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Standard Error
 
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mkparis
Posted: 31 July 2008 02:10 PM   [ Ignore ]  
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Total Posts:  8
Joined  2007-09-23

Howdy,
in computing the standard error, in the lecture we have
se(b0) = sqrt(var(b0)).
I am wondering if it is not
se(b0) = sqrt(var(b0))/sqrt(n)

Because in the CLT we have :

Let X1,X2....Xn be n random variables with a sample mean Xmean, a mean mu and a variance of sigma^2
we have.

sqrt(n) ( Xmean - mu) ==> N(0,sigma^2)

mu = Xmean +/- Z(alpha/2) x sigma/sqrt(n)

hence the standard error is sqrt of the variance and divided by the sqrt of n.

Can you please let me know where I am wrong

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David Harper, CFA, FRM, CIPM
Posted: 31 July 2008 06:38 PM   [ Ignore ]   [ # 1 ]  
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Total Posts:  1291
Joined  2006-09-24

Hi mkparis,

I assume b0 refers to estimators of parameters in linear regression (don’t i have them as b1, b2 per Gujarati?). If so, your logic is indeed good, but the sqrt(n) is already in the denominator so,

se(bm) = sqrt(var(bm)), where bm = a parameter estimator

because, to your point, the var(bm) already includes a (1/n) which becomes SQRT(1/n) in the standard error, so, i hadn’t thought of it this way, but I *think* you are correct that is indeed CLT manifesting but the formula is still the first way

David

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