In estimating vector $\hat{\beta}$ of linear regression, why does variance-covariance matrix of $\hat{\beta}$:

$$V(\hat{\beta})=E[(\hat{\beta} - E(\hat{\beta}))(\hat{\beta} - E(\hat{\beta}))^T]$$

have as the second term inside $E()$ the transpose?

Since I've found covariance to be expressed like (with no transpose):

enter image description here


3 Answers 3


You quote the definition of covariance of two variables, while the form with transpose is the definition of covariance matrix. In first case we are talking about two random variables $X$ and $Y$,

$$ \operatorname{cov}(X,Y) = \operatorname{E}{\big[(X - \operatorname{E}[X])(Y - \operatorname{E}[Y])\big]} $$

while in the second case $\mathbf{X}$ is a vector of random variables $X_1,\dots,X_n$

$$ \mathbf{X} = \begin{bmatrix}X_1 \\ \vdots \\ X_n \end{bmatrix} $$

so we are talking about covariance between multiple variables in form of covariance matrix

$$ \Sigma=\mathrm{E} \left[ \left( \mathbf{X} - \mathrm{E}[\mathbf{X}] \right) \left( \mathbf{X} - \mathrm{E}[\mathbf{X}] \right)^{\rm T} \right] $$

and the transpose appears in here because you are multiplying two vectors.


A covariance matrix is matrix-valued. Covariance of two random variables is an element in that matrix, i.e. a scalar. Hence the covariance formula that you list yields a scalar, while computing $xx^T$ for vector-valued $x$ yields a matrix.

  • $\begingroup$ So is the reason of transposing that one can do the multiplication inside $E()$? Since the two matrices to be multiplied could be $n \times m$, $n \not = m$. $\endgroup$
    – mavavilj
    Dec 6, 2016 at 17:04
  • $\begingroup$ I don't see how matrix multiplication bears on this question, since you identify $\hat{\beta}$ as a vector, and that's the only portion which bears a transpose. $\endgroup$
    – Sycorax
    Dec 6, 2016 at 17:06
  • $\begingroup$ Isn't $(\hat{\beta}-E(\hat{\beta}))$ a matrix/vector? So in order to calculate $(\hat{\beta}-E(\hat{\beta}))(\hat{\beta}-E(\hat{\beta}))$ one needs to transpose the other. $\endgroup$
    – mavavilj
    Dec 6, 2016 at 17:13
  • $\begingroup$ You're correct; that is how multiplication is defined for vectors. $\endgroup$
    – Sycorax
    Dec 6, 2016 at 18:59
  • $\begingroup$ Also the covariance matrix between X and Y is not the same as the covariance of the parameter estimates. $\endgroup$ Dec 6, 2016 at 23:38

Small example: Consider a 2x1

$$v = \begin{bmatrix} 7\\ 6 \end{bmatrix}$$

Note that

$$v'v = \begin{bmatrix} 7&6 \end{bmatrix}\begin{bmatrix} 7\\ 6 \end{bmatrix}$$

is a scalar (or 1x1 matrix?)


$$vv' = \begin{bmatrix} 7\\ 6 \end{bmatrix} \begin{bmatrix} 7&6 \end{bmatrix}$$

is a matrix


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