# Bivariate random variable and transformation

Let $$X=(X_1,X_2)$$ and $$Y=(Y_1,Y_2)$$ be non-negative absolutely continuous random vector and if $$\phi(X_j)=Y_j$$, $$j=1,2$$, are one-one transformation then $$H[Y;\phi(t_1),\phi(t_2)]=H(X;t_1,t_2)-E[\log J(X_1,X_2)|X_1>t_1,X_2>t_2]\tag 1$$ where $$J(y_1,y_2)=\left|\frac{\partial\phi_1^{-1}(y_1)}{\partial y_1}\times\frac{\partial \phi_2^{-1}(y_2)}{\partial y_2}\right|$$, and $$H(X;t_1,t_2)=-\int_{t_2}^\infty\int_{t=1}^\infty\frac{f(x_1,x_2)}{\bar{F}(t_1,t_2)}\log\left(\frac{f(x_1,x_2)}{\bar{F}(t_1,t_2)}\right)dx_1dx_2$$, $$\bar{F}(x_1,x_2)=\Pr(X_1>x_1,X_2>x_2).$$

RHS of the equation(1) can be written as

$$H(X;t_1,t_2)-E[\log J(X_1,X_2)|X_1>t_1,X_2>t_2]=-\int_{t_2}^\infty\int_{t=1}^\infty\frac{f(x_1,x_2)}{\bar{F}(t_1,t_2)}\log\left(\frac{f(x_1,x_2)}{\bar{F}(t_1,t_2)}\right)dx_1dx_2-\int_{t_2}^\infty\int_{t=1}^\infty\frac{f(x_1,x_2)J(x_1,x_2)}{\bar{F}(t_1,t_2)}dx_1dx_2.$$

I want to prove this. I am new in statistics field. I don't know some specific theorem that will be use here in order to use transformation.

I can write

\begin{align} H[Y;\phi(t_1),\phi(t_2)]&=-\int_{\phi_2(t_2)}^\infty\int_{\phi_1(t_1)}^\infty\frac{f(y_1,y_2)}{\bar{F}(\phi_1(t_1),\phi_2(t_2)}\log\left(\frac{f(y_1,y_2)}{\bar{F}(\phi_1(t_1),\phi_2(t_2))}\right)dy_1dy_2 \end{align}

\begin{align}{\bar{F}(\phi_1(t_1),\phi_2(t_2))}&=\Pr(Y_1>\phi_1(t_1),Y_2>\phi_2(t_2))\\&=\Pr(\phi_1(X_1)>\Pr(Y_1\phi_1(t_1),Y_2>\phi_2(t_2))\\ &=\Pr(X_1>t_1,X_2>t_2). \end{align} I don't know how to proceed further.

• A few typos in your question. In your double integrals, the lower bound of the inner integral should be $t_1$ (not $t=1$). Also, something went wrong in the second line of your last paragraph, where you work out the survival function of $X$ in terms of the survival function of $Y$. Feb 7 at 11:01

Notice that $$H[Y;\phi(t_1),\phi(t_2)] = -\mathbb{E}[\log f_Y(Y) | Y_1>\phi(t_1), Y_2 > \phi(t_2) ]$$.
If $$\phi$$ is a strictly increasing function (which you seem to have assumed), then $$Y_i>\phi(t_i) \Leftrightarrow X_i > t_i$$, for $$i=1,2$$.
The change-of-variable formula tells us that $$f_Y(y)=f_X(x) J(x)$$ where $$x=\phi^{-1}(y)$$ (and $$J$$ is the Jacobian determinant), so we have: $$- \mathbb{E}[\log f_Y(Y) | Y_1>\phi(t_1), Y_2 > \phi(t_2) ] = - \mathbb{E}[\log \left\{f_X(X) J(X)\right\} | X_1>t_1, X_2 > t_2 ] \\ = - \mathbb{E}[\log f_X(X) | X_1>t_1, X_2 > t_2 ] - \mathbb{E}[\log J(X) | X_1>t_1, X_2 > t_2 ]$$ where the first term is $$H(X;t_1,t_2)$$, as required.