Can someone please explain this (with further statistical assumptions).

In the standard regression model where $$Y = X\beta+u,$$

we want to show that $$ E(uu′) = \sigma^2 \cdot I \\ E(u'u)=T \cdot \sigma^2.$$

Above, I is an $n \times n$ identity matrix and $T$ is the number of observations.

  • $\begingroup$ Could you be a little more specific about what you need explained? $\endgroup$ – whuber Oct 3 '14 at 21:39
  • $\begingroup$ how do we know we have the second part? Is there a way to derive it? $\endgroup$ – Sara Oct 3 '14 at 21:50
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    $\begingroup$ @Sara I cleaned up your code. I wasn't sure what the u hats were, so if I did too much violence, please clarify what they were. $\endgroup$ – Dimitriy V. Masterov Oct 3 '14 at 22:48
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    $\begingroup$ The way to add hats is \$\hat u$. $\endgroup$ – Dimitriy V. Masterov Oct 3 '14 at 23:28
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    $\begingroup$ This looks somewhat like routine bookwork (as might be assigned for a subject), which should probably carry the self-study tag and follow the guidelines at its tag-wiki. The phrase about 'further statistical assumptions' in particular makes it sound like assigned work. $\endgroup$ – Glen_b -Reinstate Monica Oct 4 '14 at 1:54

The first assumption tells you that all the $u_i$s have the same variance, equal to $\sigma^2$. You want to calculate $$E(u'u)=E(u_1^2 + u_2^2 + ... + u_T^2)=E(u_1^2) + ... + E(u_T^2).$$ You can use the facts that $Var(u)=E(u^2)+E(u)^2$ and that usually $E(u)=0$, to do the rest.

Here's the intuition for the $E(uu')$. Suppose you just have two observations. Then

$$uu'= \begin{bmatrix} u_1\\u_2 \end{bmatrix}\begin{bmatrix} u_1 & u_2 \end{bmatrix}=\begin{bmatrix} u_1^2 & u_1 u_2\\u_1 u_2 & u_2^2 \end{bmatrix}.$$

When you take the expectation of that matrix, you can use the same logic as above, you will get the variances on the diagonal and the covariances on the off-diagonal. The first are $\sigma^2$ by identically distributed assumption, and the second are are zero by assumption of independence.

$$\begin{bmatrix} \sigma^2 & 0 \\0 & \sigma^2 \end{bmatrix}=\sigma^2 \cdot I$$

  • $\begingroup$ thank you. Can E(u′u)=T⋅σ2? Also, do you know a link where I can find out how to do that? $\endgroup$ – Sara Oct 3 '14 at 23:39
  • $\begingroup$ Var(βOLS)=E[(βOLS −β)(βOLS −β)′] = E[(X′ X)−1X′ uu′ X(X′ X)−1] = σ2(X′X)−1. $\endgroup$ – Sara Oct 3 '14 at 23:44
  • $\begingroup$ would that be the proof, and sorry for the messy notation $\endgroup$ – Sara Oct 3 '14 at 23:45
  • $\begingroup$ Yes, that would be the case. In LaTeX, you can write $\sigma^2 (X'X)^{-1}$ as \$\sigma^2 (X'X)^{-1}\$. Subscripts can be handled with \$\beta_{OLS}$. Questions that are clearly written and formatted nicely receive far more attention on these sites. $\endgroup$ – Dimitriy V. Masterov Oct 4 '14 at 18:52
  • $\begingroup$ thank you. SO just to clarify the statement is true above, based on what I wrote down? Also, how do I nicely format like you did on the site? $\endgroup$ – Sara Oct 6 '14 at 19:10

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