Mixing
Calculating the var(β) in a least square regression model
$begingroup$
The linear model that I'm working with is:
$$y_t =α +βx_t + ε_t$$
Based on my Lecture I have:
$$Var(hatβ) = Var(Σw_tε_t)$$
where ε is the error term and
$$w_t = frac{x_t-overline x}{Σ(x_t-overline x)^2}$$
That being said we then have:
$$begin{align}Var(hatβ)& = ΣVar(w_tε_t)+ΣΣCov(w_s,ε_t)\&=E[w_tε_t - E(w_tε_t)]^2\ &= E[w_tε_t]^2\&= E[w_t^2ε_t^2]\&= Σw_t^2Var(ε_t)\&=σ^2Σw_t^2 end{align}$$
What bothers me the most here is how come:
$$Var(Σw_tε_t) = ΣVar(w_tε_t)+ΣΣCov(w_s,ε_t)$$
and how do we get from this:
$$E[w_t^2ε_t^2]$$
to this:
$$Σw_t^2Var(ε_t)$$
Many thanks in advance!
regression least-squares linear-regression regression-analysis
edited Jan 23 at 22:58
Alejandro Nasif Salum
4,765118
asked Jan 23 at 22:25
FozoroFozoro
1265
$endgroup$
add a comment |
$begingroup$
The linear model that I'm working with is:
$$y_t =α +βx_t + ε_t$$
Based on my Lecture I have:
$$Var(hatβ) = Var(Σw_tε_t)$$
where ε is the error term and
$$w_t = frac{x_t-overline x}{Σ(x_t-overline x)^2}$$
That being said we then have:
$$begin{align}Var(hatβ)& = ΣVar(w_tε_t)+ΣΣCov(w_s,ε_t)\&=E[w_tε_t - E(w_tε_t)]^2\ &= E[w_tε_t]^2\&= E[w_t^2ε_t^2]\&= Σw_t^2Var(ε_t)\&=σ^2Σw_t^2 end{align}$$
What bothers me the most here is how come:
$$Var(Σw_tε_t) = ΣVar(w_tε_t)+ΣΣCov(w_s,ε_t)$$
and how do we get from this:
$$E[w_t^2ε_t^2]$$
to this:
$$Σw_t^2Var(ε_t)$$
Many thanks in advance!
regression least-squares linear-regression regression-analysis
edited Jan 23 at 22:58
Alejandro Nasif Salum
4,765118
asked Jan 23 at 22:25
FozoroFozoro
1265
$endgroup$
$begingroup$
Which is exactly the linear model you have? Just one exogenous variable? Are you assuming that the values of $x$ are predetermined (that is, non random) or not?
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 22:39
$begingroup$
Oops, very sorry I just added the linear equation. Thanks for your quick reply
$endgroup$
– Fozoro
Jan 23 at 22:43
$begingroup$
The term $w$ should not have a summation over $x - bar x$, should it?
$endgroup$
– Mark Viola
Jan 23 at 22:45
$begingroup$
@MarkViola Yes your right I put it there by accident ( just edited my question )
$endgroup$
– Fozoro
Jan 23 at 22:48
$begingroup$
Check out that there is a $w$ weight for each observation; so I added a $t$ subindex to your definition of those weights.
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 23:01
add a comment |
$begingroup$
The linear model that I'm working with is:
$$y_t =α +βx_t + ε_t$$
Based on my Lecture I have:
$$Var(hatβ) = Var(Σw_tε_t)$$
where ε is the error term and
$$w_t = frac{x_t-overline x}{Σ(x_t-overline x)^2}$$
That being said we then have:
$$begin{align}Var(hatβ)& = ΣVar(w_tε_t)+ΣΣCov(w_s,ε_t)\&=E[w_tε_t - E(w_tε_t)]^2\ &= E[w_tε_t]^2\&= E[w_t^2ε_t^2]\&= Σw_t^2Var(ε_t)\&=σ^2Σw_t^2 end{align}$$
What bothers me the most here is how come:
$$Var(Σw_tε_t) = ΣVar(w_tε_t)+ΣΣCov(w_s,ε_t)$$
and how do we get from this:
$$E[w_t^2ε_t^2]$$
to this:
$$Σw_t^2Var(ε_t)$$
Many thanks in advance!
