As I understand it, $\text{standard error} = \frac{\text{standard deviation}}{\sqrt{n}}$, where $n$ is the number of observations. I also know that $cov(\hat{\beta}) = \sigma^{2}(X^{T}X)^{-1}$; this is a pxp variance-covariance matrix. So why is the standard error of the components of the coefficient estimate vector $\boldsymbol{\hat{\beta}}$ in, for example, a normal linear model, $\sigma^{2}[(X^{T}X)^{-1}]_{ii}$ and not $\frac{1}{\sqrt{n}}\sigma^{2}[(X^{T}X)^{-1}]_{ii}$ where the subscript $ii$ denotes the $i,i$th entry of the matrix?

  • $\begingroup$ i asked a related question yesterday, using R... stats.stackexchange.com/questions/262160/… $\endgroup$
    – Haitao Du
    Feb 16, 2017 at 21:28
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    $\begingroup$ The "$n$" factor is implicit already: consider the case where $X$ has been whitened, so $X^TX=nI$. $\endgroup$
    – GeoMatt22
    Feb 16, 2017 at 21:35
  • $\begingroup$ @GeoMatt22 Hmm, yes, I can see when it is whitened it works but I can't generalise it - could you elaborate on where it is implicit? $\endgroup$ Feb 16, 2017 at 21:57
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    $\begingroup$ It's just dot products vs. covariances (sums vs. averages): $\boldsymbol{x}^T\boldsymbol{y} = \overline{(xy)}\times{n}$ $\endgroup$
    – GeoMatt22
    Feb 16, 2017 at 22:21
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    $\begingroup$ @GeoMatt22 I'm afraid I don't find it any easier to understand using the equality you presented. :l Could you provide working? $\endgroup$ Feb 16, 2017 at 22:54

1 Answer 1


The "$n$" factor is implicit already in $\boldsymbol{X}^T\boldsymbol{X}$.

This is easiest to see in the case where $\boldsymbol{X}$ has been whitened, so $\boldsymbol{X}^T\boldsymbol{X}=n\boldsymbol{I}$.

More generally we have $\boldsymbol{X}=[\boldsymbol{x}_1,\ldots\boldsymbol{x}_p]$, where $\boldsymbol{x}_p\in\mathbb{R}^n$. Then the entries of the matrix are $$\left(\boldsymbol{X}^T\boldsymbol{X}\right)_{ij}=\boldsymbol{x}_i^T\boldsymbol{x}_j$$ Now choose a pair of columns, $\boldsymbol{a}=\boldsymbol{x}_i$ and $\boldsymbol{b}=\boldsymbol{x}_j$. Then we have $$\boldsymbol{a}^T\boldsymbol{b}=\sum_{k=1}^na_kb_k=\overline{(ab)}\,n$$ So each entry of the matrix is a dot product, which can be seen as $n$ times an average.

In the case where $\boldsymbol{X}$ has been centered so the columns have zero means, the product-matrix entries are covariances.

For an overview of the general case, see here.


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