Mixing
How to show that these two alternative formulas for slope are equivalent
$begingroup$
First Formula
$$b_1=dfrac{displaystylesum_{i=1}^n(Y_iX_i-bar Y bar X)}{displaystylesum_{i=1}^n(X_i^2-bar X^2)}$$
Second Formula
$$b_1=dfrac{displaystylesum_{i=1}^nY_i(X_i- bar X)}{displaystylesum_{i=1}^n(X_i-bar X)^2}$$
I understand the derivation of the slope $b_1$ in the Ordinary Least Squares regression. However, I am having a hard time converting the initial formula to the second one using algebra.
statistics regression
asked Jan 23 at 22:09
Jac FrallJac Frall
405
$endgroup$
add a comment |
$begingroup$
First Formula
$$b_1=dfrac{displaystylesum_{i=1}^n(Y_iX_i-bar Y bar X)}{displaystylesum_{i=1}^n(X_i^2-bar X^2)}$$
Second Formula
$$b_1=dfrac{displaystylesum_{i=1}^nY_i(X_i- bar X)}{displaystylesum_{i=1}^n(X_i-bar X)^2}$$
I understand the derivation of the slope $b_1$ in the Ordinary Least Squares regression. However, I am having a hard time converting the initial formula to the second one using algebra.
statistics regression
asked Jan 23 at 22:09
Jac FrallJac Frall
405
$endgroup$
$begingroup$
The numerators are obviously equal, thus you only need to show that the denominators are equal.
$endgroup$
– lightxbulb
Jan 23 at 22:18
add a comment |
$begingroup$
First Formula
$$b_1=dfrac{displaystylesum_{i=1}^n(Y_iX_i-bar Y bar X)}{displaystylesum_{i=1}^n(X_i^2-bar X^2)}$$
Second Formula
$$b_1=dfrac{displaystylesum_{i=1}^nY_i(X_i- bar X)}{displaystylesum_{i=1}^n(X_i-bar X)^2}$$
I understand the derivation of the slope $b_1$ in the Ordinary Least Squares regression. However, I am having a hard time converting the initial formula to the second one using algebra.
statistics regression
asked Jan 23 at 22:09
Jac FrallJac Frall
405
$endgroup$
First Formula
$$b_1=dfrac{displaystylesum_{i=1}^n(Y_iX_i-bar Y bar X)}{displaystylesum_{i=1}^n(X_i^2-bar X^2)}$$
Second Formula
$$b_1=dfrac{displaystylesum_{i=1}^nY_i(X_i- bar X)}{displaystylesum_{i=1}^n(X_i-bar X)^2}$$
I understand the derivation of the slope $b_1$ in the Ordinary Least Squares regression. However, I am having a hard time converting the initial formula to the second one using algebra.
statistics regression
statistics regression
asked Jan 23 at 22:09
Jac FrallJac Frall
405
asked Jan 23 at 22:09
Jac FrallJac Frall
405
asked Jan 23 at 22:09
Jac FrallJac Frall
405
asked Jan 23 at 22:09
Jac FrallJac Frall
405
asked Jan 23 at 22:09
Jac FrallJac Frall
405
405
$begingroup$
The numerators are obviously equal, thus you only need to show that the denominators are equal.
$endgroup$
– lightxbulb
Jan 23 at 22:18
add a comment |
$begingroup$
The numerators are obviously equal, thus you only need to show that the denominators are equal.
$endgroup$
– lightxbulb
Jan 23 at 22:18
$begingroup$
The numerators are obviously equal, thus you only need to show that the denominators are equal.
$endgroup$
– lightxbulb
Jan 23 at 22:18
$begingroup$
The numerators are obviously equal, thus you only need to show that the denominators are equal.
$endgroup$
– lightxbulb
Jan 23 at 22:18
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Both the numerator and the denominator can be adjusted using the two tricks that $sum_{i=1}^n C = nC$ where $C$ is a constant, and $nbar{Y} = sum_{i=1}^n Y_i$.