regression least-squares linear-regression regression-analysis
edited Jan 23 at 22:58
Alejandro Nasif Salum
4,765118
asked Jan 23 at 22:25
FozoroFozoro
1265
$endgroup$
The linear model that I'm working with is:
$$y_t =α +βx_t + ε_t$$
Based on my Lecture I have:
$$Var(hatβ) = Var(Σw_tε_t)$$
where ε is the error term and
$$w_t = frac{x_t-overline x}{Σ(x_t-overline x)^2}$$
That being said we then have:
$$begin{align}Var(hatβ)& = ΣVar(w_tε_t)+ΣΣCov(w_s,ε_t)\&=E[w_tε_t - E(w_tε_t)]^2\ &= E[w_tε_t]^2\&= E[w_t^2ε_t^2]\&= Σw_t^2Var(ε_t)\&=σ^2Σw_t^2 end{align}$$
What bothers me the most here is how come:
$$Var(Σw_tε_t) = ΣVar(w_tε_t)+ΣΣCov(w_s,ε_t)$$
and how do we get from this:
$$E[w_t^2ε_t^2]$$
to this:
$$Σw_t^2Var(ε_t)$$
Many thanks in advance!
regression least-squares linear-regression regression-analysis
regression least-squares linear-regression regression-analysis
edited Jan 23 at 22:58
Alejandro Nasif Salum
4,765118
asked Jan 23 at 22:25
FozoroFozoro
1265
edited Jan 23 at 22:58
Alejandro Nasif Salum
4,765118
asked Jan 23 at 22:25
FozoroFozoro
1265
edited Jan 23 at 22:58
Alejandro Nasif Salum
4,765118
edited Jan 23 at 22:58
Alejandro Nasif Salum
4,765118
edited Jan 23 at 22:58
Alejandro Nasif Salum
4,765118
4,765118
asked Jan 23 at 22:25
FozoroFozoro
1265
asked Jan 23 at 22:25
FozoroFozoro
1265
asked Jan 23 at 22:25
FozoroFozoro
1265
1265
$begingroup$
Which is exactly the linear model you have? Just one exogenous variable? Are you assuming that the values of $x$ are predetermined (that is, non random) or not?
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 22:39
$begingroup$
Oops, very sorry I just added the linear equation. Thanks for your quick reply
$endgroup$
– Fozoro
Jan 23 at 22:43
$begingroup$
The term $w$ should not have a summation over $x - bar x$, should it?
$endgroup$
– Mark Viola
Jan 23 at 22:45
$begingroup$
@MarkViola Yes your right I put it there by accident ( just edited my question )
$endgroup$
– Fozoro
Jan 23 at 22:48
$begingroup$
Check out that there is a $w$ weight for each observation; so I added a $t$ subindex to your definition of those weights.
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 23:01
add a comment |
$begingroup$
Which is exactly the linear model you have? Just one exogenous variable? Are you assuming that the values of $x$ are predetermined (that is, non random) or not?
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 22:39
$begingroup$
Oops, very sorry I just added the linear equation. Thanks for your quick reply
$endgroup$
– Fozoro
Jan 23 at 22:43
$begingroup$
The term $w$ should not have a summation over $x - bar x$, should it?
$endgroup$
– Mark Viola
Jan 23 at 22:45
$begingroup$
@MarkViola Yes your right I put it there by accident ( just edited my question )
$endgroup$
– Fozoro
Jan 23 at 22:48
$begingroup$
Check out that there is a $w$ weight for each observation; so I added a $t$ subindex to your definition of those weights.
