How to show $operatornameCov(b_0,b_1)=-fracsigmabarxS_xx$
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Consider the equation $y_i=beta_0+beta_1x_i+epsilon_i$ for $i=1, dotsc, n$.
We have unbiased estimators $b_0$ and $b_1$ for $beta_0$ and $beta_1$ respectively, where $b_0=bary-b_1barx$ and $b_1= S_xy / S_xx$.
How does one show that $operatornameCov(b_0,b_1)=-fracsigmabarxS_xx$
I tried using $operatornameCov(b_0,b_1)=E(b_0b_1)-E(b_0)E(b_1)$ to no avail as it just equals $0$ when I try and do that. Thanks!
probability probability-theory statistics regression
edited Sep 2 at 12:21
Bernard
112k635104
asked Sep 2 at 12:17
Programmer
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up vote
2
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Consider the equation $y_i=beta_0+beta_1x_i+epsilon_i$ for $i=1, dotsc, n$.
We have unbiased estimators $b_0$ and $b_1$ for $beta_0$ and $beta_1$ respectively, where $b_0=bary-b_1barx$ and $b_1= S_xy / S_xx$.
How does one show that $operatornameCov(b_0,b_1)=-fracsigmabarxS_xx$
I tried using $operatornameCov(b_0,b_1)=E(b_0b_1)-E(b_0)E(b_1)$ to no avail as it just equals $0$ when I try and do that. Thanks!
probability probability-theory statistics regression
edited Sep 2 at 12:21
Bernard
112k635104
asked Sep 2 at 12:17
Programmer
392112
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider the equation $y_i=beta_0+beta_1x_i+epsilon_i$ for $i=1, dotsc, n$.
We have unbiased estimators $b_0$ and $b_1$ for $beta_0$ and $beta_1$ respectively, where $b_0=bary-b_1barx$ and $b_1= S_xy / S_xx$.
How does one show that $operatornameCov(b_0,b_1)=-fracsigmabarxS_xx$
I tried using $operatornameCov(b_0,b_1)=E(b_0b_1)-E(b_0)E(b_1)$ to no avail as it just equals $0$ when I try and do that. Thanks!
probability probability-theory statistics regression
edited Sep 2 at 12:21
Bernard
112k635104
asked Sep 2 at 12:17
Programmer
392112
Consider the equation $y_i=beta_0+beta_1x_i+epsilon_i$ for $i=1, dotsc, n$.
We have unbiased estimators $b_0$ and $b_1$ for $beta_0$ and $beta_1$ respectively, where $b_0=bary-b_1barx$ and $b_1= S_xy / S_xx$.
How does one show that $operatornameCov(b_0,b_1)=-fracsigmabarxS_xx$
I tried using $operatornameCov(b_0,b_1)=E(b_0b_1)-E(b_0)E(b_1)$ to no avail as it just equals $0$ when I try and do that. Thanks!
probability probability-theory statistics regression
probability probability-theory statistics regression
edited Sep 2 at 12:21
Bernard
112k635104
asked Sep 2 at 12:17
Programmer
392112
edited Sep 2 at 12:21
Bernard
112k635104
asked Sep 2 at 12:17
Programmer
392112
edited Sep 2 at 12:21
Bernard
112k635104
edited Sep 2 at 12:21
Bernard
112k635104
edited Sep 2 at 12:21
Bernard
112k635104
112k635104
asked Sep 2 at 12:17
Programmer
392112
asked Sep 2 at 12:17
Programmer
392112
asked Sep 2 at 12:17
Programmer
392112
392112
add a comment |Â
add a comment |Â
1 Answer
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1
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You have not defined your symbols, so presumably $$S_xx=sum (x_i-bar x)^2 quadtext and quad S_xy=sum (x_i-bar x)(y_i-bar y)$$
Use the bilinearity of covariance.
We have
beginalign
operatornameCov(b_0,b_1)&=operatornameCov(bar y-b_1bar x,b_1)
\&=operatornameCov(bar y,b_1)-bar xoperatornameVar(b_1)
endalign
Recall that $bar y$ and $b_1$ are independently distributed, so their covariance vanishes.
Moreover, the exact distribution of $b_1$ is $$b_1simmathcal Nleft(beta_1,fracsigma^2S_xxright)$$
Hence you finally get as the covariance between the least square estimates $$operatornameCov(b_0,b_1)=-fracbar xsigma^2S_xx$$
answered Sep 2 at 12:55
StubbornAtom
4,09511135
Thanks so much! And sorry for not defining the sum of squares!