In the numerator, you just have to expand:
$sum bar{X} bar{Y} = bar{X} left(sum bar{Y}right) = bar{X} left(n bar{Y}right) = bar{X} sum Y_i$
and you can then combine that with the other term and factorise the common $bar{X}$
In the denominator, you need to expand the square, and then use the tricks to make two of the terms partially cancel out.
answered Jan 23 at 22:21
ConManConMan
7,9021324
$endgroup$
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$begingroup$
$$sum_{i=1}^n(X_i-bar X)^2=sum_{i=1}^n(X_i^2-2X_ibar X+(bar X)^2)=sum_{i=1}^nX_i^2-2bar Xsum_{i=1}^nX_i+bar X^2sum_{i=1}^n1$$
Now $sum_{i=1}^nX_i=nbar X=bar Xsum_{i=1}^n1$. I will replace this in the middle term. I get $$sum_{i=1}^nX_i^2-2bar Xbar Xsum_{i=1}^n1+bar X^2sum_{i=1}^n1=sum_{i=1}^nX_i^2-bar X^2sum_{i=1}^n1=sum_{i=1}^n(X_i^2-bar X^2)$$
You can proceed similarly for the numerator
answered Jan 23 at 22:23
AndreiAndrei
13.1k21230
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add a comment |
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2 Answers
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2 Answers
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active
oldest
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active
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votes
$begingroup$
Both the numerator and the denominator can be adjusted using the two tricks that $sum_{i=1}^n C = nC$ where $C$ is a constant, and $nbar{Y} = sum_{i=1}^n Y_i$.
In the numerator, you just have to expand:
$sum bar{X} bar{Y} = bar{X} left(sum bar{Y}right) = bar{X} left(n bar{Y}right) = bar{X} sum Y_i$
and you can then combine that with the other term and factorise the common $bar{X}$
In the denominator, you need to expand the square, and then use the tricks to make two of the terms partially cancel out.
answered Jan 23 at 22:21
ConManConMan
7,9021324
$endgroup$
add a comment |
$begingroup$
Both the numerator and the denominator can be adjusted using the two tricks that $sum_{i=1}^n C = nC$ where $C$ is a constant, and $nbar{Y} = sum_{i=1}^n Y_i$.
In the numerator, you just have to expand:
$sum bar{X} bar{Y} = bar{X} left(sum bar{Y}right) = bar{X} left(n bar{Y}right) = bar{X} sum Y_i$
and you can then combine that with the other term and factorise the common $bar{X}$
In the denominator, you need to expand the square, and then use the tricks to make two of the terms partially cancel out.
answered Jan 23 at 22:21
ConManConMan
7,9021324
$endgroup$
add a comment |
$begingroup$
Both the numerator and the denominator can be adjusted using the two tricks that $sum_{i=1}^n C = nC$ where $C$ is a constant, and $nbar{Y} = sum_{i=1}^n Y_i$.
In the numerator, you just have to expand:
$sum bar{X} bar{Y} = bar{X} left(sum bar{Y}right) = bar{X} left(n bar{Y}right) = bar{X} sum Y_i$
and you can then combine that with the other term and factorise the common $bar{X}$
In the denominator, you need to expand the square, and then use the tricks to make two of the terms partially cancel out.
answered Jan 23 at 22:21
ConManConMan
7,9021324
$endgroup$
Both the numerator and the denominator can be adjusted using the two tricks that $sum_{i=1}^n C = nC$ where $C$ is a constant, and $nbar{Y} = sum_{i=1}^n Y_i$.