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 23:01
$begingroup$
Which is exactly the linear model you have? Just one exogenous variable? Are you assuming that the values of $x$ are predetermined (that is, non random) or not?
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 22:39
$begingroup$
Which is exactly the linear model you have? Just one exogenous variable? Are you assuming that the values of $x$ are predetermined (that is, non random) or not?
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 22:39
$begingroup$
Oops, very sorry I just added the linear equation. Thanks for your quick reply
$endgroup$
– Fozoro
Jan 23 at 22:43
$begingroup$
Oops, very sorry I just added the linear equation. Thanks for your quick reply
$endgroup$
– Fozoro
Jan 23 at 22:43
$begingroup$
The term $w$ should not have a summation over $x - bar x$, should it?
$endgroup$
– Mark Viola
Jan 23 at 22:45
$begingroup$
The term $w$ should not have a summation over $x - bar x$, should it?
$endgroup$
– Mark Viola
Jan 23 at 22:45
$begingroup$
@MarkViola Yes your right I put it there by accident ( just edited my question )
$endgroup$
– Fozoro
Jan 23 at 22:48
$begingroup$
@MarkViola Yes your right I put it there by accident ( just edited my question )
$endgroup$
– Fozoro
Jan 23 at 22:48
$begingroup$
Check out that there is a $w$ weight for each observation; so I added a $t$ subindex to your definition of those weights.
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 23:01
$begingroup$
Check out that there is a $w$ weight for each observation; so I added a $t$ subindex to your definition of those weights.
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 23:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you're not assuming that the $x_t$ values are random, and if you assume no serial correlation and homoskedasticty (that is
$$text{var}(e_t)=sigma^2$$
and
$$text{cov}(e_t,e_s)=0,quad tneq s$$
for any $t$ and $s$), then
$$text{var}(hat beta)=text{var}left(sum w_t e_tright)=sum w_t^2 text{var}(e_t)=sigma^2sum w_t^2.$$
and you can check that this amounts to
$$text{var}(hatbeta)=frac{sigma^2}{sum(x_t-bar x)^2}.$$
The formula with the covariance you give is just wrong. It is true although that
$$text{var}left(sum w_t e_tright)=sum text{var}(w_t e_t)+sum_{tneq s} text{cov}(w_te_t,w_s e_s)=$$
$$=sum w_t^2text{var}(e_t)+sum_{tneq s} w_t w_stext{cov}(e_t,e_s),$$
but our assumptions imply that the second term is zero anyway.
answered Jan 23 at 22:55
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
$begingroup$
Thank you very much for your answer. That beeing said I don't quite understand how we went from $E[w_t^2ε_t^2]$ to $Σw_t^2Var(ε_t)$ did we use some kind of formula?
$endgroup$
– Fozoro
Jan 23 at 23:06
$begingroup$
also how come we take out $w$ from the variance. Isn't w a variable?
$endgroup$
– Fozoro
Jan 23 at 23:14
$begingroup$
If $a$ is not random, then $$text{var}(aX)=a^2text{var}(X)$$ and if $text{cov}(X,Y)=0$ (in particular, if $X$ and $Y$ are independent), then $$text{var}(X+Y)=text{var}(X)+text{var}(Y).$$
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 23:17
1
$begingroup$
I'm making the classical assumption that the $x_t$ values are not random but predetermined before measuring $Y_t$ (check if this is the background with which you're supposed to be working). In that case, since the $w_t$ variables only depend on the $x_t$ values, they're not random either and can be thought of as constants (in the probabilistic sense).
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 23:19
1
$begingroup$
Great, I get it now thank you very much for your help!
