystyle \frac {\partial g} {\partial y_2} }\end{bmatrix} \right| \)
4. Finally \( f_{Y_1, Y_2}(y_1, y_2) = f_{X_1, X_2}(g(y_1, y_2), h(y_1, y_2))|J| \)
1. Write down the joint density \( f_{X_1, X_2}(x_1, x_2) \)
2. In the question you are given that \( Y_1 = \frac {X_1 + X_2} {2}, Y_2 = \frac {X_2} {2} \). Since this transformation is bijective, you just need to find out the inverse transform - i.e. find out the function \( g, h \) such that \( X_1 = g(Y_1, Y_2) \) and \( X_2 = h(Y_1, Y_2) \). As this is just a simple linear transformation, the inverse is very easy to find out.
3. Compute the absolute value of the Jacobian determinant
\( |J| = \left|\det\begin{bmatrix} {\displaystyle \frac {\partial g} {\partial y_1} } && {\displaystyle \frac {\partial g} {\partial y_2} }\\ {\displaystyle \frac {\partial h} {\partial y_1}} && {\displaystyle \frac {\partial g} {\partial y_2} }\end{bmatrix} \right| \)
4. Finally \( f_{Y_1, Y_2}(y_1, y_2) = f_{X_1, X_2}(g(y_1, y_2), h(y_1, y_2))|J| \)