Mixing
Property of least squares estimates question
$begingroup$
The assumption is the following.
$$E[Y_i]=beta _0 +beta _{1}X_i, Var[Y_i]=sigma ^2, Cov[Y_i,Y_j]=0, forall ine j $$
Where $hat beta_0$ and $hat beta_1 $ are the least squares estimators.
I want to prove that $E[hat beta_1]=beta_1$ and I am looking at my professor's work,
I get the first two rows, but I do not understand how he got the third row.
understand that
$$E[Sigma (x_i-bar{x})y_i]=Sigma left( E[(x_i-bar{x})y_i] right)$$
but I don't know how he got rid of the Expected value part.
May I get some help?
statistics regression expected-value
asked Feb 3 at 6:22
hyg17hyg17
2,00422044
$endgroup$
add a comment |
$begingroup$
The assumption is the following.
$$E[Y_i]=beta _0 +beta _{1}X_i, Var[Y_i]=sigma ^2, Cov[Y_i,Y_j]=0, forall ine j $$
Where $hat beta_0$ and $hat beta_1 $ are the least squares estimators.
I want to prove that $E[hat beta_1]=beta_1$ and I am looking at my professor's work,
I get the first two rows, but I do not understand how he got the third row.
understand that
$$E[Sigma (x_i-bar{x})y_i]=Sigma left( E[(x_i-bar{x})y_i] right)$$
but I don't know how he got rid of the Expected value part.
May I get some help?
statistics regression expected-value
asked Feb 3 at 6:22
hyg17hyg17
2,00422044
$endgroup$
$begingroup$
Because $x$ is non-random and $y$ is random( the denominator in $hatbeta_1$ is just a constant).
$endgroup$
– StubbornAtom
Feb 3 at 6:26
$begingroup$
Can you elaborate on that, please? How can you tell that x is non-random and y is?
$endgroup$
– hyg17
Feb 3 at 6:46
$begingroup$
That part must have been mentioned right at the start, because otherwise how can you proceed?
$endgroup$
– StubbornAtom
Feb 3 at 6:47
add a comment |
$begingroup$
The assumption is the following.
$$E[Y_i]=beta _0 +beta _{1}X_i, Var[Y_i]=sigma ^2, Cov[Y_i,Y_j]=0, forall ine j $$
Where $hat beta_0$ and $hat beta_1 $ are the least squares estimators.
I want to prove that $E[hat beta_1]=beta_1$ and I am looking at my professor's work,
I get the first two rows, but I do not understand how he got the third row.
understand that
$$E[Sigma (x_i-bar{x})y_i]=Sigma left( E[(x_i-bar{x})y_i] right)$$
but I don't know how he got rid of the Expected value part.
May I get some help?
statistics regression expected-value
asked Feb 3 at 6:22
hyg17hyg17
2,00422044
$endgroup$
The assumption is the following.
$$E[Y_i]=beta _0 +beta _{1}X_i, Var[Y_i]=sigma ^2, Cov[Y_i,Y_j]=0, forall ine j $$
Where $hat beta_0$ and $hat beta_1 $ are the least squares estimators.
I want to prove that $E[hat beta_1]=beta_1$ and I am looking at my professor's work,
I get the first two rows, but I do not understand how he got the third row.
understand that
$$E[Sigma (x_i-bar{x})y_i]=Sigma left( E[(x_i-bar{x})y_i] right)$$
but I don't know how he got rid of the Expected value part.
May I get some help?
statistics regression expected-value
statistics regression expected-value
asked Feb 3 at 6:22
hyg17hyg17
2,00422044
asked Feb 3 at 6:22
hyg17hyg17
2,00422044
asked Feb 3 at 6:22
hyg17hyg17
2,00422044
asked Feb 3 at 6:22
hyg17hyg17
2,00422044
asked Feb 3 at 6:22
hyg17hyg17
2,00422044
2,00422044
$begingroup$
Because $x$ is non-random and $y$ is random( the denominator in $hatbeta_1$ is just a constant).
$endgroup$
– StubbornAtom
Feb 3 at 6:26
$begingroup$
Can you elaborate on that, please? How can you tell that x is non-random and y is?
$endgroup$
– hyg17
Feb 3 at 6:46
$begingroup$
That part must have been mentioned right at the start, because otherwise how can you proceed?
$endgroup$
– StubbornAtom
Feb 3 at 6:47
add a comment |
$begingroup$
Because $x$ is non-random and $y$ is random( the denominator in $hatbeta_1$ is just a constant).
$endgroup$
– StubbornAtom
Feb 3 at 6:26
$begingroup$
Can you elaborate on that, please? How can you tell that x is non-random and y is?
$endgroup$
– hyg17
Feb 3 at 6:46
$begingroup$
That part must have been mentioned right at the start, because otherwise how can you proceed?
$endgroup$
– StubbornAtom
Feb 3 at 6:47
$begingroup$
Because $x$ is non-random and $y$ is random( the denominator in $hatbeta_1$ is just a constant).