$endgroup$
– Fozoro
Jan 23 at 23:22
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you're not assuming that the $x_t$ values are random, and if you assume no serial correlation and homoskedasticty (that is
$$text{var}(e_t)=sigma^2$$
and
$$text{cov}(e_t,e_s)=0,quad tneq s$$
for any $t$ and $s$), then
$$text{var}(hat beta)=text{var}left(sum w_t e_tright)=sum w_t^2 text{var}(e_t)=sigma^2sum w_t^2.$$
and you can check that this amounts to
$$text{var}(hatbeta)=frac{sigma^2}{sum(x_t-bar x)^2}.$$
The formula with the covariance you give is just wrong. It is true although that
$$text{var}left(sum w_t e_tright)=sum text{var}(w_t e_t)+sum_{tneq s} text{cov}(w_te_t,w_s e_s)=$$
$$=sum w_t^2text{var}(e_t)+sum_{tneq s} w_t w_stext{cov}(e_t,e_s),$$
but our assumptions imply that the second term is zero anyway.
answered Jan 23 at 22:55
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
$begingroup$
Thank you very much for your answer. That beeing said I don't quite understand how we went from $E[w_t^2ε_t^2]$ to $Σw_t^2Var(ε_t)$ did we use some kind of formula?
$endgroup$
– Fozoro
Jan 23 at 23:06
$begingroup$
also how come we take out $w$ from the variance. Isn't w a variable?
$endgroup$
– Fozoro
Jan 23 at 23:14
$begingroup$
If $a$ is not random, then $$text{var}(aX)=a^2text{var}(X)$$ and if $text{cov}(X,Y)=0$ (in particular, if $X$ and $Y$ are independent), then $$text{var}(X+Y)=text{var}(X)+text{var}(Y).$$
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 23:17
1
$begingroup$
I'm making the classical assumption that the $x_t$ values are not random but predetermined before measuring $Y_t$ (check if this is the background with which you're supposed to be working). In that case, since the $w_t$ variables only depend on the $x_t$ values, they're not random either and can be thought of as constants (in the probabilistic sense).
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 23:19
1
$begingroup$
Great, I get it now thank you very much for your help!
$endgroup$
– Fozoro
Jan 23 at 23:22
add a comment |
$begingroup$
If you're not assuming that the $x_t$ values are random, and if you assume no serial correlation and homoskedasticty (that is
$$text{var}(e_t)=sigma^2$$
and
$$text{cov}(e_t,e_s)=0,quad tneq s$$
for any $t$ and $s$), then
$$text{var}(hat beta)=text{var}left(sum w_t e_tright)=sum w_t^2 text{var}(e_t)=sigma^2sum w_t^2.$$
and you can check that this amounts to
$$text{var}(hatbeta)=frac{sigma^2}{sum(x_t-bar x)^2}.$$
The formula with the covariance you give is just wrong. It is true although that
$$text{var}left(sum w_t e_tright)=sum text{var}(w_t e_t)+sum_{tneq s} text{cov}(w_te_t,w_s e_s)=$$
$$=sum w_t^2text{var}(e_t)+sum_{tneq s} w_t w_stext{cov}(e_t,e_s),$$
but our assumptions imply that the second term is zero anyway.
answered Jan 23 at 22:55
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
$begingroup$
Thank you very much for your answer. That beeing said I don't quite understand how we went from $E[w_t^2ε_t^2]$ to $Σw_t^2Var(ε_t)$ did we use some kind of formula?
$endgroup$
– Fozoro
Jan 23 at 23:06
$begingroup$
also how come we take out $w$ from the variance. Isn't w a variable?
$endgroup$
– Fozoro
Jan 23 at 23:14
$begingroup$
If $a$ is not random, then $$text{var}(aX)=a^2text{var}(X)$$ and if $text{cov}(X,Y)=0$ (in particular, if $X$ and $Y$ are independent), then $$text{var}(X+Y)=text{var}(X)+text{var}(Y).$$
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 23:17
1
$begingroup$
I'm making the classical assumption that the $x_t$ values are not random but predetermined before measuring $Y_t$ (check if this is the background with which you're supposed to be working). In that case, since the $w_t$ variables only depend on the $x_t$ values, they're not random either and can be thought of as constants (in the probabilistic sense).