– Programmer
Sep 2 at 13:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have not defined your symbols, so presumably $$S_xx=sum (x_i-bar x)^2 quadtext and quad S_xy=sum (x_i-bar x)(y_i-bar y)$$
Use the bilinearity of covariance.
We have
beginalign
operatornameCov(b_0,b_1)&=operatornameCov(bar y-b_1bar x,b_1)
\&=operatornameCov(bar y,b_1)-bar xoperatornameVar(b_1)
endalign
Recall that $bar y$ and $b_1$ are independently distributed, so their covariance vanishes.
Moreover, the exact distribution of $b_1$ is $$b_1simmathcal Nleft(beta_1,fracsigma^2S_xxright)$$
Hence you finally get as the covariance between the least square estimates $$operatornameCov(b_0,b_1)=-fracbar xsigma^2S_xx$$
answered Sep 2 at 12:55
StubbornAtom
4,09511135
Thanks so much! And sorry for not defining the sum of squares!
– Programmer
Sep 2 at 13:07
add a comment |Â
up vote
1
down vote
accepted
You have not defined your symbols, so presumably $$S_xx=sum (x_i-bar x)^2 quadtext and quad S_xy=sum (x_i-bar x)(y_i-bar y)$$
Use the bilinearity of covariance.
We have
beginalign
operatornameCov(b_0,b_1)&=operatornameCov(bar y-b_1bar x,b_1)
\&=operatornameCov(bar y,b_1)-bar xoperatornameVar(b_1)
endalign
Recall that $bar y$ and $b_1$ are independently distributed, so their covariance vanishes.
Moreover, the exact distribution of $b_1$ is $$b_1simmathcal Nleft(beta_1,fracsigma^2S_xxright)$$
Hence you finally get as the covariance between the least square estimates $$operatornameCov(b_0,b_1)=-fracbar xsigma^2S_xx$$
answered Sep 2 at 12:55
StubbornAtom
4,09511135
Thanks so much! And sorry for not defining the sum of squares!
– Programmer
Sep 2 at 13:07
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have not defined your symbols, so presumably $$S_xx=sum (x_i-bar x)^2 quadtext and quad S_xy=sum (x_i-bar x)(y_i-bar y)$$
Use the bilinearity of covariance.
We have
beginalign
operatornameCov(b_0,b_1)&=operatornameCov(bar y-b_1bar x,b_1)
\&=operatornameCov(bar y,b_1)-bar xoperatornameVar(b_1)
endalign
Recall that $bar y$ and $b_1$ are independently distributed, so their covariance vanishes.
Moreover, the exact distribution of $b_1$ is $$b_1simmathcal Nleft(beta_1,fracsigma^2S_xxright)$$
Hence you finally get as the covariance between the least square estimates $$operatornameCov(b_0,b_1)=-fracbar xsigma^2S_xx$$
answered Sep 2 at 12:55
StubbornAtom
4,09511135
You have not defined your symbols, so presumably $$S_xx=sum (x_i-bar x)^2 quadtext and quad S_xy=sum (x_i-bar x)(y_i-bar y)$$
Use the bilinearity of covariance.
We have
beginalign
operatornameCov(b_0,b_1)&=operatornameCov(bar y-b_1bar x,b_1)
\&=operatornameCov(bar y,b_1)-bar xoperatornameVar(b_1)
endalign
Recall that $bar y$ and $b_1$ are independently distributed, so their covariance vanishes.
Moreover, the exact distribution of $b_1$ is $$b_1simmathcal Nleft(beta_1,fracsigma^2S_xxright)$$
Hence you finally get as the covariance between the least square estimates $$operatornameCov(b_0,b_1)=-fracbar xsigma^2S_xx$$
answered Sep 2 at 12:55
StubbornAtom
4,09511135
answered Sep 2 at 12:55
StubbornAtom
4,09511135
answered Sep 2 at 12:55
StubbornAtom
4,09511135
answered Sep 2 at 12:55
StubbornAtom
4,09511135
4,09511135
Thanks so much! And sorry for not defining the sum of squares!
– Programmer
Sep 2 at 13:07
add a comment |Â
Thanks so much! And sorry for not defining the sum of squares!
– Programmer
Sep 2 at 13:07
Thanks so much! And sorry for not defining the sum of squares!
– Programmer
Sep 2 at 13:07
Thanks so much! And sorry for not defining the sum of squares!
– Programmer
Sep 2 at 13:07
add a comment |Â
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