Bivariate transformation problem

Question

Let $X = (X_1, X_2)$ be a non-negative bivariate random vector. Let $Y_i = \phi _i(X_i), i = 1, 2,$ be one-to-one transformations with $\phi_i(X_i)$ differentiable functions, and let $Y = (Y_1, Y_2)$ be a non-negative bivariate random vector connected with $X$ by the relation $Y_i = \phi_i(X_i)$.

I want to transform this integral : $$K(Y_1,Y_2)=\int_0^\infty \int_0^\infty f^2(y_1,y_2)\,dy_2\,dy_1,$$ where $f(y_1,y_2)$ is the density fucntion of $Y=(Y_1,Y_2)$.

According to the Jacobian transformation theorem, if $ Y = \phi(X) $, then the joint density function of ( Y ) can be expressed as:

$ f(y_1, y_2) = f(x_1, x_2) \left| \frac{\partial(x_1, x_2)}{\partial(y_1, y_2)} \right|=f(x_1, x_2) |J|, $

Since $Y_i = \phi_i(X_i) $, we can express the transformation as:

$ y_1 = \phi_1(x_1), $ $ y_2 = \phi_2(x_2)$

So, the Jacobian determinant simplifies to:

\begin{align*} \left| \frac{\partial(x_1, x_2)}{\partial(y_1, y_2)} \right| &= \left| \begin{array}{cc} \frac{\partial}{\partial y_1} \phi_1^{-1}(y_1) & 0 \\ 0 & \frac{\partial}{\partial y_2} \phi_2^{-1}(y_2) \end{array} \right| \\ & = \left| \frac{\partial}{\partial y_1} \phi_1^{-1}(y_1) \cdot \frac{\partial}{\partial y_2} \phi_2^{-1}(y_2) \right| =|J| \end{align*} So we have: $$K(Y_1,Y_2)=\int_0^\infty \int_0^\infty f^2(x_1,x_2) |J|^3 \,dx_2\,dx_1,$$

Is it correct? Thanks beforehand!

Answer 1

Let us write everything with details to avoid mistakes. First we have

$$ f_\color{blue}{{Y_1,Y_2}}(y_1, y_2) = \left| \frac{\partial}{\partial y_1} \phi_1^{-1}(y_1) \cdot \frac{\partial}{\partial y_2} \phi_2^{-1}(y_2) \right| f_\color{blue}{{X_1,X_2}}(\color{blue}{\phi_1^{-1}(y_1)}, \color{blue}{\phi_2^{-1}(y_2)})=\left| (\phi_1^{-1})'[y_1] \cdot (\phi_2^{-1})'[y_2] \right| f_{X_1,X_2}(\phi_1^{-1}(y_1), \phi_2^{-1}(y_2)). $$

Then, we obtain

$$K(Y_1,Y_2)=\int_0^\infty \int_0^\infty f^2_{X_1,X_2}(\phi_1^{-1}(y_1), \phi_1^{-1}(y_2)) \left( (\phi_1^{-1})'[y_1] \cdot (\phi_2^{-1})'[y_2] \right)^2 \,dy_2\,dy_1.$$

By change of variables $y_i = \phi _i(x_i), i = 1, 2,$, this can be written as

$$K(Y_1,Y_2)=\int_0^\infty \int_0^\infty f^2_{{X_1,X_2}}(x_1, x_2) \left( (\phi_1^{-1})'[\phi _1(x_1)] \cdot (\phi_2^{-1})'[\phi _2(x_2)] \right)^2 \phi' _1(x_1) \phi' _2(x_2)\,dx_2\,dx_1$$

where above for simplicity I assume $\phi _i(x_i), i = 1, 2,$ are increasing with $\phi _1(0)=\phi _2(0)=0$ and $\phi _1(x)=\phi _2(x) \to \infty$ as $x \to \infty$ (other cases are slightly different).

Finally by using the known relation

$$(\phi_i^{-1})'[\phi _i(x)] \phi' _i(x)=1, i=1,2$$

we get

$$K(Y_1,Y_2)=\int_0^\infty \int_0^\infty f^2_{{X_1,X_2}}(x_1, x_2) \left| (\phi_1^{-1})'[\phi _1(x_1)] \cdot (\phi_2^{-1})'[\phi _2(x_2)] \right|\,dx_2\,dx_1 \\=\int_0^\infty \int_0^\infty f^2_{{X_1,X_2}}(x_1, x_2) \left| \frac{1}{\phi' _1(x_1)\phi' _2(x_2)} \right|\,dx_2\,dx_1.$$

Note that you can drop $| \cdot |$ when both $\phi _i(x_i), i = 1, 2,$ are either increasing or decreasing.