Bivariate Transformation Method
Given random variables $Y_1$ and $Y_2$ with some joint distribution function $f_{Y_1,Y_2}(y_1,y_2)$ and some transformative functions of $y_1$ and $y_2$ such that
\[\begin{align*} u_1 &= h_1(y_1, y_2) & u_2 &= h_2(y_1,y_2) \end{align*}\]then there are three simple steps to find the joint distribution of $u_1$ and $u_2$
- Find the inverse functions,
- Find the Jacobian determinant,
- Find the joint distribution $f_{U_1,U_2}(u_1,u_2)$.
Throughout this, I’ll show the general case and the work for an example given below.
Example Variables
There are two random variables $Y_1$ and $Y_2$ with joint distribution
\[\begin{equation*} f_{Y_1,Y_2}(y_1,y_2) = \begin{cases} 2(1-y_1), & 0 \le y_1 \le 1,\; 0 \le y_2 \le 1\\ 0, & \textit{otherwise} \end{cases} \end{equation*}\]and we want to find the distribution of the transformation such that $U_1 = Y_1$ and $U_2 = Y_1Y_2$
Inverse functions
Given $u_1 = h_1(y_1, y_2)$ and $u_2 = h_2(y_1,y_2)$, we need to find the inverse of these functions such that
\[\begin{align*} y_1 &= h_1^{-1}(u_1, u_2) & y_2 &= h_2^{-1}(u_1,u_2) \end{align*}\]For our example, we would first find the inverse of $u_1$ which is pretty easy.
\( y_1 = y_1 = h_1(y_1,y_2) \implies y_1 = u_1 = h_1^{-1}(u_1,u_2) \)
Then we can use the fact that $y_1 = u_1$ to write our second inverse function:
\( u_2 = y_1y_2 = h_2(y_1,y_2) \implies y_2 = \frac{u_2}{u_1} = h_2^{-1}(u_1,u_2) \)
Jacobian Determinant
Next we need to find the Jacobian determinant of these inverse functions. The Jacobian matrix is simply the partials of $h_1^{-1}(u_1,u_2)$ and $h_2^{-1}(u_1,u_2)$ with respect to both $u_1$ and $u_2$. It looks as follows:
\[\begin{equation*} \begin{bmatrix} \frac{\partial h_1^{-1}}{\partial u_1} & \frac{\partial h_1^{-1}}{\partial u_2}\\ \frac{\partial h_2^{-1}}{\partial u_1} & \frac{\partial h_2^{-1}}{\partial u_2}\\ \end{bmatrix} \end{equation*}\]The determinant of this matrix is then the difference of the diagonals:
\[\begin{equation*} \frac{\partial h_1^{-1}}{\partial u_1}\frac{\partial h_2^{-1}}{\partial u_2} - \frac{\partial h_1^{-1}}{\partial u_2}\frac{\partial h_2^{-1}}{\partial u_1} \end{equation*}\]For our example, the Jacobian matrix would be as follows:
\[\begin{equation*} \begin{bmatrix} \frac{\partial (u_1)}{\partial u_1} & \frac{\partial (u_1)}{\partial u_2}\\ \frac{\partial (u_2/u_1)}{\partial u_1} & \frac{\partial (\frac{u_2}{u_1})}{\partial u_2}\\ \end{bmatrix} = \begin{bmatrix} 1 & 0\\ -u_2/u_1^2 & 1/u_1\\ \end{bmatrix} \end{equation*}\]and the determinant would simply be $1(1/u_1)-0(-u_2/u_1^2)=1/u_1$.
Joint Distribution
Finally we can find the joint distribution of $u_1$ and $u_2$ . The general equation for this distribution is
\( f_{U_1,U_2}(u_1,u_2) = f_{Y_1,Y_2}(h_1^{-1}(u_1,u_2),h_2^{-1}(u_1,u_2))\cdot\lvert J\rvert \)
where $\lvert J\rvert$ is the absolute value of the Jacobian matrix.
In our example, this would work out to be
\( f_{U_1,U_2}(u_1,u_2) = 2(1-u_1)\cdot \lvert\frac{1}{u_1}\rvert \)
for $0 \le u_2 \le u_1 \le 1$ and $0$ otherwise.
Note
Often you will only have one $U$ representing a transformation. This seems to pose a problem with using this method right? Well actually you can just use a helper variable to manipulate the transformation.
Let’s say the example above instead said
We want to find the distribution of the transformation such that $U = Y_1Y_2$
Then we need to define some $u_1 = h_1(y_1,y_2)$ such that we can find the inverse of $U = U_2 = Y_1Y_2$.
To find this helper variable $u_2$, we should first try to find the inverse of $u_2$ to see what is missing. If $u_2 = y_1y_2$, then $y_2 = u_2/y_1$. Now $y_2$ cannot be a function of both $u_2$ and $y_1$ for this method to work, so we need to replace $y_1$ with some function of $u$s. The easiest choice would be $u_1 = y_1$ and thus $y_1= u_1$. Then we get $y_2 = u_2/u_1$.
From here we can perform all the steps above in the same way until the end. At that point, we have a joint distribution function of $f_{U_1,U_2}(u_1,u_2)$, but we are just looking for the distribution for $U_2 = U$. So, all we need to do is find the marginal distribution by integrating with respect to $U_1$ to get
\[\begin{equation*} f_{U}(u) = \begin{cases} 2(u_2-\ln u_2 - 1), & 0 \le u_2 \le 1\\ 0, & \textit{otherwise} \end{cases} \end{equation*}\]