$endgroup$
– StubbornAtom
Feb 3 at 6:26
$begingroup$
Because $x$ is non-random and $y$ is random( the denominator in $hatbeta_1$ is just a constant).
$endgroup$
– StubbornAtom
Feb 3 at 6:26
$begingroup$
Can you elaborate on that, please? How can you tell that x is non-random and y is?
$endgroup$
– hyg17
Feb 3 at 6:46
$begingroup$
Can you elaborate on that, please? How can you tell that x is non-random and y is?
$endgroup$
– hyg17
Feb 3 at 6:46
$begingroup$
That part must have been mentioned right at the start, because otherwise how can you proceed?
$endgroup$
– StubbornAtom
Feb 3 at 6:47
$begingroup$
That part must have been mentioned right at the start, because otherwise how can you proceed?
$endgroup$
– StubbornAtom
Feb 3 at 6:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$X$ is a variable, but not a random variable. To be more rigorous, I would prefer conditioning on $X$, i.e,
begin{align}
mathbb{E}[hat{beta}_1|X] &=mathbb{E}left(frac{sum(X_i - bar{X})Y_i}{sum (X_i - bar{X})^2} |X=mathrm{x} right)\
&=
left(frac{sum(X_i - bar{X})(beta_0 + beta_1 X_i)}{sum (X_i - bar{X})^2} |X=mathrm{x} right)\
&=beta_1
end{align}
answered Feb 3 at 20:15
V. VancakV. Vancak
11.4k31026
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
oldest
votes
$begingroup$
$X$ is a variable, but not a random variable. To be more rigorous, I would prefer conditioning on $X$, i.e,
begin{align}
mathbb{E}[hat{beta}_1|X] &=mathbb{E}left(frac{sum(X_i - bar{X})Y_i}{sum (X_i - bar{X})^2} |X=mathrm{x} right)\
&=
left(frac{sum(X_i - bar{X})(beta_0 + beta_1 X_i)}{sum (X_i - bar{X})^2} |X=mathrm{x} right)\
&=beta_1
end{align}
answered Feb 3 at 20:15
V. VancakV. Vancak
11.4k31026
$endgroup$
add a comment |
$begingroup$
$X$ is a variable, but not a random variable. To be more rigorous, I would prefer conditioning on $X$, i.e,
begin{align}
mathbb{E}[hat{beta}_1|X] &=mathbb{E}left(frac{sum(X_i - bar{X})Y_i}{sum (X_i - bar{X})^2} |X=mathrm{x} right)\
&=
left(frac{sum(X_i - bar{X})(beta_0 + beta_1 X_i)}{sum (X_i - bar{X})^2} |X=mathrm{x} right)\
&=beta_1
end{align}
answered Feb 3 at 20:15
V. VancakV. Vancak
11.4k31026
$endgroup$
add a comment |
$begingroup$
$X$ is a variable, but not a random variable. To be more rigorous, I would prefer conditioning on $X$, i.e,
begin{align}
mathbb{E}[hat{beta}_1|X] &=mathbb{E}left(frac{sum(X_i - bar{X})Y_i}{sum (X_i - bar{X})^2} |X=mathrm{x} right)\
&=
left(frac{sum(X_i - bar{X})(beta_0 + beta_1 X_i)}{sum (X_i - bar{X})^2} |X=mathrm{x} right)\
&=beta_1
end{align}
answered Feb 3 at 20:15
V. VancakV. Vancak
11.4k31026
$endgroup$
$X$ is a variable, but not a random variable. To be more rigorous, I would prefer conditioning on $X$, i.e,
begin{align}
mathbb{E}[hat{beta}_1|X] &=mathbb{E}left(frac{sum(X_i - bar{X})Y_i}{sum (X_i - bar{X})^2} |X=mathrm{x} right)\
&=
left(frac{sum(X_i - bar{X})(beta_0 + beta_1 X_i)}{sum (X_i - bar{X})^2} |X=mathrm{x} right)\
&=beta_1
end{align}
answered Feb 3 at 20:15
V. VancakV. Vancak
11.4k31026
answered Feb 3 at 20:15
V. VancakV. Vancak
11.4k31026
answered Feb 3 at 20:15
V. VancakV. Vancak
11.4k31026
answered Feb 3 at 20:15
V. VancakV. Vancak
11.4k31026
11.4k31026
add a comment |
add a comment |
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$begingroup$
Because $x$ is non-random and $y$ is random( the denominator in $hatbeta_1$ is just a constant).
$endgroup$
– StubbornAtom
Feb 3 at 6:26
$begingroup$
Can you elaborate on that, please? How can you tell that x is non-random and y is?
$endgroup$
– hyg17
Feb 3 at 6:46
$begingroup$
That part must have been mentioned right at the start, because otherwise how can you proceed?
$endgroup$
– StubbornAtom
Feb 3 at 6:47