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I am trying to prove, under the assumption $E[u_t^2x_t^Tx_t]=\underbrace{E[u_t^2]}_{\sigma^2}E[x_t^Tx_t]$, that the

$$AVar[\beta_{POLS}]=\sigma^2 E[x_t^Tx_t]^{-1}$$ My result: $$\begin{eqnarray}AVar[\beta_{POLS}]&=&E[(\beta_{POLS}-\beta)(\beta_{POLS}-\beta)^T]\\ &=&E[((x_t^Tx_t)^{-1}x_t^T u_t)(u_t^Tx_t(x^T_tx_t)^{-1})]\\ &=&E[E[((x_t^Tx_t)^{-1}x_t^T u_t)(u_t^Tx_t(x^T_tx_t)^{-1})|x_t]]\\ &=&E[(x_t^Tx_t)^{-1}x_t^TE[u_tu_t^T|x_t]x_t(x_t^Tx_t)^{-1}]\\ &=&\sigma^2E[(x_t^Tx_t)^{-1}]\neq\sigma^2E[(x_t^Tx_t)]^{-1} \end{eqnarray} $$

Could anyone point out where I made the mistake?

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  • $\begingroup$ You want to calculate the asymptotic (limiting) variance of the estimator, but your operations pertain to the finite-sample variance. $\endgroup$ – Alecos Papadopoulos Apr 9 '18 at 2:21
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They are different simply beacuse the LHS is the variance of the estimator, while your RHS $\sigma^2 E[X'X]^{-1}$ is the asymptotic variance, or the variance of the asymptotic (normal) distribution of $ \beta_{POLS} $ and can be found using the central limit theorem and and the slutsky theorem, starting from:

$$ \sqrt{n}( \beta_{POLS} - \beta) = \left(n^{-1}\sum^n_{i=1}X'_iX_i^{-1} \right)^{-1}\left((n^{-1/2}\sum^n_{i=1}X'_iu_i \right)$$

Note also that the asympotic variance has that form iff:

$$ E[u^{2}x'_tx_t] = \sigma^2E[x'_tx_t] \ \ \text{for } t=1,2,\dots,T \ \ \text{(and for all } i )$$

and

$$E[u_tu_sx'_tx_s] = 0 \ \ \text{for } t\ne s, \ \ t,s=1,2,\dots,T\ \ \text{(and for all } i ). $$

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