How to show that the simple least square estimators minimizes the SSE using the second partial derivative...
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I am trying to apply the second partial derivative test to show that the simple least square estimators $hatbeta_0$ and $hatbeta_1$ does minimize the sum of the squared errors based on page 3 of this lecture notes. Based on the second last equations on page 3, I found that:
$frac{partial^2 SSE}{partial hatbeta_0^2} = 2n > 0 $, which should show that this is a minimum if the second condition holds true. However, in trying to evaluate the second condition, I ended up with:
$2n*2{sum}x^2 - 4({sum}x)^2$. I'm unsure if I did something wrong and if not, how do I prove that this is greater than zero?
partial-derivative least-squares
asked Dec 29 '18 at 20:08
YandleYandle
1254
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add a comment |
$begingroup$
I am trying to apply the second partial derivative test to show that the simple least square estimators $hatbeta_0$ and $hatbeta_1$ does minimize the sum of the squared errors based on page 3 of this lecture notes. Based on the second last equations on page 3, I found that:
$frac{partial^2 SSE}{partial hatbeta_0^2} = 2n > 0 $, which should show that this is a minimum if the second condition holds true. However, in trying to evaluate the second condition, I ended up with:
$2n*2{sum}x^2 - 4({sum}x)^2$. I'm unsure if I did something wrong and if not, how do I prove that this is greater than zero?
partial-derivative least-squares
asked Dec 29 '18 at 20:08
YandleYandle
1254
$endgroup$
add a comment |
$begingroup$
I am trying to apply the second partial derivative test to show that the simple least square estimators $hatbeta_0$ and $hatbeta_1$ does minimize the sum of the squared errors based on page 3 of this lecture notes. Based on the second last equations on page 3, I found that:
$frac{partial^2 SSE}{partial hatbeta_0^2} = 2n > 0 $, which should show that this is a minimum if the second condition holds true. However, in trying to evaluate the second condition, I ended up with:
$2n*2{sum}x^2 - 4({sum}x)^2$. I'm unsure if I did something wrong and if not, how do I prove that this is greater than zero?
partial-derivative least-squares
asked Dec 29 '18 at 20:08
YandleYandle
1254
$endgroup$
I am trying to apply the second partial derivative test to show that the simple least square estimators $hatbeta_0$ and $hatbeta_1$ does minimize the sum of the squared errors based on page 3 of this lecture notes. Based on the second last equations on page 3, I found that:
$frac{partial^2 SSE}{partial hatbeta_0^2} = 2n > 0 $, which should show that this is a minimum if the second condition holds true. However, in trying to evaluate the second condition, I ended up with:
$2n*2{sum}x^2 - 4({sum}x)^2$. I'm unsure if I did something wrong and if not, how do I prove that this is greater than zero?
partial-derivative least-squares
partial-derivative least-squares
asked Dec 29 '18 at 20:08
YandleYandle
1254
asked Dec 29 '18 at 20:08
YandleYandle
1254
asked Dec 29 '18 at 20:08
YandleYandle
1254
asked Dec 29 '18 at 20:08
YandleYandle
1254
asked Dec 29 '18 at 20:08
YandleYandle
1254
1254
add a comment |
add a comment |
1 Answer
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It seems that you are on the right track. You can divide both sides of the equation by 4 and $n^2$.
