Bivariate transformation of random variables: brute force algebra?











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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:



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










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    up vote
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    down vote

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    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:



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










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      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:



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










      share|cite|improve this question













      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:



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







      algebra-precalculus random-variables transformation






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      asked 2 days ago









      yqc

      936819




      936819






















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          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 $Z_2$






          share|cite|improve this answer





















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            up vote
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            down vote



            accepted










            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 $Z_2$






            share|cite|improve this answer

























              up vote
              0
              down vote



              accepted










              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 $Z_2$






              share|cite|improve this answer























                up vote
                0
                down vote



                accepted







                up vote
                0
                down vote



                accepted






                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 $Z_2$






                share|cite|improve this 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 $Z_2$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Makina

                861113




                861113






























                     

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