Bivariate transformation of random variables: brute force algebra?

36 Views Asked by At

Suppose I have 2 random variables, $Z_1$ and $Z_2$. I then define the following bivariate transformations,

\begin{equation} X = a_xZ_1 + b_xZ_2 + c_x \end{equation} $$Y = a_yZ_1 + b_yZ_2 + c_y$$

where $a_x, a_y, b_x, b_y, c_x, c_y$ are constants.

I want to write $Z_1$ and $Z_2$ in terms of $X$ and $Y$. Typically, this involves isolating one of $Z_1$, $Z_2$ and then plugging it into the second equation, i.e.

$$Z_2 = \frac{X-c_x-a_xZ_1}{b_x}$$ and then plug this expression into $Y = a_yZ_1 + b_yZ_2 + c_y$ to obtain an expression where I can write $Z_1$ as a function of $X$ and $Y$.

Typically, this involves some brute force algebra, and I'm wondering if there are any tips/tricks when it comes to dealing with the algebra so that I can quickly arrive at these expressions:

$$Z_1 = \frac{b_y(X - c_x) - b_x(Y-c_y)}{a_xb_y-a_yb_x}$$ $$Z_2 = \frac{a_y(X - c_x) - a_x (Y-c_y)}{a_yb_x-a_xb_y}$$

1

There are 1 best solutions below

0
On BEST ANSWER

Multiply both sides of the first equation by $b_y$ and both sides of the second equation by $b_x$ and then subtract second equation from the first one. You end up with $Z_1, X$ and $Y$ since $Z_2$ will cancel out from the subtraction.

Can do the same type of action for $Z_2$