# Need help with joint distribution involving cauchy distributions

#### iiiiaann

##### New Member

Okay so I know that help is only given to those who show effort, but I'm totally stuck on this one, I don't even know where to start. I know what the Cauchy distribution is and i could plug it in for the Y variables, but I don't know where that would get me. Please help me out!

#### BGM

##### TS Contributor
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 $$(Y_1, Y_2)$$

#### iiiiaann

##### New Member
your words just went right over my head my friend....I'm sorry I'm not familiar with a lot of the terminology

#### BGM

##### TS Contributor
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 $$f_{Y_1, Y_2}(y_1, y_2) = f_{X_1, X_2}(g(y_1, y_2), h(y_1, y_2))|J|$$