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:
$$Z_1 = frac{b_y(X - c_x) - b_x(Y-c_y)}{a_xb_y-a_yb_x}$$
$$Z_2 = frac{a_y(X - c_x) - a_x (Y-c_y)}{a_yb_x-a_xb_y}$$
algebra-precalculus random-variables transformation
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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:
$$Z_1 = frac{b_y(X - c_x) - b_x(Y-c_y)}{a_xb_y-a_yb_x}$$
$$Z_2 = frac{a_y(X - c_x) - a_x (Y-c_y)}{a_yb_x-a_xb_y}$$
algebra-precalculus random-variables transformation
add a comment |
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:
$$Z_1 = frac{b_y(X - c_x) - b_x(Y-c_y)}{a_xb_y-a_yb_x}$$
$$Z_2 = frac{a_y(X - c_x) - a_x (Y-c_y)}{a_yb_x-a_xb_y}$$
algebra-precalculus random-variables transformation
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
algebra-precalculus random-variables transformation
asked 2 days ago
yqc
936819
936819
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1 Answer
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up vote
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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 $Z_2$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $Z_2$
add a comment |
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 $Z_2$
add a comment |
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 $Z_2$
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 $Z_2$
answered 2 days ago
Makina
861113
861113
add a comment |
add a comment |
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