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 23:19
1
$begingroup$
Great, I get it now thank you very much for your help!
$endgroup$
– Fozoro
Jan 23 at 23:22
add a comment |
$begingroup$
If you're not assuming that the $x_t$ values are random, and if you assume no serial correlation and homoskedasticty (that is
$$text{var}(e_t)=sigma^2$$
and
$$text{cov}(e_t,e_s)=0,quad tneq s$$
for any $t$ and $s$), then
$$text{var}(hat beta)=text{var}left(sum w_t e_tright)=sum w_t^2 text{var}(e_t)=sigma^2sum w_t^2.$$
and you can check that this amounts to
$$text{var}(hatbeta)=frac{sigma^2}{sum(x_t-bar x)^2}.$$
The formula with the covariance you give is just wrong. It is true although that
$$text{var}left(sum w_t e_tright)=sum text{var}(w_t e_t)+sum_{tneq s} text{cov}(w_te_t,w_s e_s)=$$
$$=sum w_t^2text{var}(e_t)+sum_{tneq s} w_t w_stext{cov}(e_t,e_s),$$
but our assumptions imply that the second term is zero anyway.
answered Jan 23 at 22:55
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
If you're not assuming that the $x_t$ values are random, and if you assume no serial correlation and homoskedasticty (that is
$$text{var}(e_t)=sigma^2$$
and
$$text{cov}(e_t,e_s)=0,quad tneq s$$
for any $t$ and $s$), then
$$text{var}(hat beta)=text{var}left(sum w_t e_tright)=sum w_t^2 text{var}(e_t)=sigma^2sum w_t^2.$$
and you can check that this amounts to
$$text{var}(hatbeta)=frac{sigma^2}{sum(x_t-bar x)^2}.$$
The formula with the covariance you give is just wrong. It is true although that
$$text{var}left(sum w_t e_tright)=sum text{var}(w_t e_t)+sum_{tneq s} text{cov}(w_te_t,w_s e_s)=$$
$$=sum w_t^2text{var}(e_t)+sum_{tneq s} w_t w_stext{cov}(e_t,e_s),$$
but our assumptions imply that the second term is zero anyway.
answered Jan 23 at 22:55
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
edited Jan 23 at 23:08
answered Jan 23 at 22:55
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Jan 23 at 22:55
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Jan 23 at 22:55
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
4,765118
$begingroup$
Thank you very much for your answer. That beeing said I don't quite understand how we went from $E[w_t^2ε_t^2]$ to $Σw_t^2Var(ε_t)$ did we use some kind of formula?
$endgroup$
– Fozoro
Jan 23 at 23:06
$begingroup$
also how come we take out $w$ from the variance. Isn't w a variable?
$endgroup$
– Fozoro
Jan 23 at 23:14
$begingroup$
If $a$ is not random, then $$text{var}(aX)=a^2text{var}(X)$$ and if $text{cov}(X,Y)=0$ (in particular, if $X$ and $Y$ are independent), then $$text{var}(X+Y)=text{var}(X)+text{var}(Y).$$
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 23:17
1
$begingroup$
I'm making the classical assumption that the $x_t$ values are not random but predetermined before measuring $Y_t$ (check if this is the background with which you're supposed to be working). In that case, since the $w_t$ variables only depend on the $x_t$ values, they're not random either and can be thought of as constants (in the probabilistic sense).
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 23:19
1
$begingroup$
Great, I get it now thank you very much for your help!
$endgroup$
– Fozoro
Jan 23 at 23:22
add a comment |
$begingroup$
Thank you very much for your answer. That beeing said I don't quite understand how we went from $E[w_t^2ε_t^2]$ to $Σw_t^2Var(ε_t)$ did we use some kind of formula?
$endgroup$
– Fozoro
Jan 23 at 23:06
$begingroup$
also how come we take out $w$ from the variance. Isn't w a variable?