$$frac1n{sum_{i=1}^n}x_i^2 - frac1{n^2}left({sum_{i=1}^n}x_iright)^2$$
This is the definition of the (empirical) $text{variance}$ for $n$ values, which is always larger than $0$ (at least one $x_ineq overline x$). It can be further transformed to get a more familiar expression. We use that $frac1{n^2}left({sum_{i=1}^n}x_iright)^2=left(frac1{n}{sum_{i=1}^n}x_iright)^2=overline x^2$
$$frac1n{sum_{i=1}^n}x_i^2 - overline x^2$$
Adding at subtracting $2overline x^2$
$$frac1n{sum_{i=1}^n}x_i^2 underbrace{- 2overline x^2+2overline x^2}_{=0}- overline x^2$$
$$frac1n{sum_{i=1}^n}x_i^2 - 2overline x^2+overline x^2$$
$$frac1n{sum_{i=1}^n}x_i^2 - 2left(frac1n sum_{i=1}^{n} x_iright) overline x+overline x^2$$
$$frac1n{sum_{i=1}^n}x_i^2 - 2frac1n sum_{i=1}^{n} x_ioverline x++overline x^2$$
$$frac1nsum_{i=1}^{n} left(x_i^2-2x_ioverline x+overline x^2right)$$
Using the first binomial fomumla we get another common version of the variance
$$frac1nsum_{i=1}^{n} left(x_i-overline xright)^2>0$$
A square of a number is always positive if the squared number is not $0$. Again $x_ineq overline x$ in at least one case.
answered Dec 29 '18 at 21:20
callculuscallculus
17.9k31427
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1 Answer
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1 Answer
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$begingroup$
It seems that you are on the right track. You can divide both sides of the equation by 4 and $n^2$.
$$frac1n{sum_{i=1}^n}x_i^2 - frac1{n^2}left({sum_{i=1}^n}x_iright)^2$$
This is the definition of the (empirical) $text{variance}$ for $n$ values, which is always larger than $0$ (at least one $x_ineq overline x$). It can be further transformed to get a more familiar expression. We use that $frac1{n^2}left({sum_{i=1}^n}x_iright)^2=left(frac1{n}{sum_{i=1}^n}x_iright)^2=overline x^2$
$$frac1n{sum_{i=1}^n}x_i^2 - overline x^2$$
Adding at subtracting $2overline x^2$
$$frac1n{sum_{i=1}^n}x_i^2 underbrace{- 2overline x^2+2overline x^2}_{=0}- overline x^2$$
$$frac1n{sum_{i=1}^n}x_i^2 - 2overline x^2+overline x^2$$
$$frac1n{sum_{i=1}^n}x_i^2 - 2left(frac1n sum_{i=1}^{n} x_iright) overline x+overline x^2$$
$$frac1n{sum_{i=1}^n}x_i^2 - 2frac1n sum_{i=1}^{n} x_ioverline x++overline x^2$$
$$frac1nsum_{i=1}^{n} left(x_i^2-2x_ioverline x+overline x^2right)$$
Using the first binomial fomumla we get another common version of the variance
$$frac1nsum_{i=1}^{n} left(x_i-overline xright)^2>0$$
A square of a number is always positive if the squared number is not $0$. Again $x_ineq overline x$ in at least one case.
answered Dec 29 '18 at 21:20
callculuscallculus
17.9k31427
$endgroup$
add a comment |
$begingroup$
It seems that you are on the right track. You can divide both sides of the equation by 4 and $n^2$.
$$frac1n{sum_{i=1}^n}x_i^2 - frac1{n^2}left({sum_{i=1}^n}x_iright)^2$$
This is the definition of the (empirical) $text{variance}$ for $n$ values, which is always larger than $0$ (at least one $x_ineq overline x$). It can be further transformed to get a more familiar expression. We use that $frac1{n^2}left({sum_{i=1}^n}x_iright)^2=left(frac1{n}{sum_{i=1}^n}x_iright)^2=overline x^2$
$$frac1n{sum_{i=1}^n}x_i^2 - overline x^2$$
Adding at subtracting $2overline x^2$
$$frac1n{sum_{i=1}^n}x_i^2 underbrace{- 2overline x^2+2overline x^2}_{=0}- overline x^2$$
$$frac1n{sum_{i=1}^n}x_i^2 - 2overline x^2+overline x^2$$
$$frac1n{sum_{i=1}^n}x_i^2 - 2left(frac1n sum_{i=1}^{n} x_iright) overline x+overline x^2$$
$$frac1n{sum_{i=1}^n}x_i^2 - 2frac1n sum_{i=1}^{n} x_ioverline x++overline x^2$$
$$frac1nsum_{i=1}^{n} left(x_i^2-2x_ioverline x+overline x^2right)$$
Using the first binomial fomumla we get another common version of the variance
$$frac1nsum_{i=1}^{n} left(x_i-overline xright)^2>0$$
A square of a number is always positive if the squared number is not $0$. Again $x_ineq overline x$ in at least one case.