In the numerator, you just have to expand:
$sum bar{X} bar{Y} = bar{X} left(sum bar{Y}right) = bar{X} left(n bar{Y}right) = bar{X} sum Y_i$
and you can then combine that with the other term and factorise the common $bar{X}$
In the denominator, you need to expand the square, and then use the tricks to make two of the terms partially cancel out.
answered Jan 23 at 22:21
ConManConMan
7,9021324
answered Jan 23 at 22:21
ConManConMan
7,9021324
answered Jan 23 at 22:21
ConManConMan
7,9021324
answered Jan 23 at 22:21
ConManConMan
7,9021324
7,9021324
add a comment |
add a comment |
$begingroup$
$$sum_{i=1}^n(X_i-bar X)^2=sum_{i=1}^n(X_i^2-2X_ibar X+(bar X)^2)=sum_{i=1}^nX_i^2-2bar Xsum_{i=1}^nX_i+bar X^2sum_{i=1}^n1$$
Now $sum_{i=1}^nX_i=nbar X=bar Xsum_{i=1}^n1$. I will replace this in the middle term. I get $$sum_{i=1}^nX_i^2-2bar Xbar Xsum_{i=1}^n1+bar X^2sum_{i=1}^n1=sum_{i=1}^nX_i^2-bar X^2sum_{i=1}^n1=sum_{i=1}^n(X_i^2-bar X^2)$$
You can proceed similarly for the numerator
answered Jan 23 at 22:23
AndreiAndrei
13.1k21230
$endgroup$
add a comment |
$begingroup$
$$sum_{i=1}^n(X_i-bar X)^2=sum_{i=1}^n(X_i^2-2X_ibar X+(bar X)^2)=sum_{i=1}^nX_i^2-2bar Xsum_{i=1}^nX_i+bar X^2sum_{i=1}^n1$$
Now $sum_{i=1}^nX_i=nbar X=bar Xsum_{i=1}^n1$. I will replace this in the middle term. I get $$sum_{i=1}^nX_i^2-2bar Xbar Xsum_{i=1}^n1+bar X^2sum_{i=1}^n1=sum_{i=1}^nX_i^2-bar X^2sum_{i=1}^n1=sum_{i=1}^n(X_i^2-bar X^2)$$
You can proceed similarly for the numerator
answered Jan 23 at 22:23
AndreiAndrei
13.1k21230
$endgroup$
add a comment |
$begingroup$
$$sum_{i=1}^n(X_i-bar X)^2=sum_{i=1}^n(X_i^2-2X_ibar X+(bar X)^2)=sum_{i=1}^nX_i^2-2bar Xsum_{i=1}^nX_i+bar X^2sum_{i=1}^n1$$
Now $sum_{i=1}^nX_i=nbar X=bar Xsum_{i=1}^n1$. I will replace this in the middle term. I get $$sum_{i=1}^nX_i^2-2bar Xbar Xsum_{i=1}^n1+bar X^2sum_{i=1}^n1=sum_{i=1}^nX_i^2-bar X^2sum_{i=1}^n1=sum_{i=1}^n(X_i^2-bar X^2)$$
You can proceed similarly for the numerator
answered Jan 23 at 22:23
AndreiAndrei
13.1k21230
$endgroup$
$$sum_{i=1}^n(X_i-bar X)^2=sum_{i=1}^n(X_i^2-2X_ibar X+(bar X)^2)=sum_{i=1}^nX_i^2-2bar Xsum_{i=1}^nX_i+bar X^2sum_{i=1}^n1$$
Now $sum_{i=1}^nX_i=nbar X=bar Xsum_{i=1}^n1$. I will replace this in the middle term. I get $$sum_{i=1}^nX_i^2-2bar Xbar Xsum_{i=1}^n1+bar X^2sum_{i=1}^n1=sum_{i=1}^nX_i^2-bar X^2sum_{i=1}^n1=sum_{i=1}^n(X_i^2-bar X^2)$$
You can proceed similarly for the numerator
answered Jan 23 at 22:23
AndreiAndrei
13.1k21230
answered Jan 23 at 22:23
AndreiAndrei
13.1k21230
answered Jan 23 at 22:23
AndreiAndrei
13.1k21230
answered Jan 23 at 22:23
AndreiAndrei
13.1k21230
13.1k21230
add a comment |
add a comment |
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$begingroup$
The numerators are obviously equal, thus you only need to show that the denominators are equal.
$endgroup$
– lightxbulb
Jan 23 at 22:18