$endgroup$
– Fozoro
Jan 23 at 23:14
$begingroup$
If $a$ is not random, then $$text{var}(aX)=a^2text{var}(X)$$ and if $text{cov}(X,Y)=0$ (in particular, if $X$ and $Y$ are independent), then $$text{var}(X+Y)=text{var}(X)+text{var}(Y).$$
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 23:17
1
$begingroup$
I'm making the classical assumption that the $x_t$ values are not random but predetermined before measuring $Y_t$ (check if this is the background with which you're supposed to be working). In that case, since the $w_t$ variables only depend on the $x_t$ values, they're not random either and can be thought of as constants (in the probabilistic sense).
$endgroup$
– Alejandro Nasif Salum
Jan 23 at 23:19
1
$begingroup$
Great, I get it now thank you very much for your help!
$endgroup$
– Fozoro
Jan 23 at 23:22
$begingroup$
Thank you very much for your answer. That beeing said I don't quite understand how we went from $E[w_t^2ε_t^2]$ to $Σw_t^2Var(ε_t)$ did we use some kind of formula?
$endgroup$
– Fozoro
Jan 23 at 23:06
$begingroup$
Thank you very much for your answer. That beeing said I don't quite understand how we went from $E[w_t^2ε_t^2]$ to $Σw_t^2Var(ε_t)$ did we use some kind of formula?
$endgroup$
– Fozoro
Jan 23 at 23:06
$begingroup$
also how come we take out $w$ from the variance. Isn't w a variable?
$endgroup$
– Fozoro
Jan 23 at 23:14
$begingroup$
also how come we take out $w$ from the variance. Isn't w a variable?
$endgroup$
– Fozoro
Jan 23 at 23:14
$begingroup$
If $a$ is not random, then $$text{var}(aX)=a^2text{var}(X)$$ and if $text{cov}(X,Y)=0$ (in particular, if $X$ and $Y$ are independent), then $$text{var}(X+Y)=text{var}(X)+text{var}(Y).$$
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– Alejandro Nasif Salum
Jan 23 at 23:17
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If $a$ is not random, then $$text{var}(aX)=a^2text{var}(X)$$ and if $text{cov}(X,Y)=0$ (in particular, if $X$ and $Y$ are independent), then $$text{var}(X+Y)=text{var}(X)+text{var}(Y).$$
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– Alejandro Nasif Salum
Jan 23 at 23:17
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I'm making the classical assumption that the $x_t$ values are not random but predetermined before measuring $Y_t$ (check if this is the background with which you're supposed to be working). In that case, since the $w_t$ variables only depend on the $x_t$ values, they're not random either and can be thought of as constants (in the probabilistic sense).
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– Alejandro Nasif Salum
Jan 23 at 23:19
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I'm making the classical assumption that the $x_t$ values are not random but predetermined before measuring $Y_t$ (check if this is the background with which you're supposed to be working). In that case, since the $w_t$ variables only depend on the $x_t$ values, they're not random either and can be thought of as constants (in the probabilistic sense).
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– Alejandro Nasif Salum
Jan 23 at 23:19
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Great, I get it now thank you very much for your help!
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– Fozoro
Jan 23 at 23:22
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Great, I get it now thank you very much for your help!
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– Fozoro
Jan 23 at 23:22
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Which is exactly the linear model you have? Just one exogenous variable? Are you assuming that the values of $x$ are predetermined (that is, non random) or not?
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– Alejandro Nasif Salum
Jan 23 at 22:39
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Oops, very sorry I just added the linear equation. Thanks for your quick reply
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– Fozoro
Jan 23 at 22:43
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The term $w$ should not have a summation over $x - bar x$, should it?
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– Mark Viola
Jan 23 at 22:45
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@MarkViola Yes your right I put it there by accident ( just edited my question )
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– Fozoro
Jan 23 at 22:48
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Check out that there is a $w$ weight for each observation; so I added a $t$ subindex to your definition of those weights.
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– Alejandro Nasif Salum
Jan 23 at 23:01