answered Dec 29 '18 at 21:20
callculuscallculus
17.9k31427
$endgroup$
add a comment |
$begingroup$
It seems that you are on the right track. You can divide both sides of the equation by 4 and $n^2$.
$$frac1n{sum_{i=1}^n}x_i^2 - frac1{n^2}left({sum_{i=1}^n}x_iright)^2$$
This is the definition of the (empirical) $text{variance}$ for $n$ values, which is always larger than $0$ (at least one $x_ineq overline x$). It can be further transformed to get a more familiar expression. We use that $frac1{n^2}left({sum_{i=1}^n}x_iright)^2=left(frac1{n}{sum_{i=1}^n}x_iright)^2=overline x^2$
$$frac1n{sum_{i=1}^n}x_i^2 - overline x^2$$
Adding at subtracting $2overline x^2$
$$frac1n{sum_{i=1}^n}x_i^2 underbrace{- 2overline x^2+2overline x^2}_{=0}- overline x^2$$
$$frac1n{sum_{i=1}^n}x_i^2 - 2overline x^2+overline x^2$$
$$frac1n{sum_{i=1}^n}x_i^2 - 2left(frac1n sum_{i=1}^{n} x_iright) overline x+overline x^2$$
$$frac1n{sum_{i=1}^n}x_i^2 - 2frac1n sum_{i=1}^{n} x_ioverline x++overline x^2$$
$$frac1nsum_{i=1}^{n} left(x_i^2-2x_ioverline x+overline x^2right)$$
Using the first binomial fomumla we get another common version of the variance
$$frac1nsum_{i=1}^{n} left(x_i-overline xright)^2>0$$
A square of a number is always positive if the squared number is not $0$. Again $x_ineq overline x$ in at least one case.
answered Dec 29 '18 at 21:20
callculuscallculus
17.9k31427
$endgroup$
It seems that you are on the right track. You can divide both sides of the equation by 4 and $n^2$.
$$frac1n{sum_{i=1}^n}x_i^2 - frac1{n^2}left({sum_{i=1}^n}x_iright)^2$$
This is the definition of the (empirical) $text{variance}$ for $n$ values, which is always larger than $0$ (at least one $x_ineq overline x$). It can be further transformed to get a more familiar expression. We use that $frac1{n^2}left({sum_{i=1}^n}x_iright)^2=left(frac1{n}{sum_{i=1}^n}x_iright)^2=overline x^2$
$$frac1n{sum_{i=1}^n}x_i^2 - overline x^2$$
Adding at subtracting $2overline x^2$
$$frac1n{sum_{i=1}^n}x_i^2 underbrace{- 2overline x^2+2overline x^2}_{=0}- overline x^2$$
$$frac1n{sum_{i=1}^n}x_i^2 - 2overline x^2+overline x^2$$
$$frac1n{sum_{i=1}^n}x_i^2 - 2left(frac1n sum_{i=1}^{n} x_iright) overline x+overline x^2$$
$$frac1n{sum_{i=1}^n}x_i^2 - 2frac1n sum_{i=1}^{n} x_ioverline x++overline x^2$$
$$frac1nsum_{i=1}^{n} left(x_i^2-2x_ioverline x+overline x^2right)$$
Using the first binomial fomumla we get another common version of the variance
$$frac1nsum_{i=1}^{n} left(x_i-overline xright)^2>0$$
A square of a number is always positive if the squared number is not $0$. Again $x_ineq overline x$ in at least one case.
answered Dec 29 '18 at 21:20
callculuscallculus
17.9k31427
answered Dec 29 '18 at 21:20
callculuscallculus
17.9k31427
answered Dec 29 '18 at 21:20
callculuscallculus
17.9k31427
answered Dec 29 '18 at 21:20
callculuscallculus
17.9k31427
17.9k31427
add a comment |
add a comment |
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