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I'm following the "Time Series Analysis and Its Applications With R Examples" from Shumway and Stoffer. In chapter 3.8 they talk about Regression with Autocorrelated Errors.

"They use the model of the form: $$y=Z*\beta+x$$

Where $y=(y_1,...,y_n)'$ and $x=(x_1,...,x_n)'$ are $n \times 1$ vectors, $\beta=(\beta_1,...,\beta_n)'$ is $r \times 1$, and $Z=[z_1|z_2|...|z_n]'$ is the $n \times r$ matrix composed of the input variables. Let $\Gamma={\gamma_x(s,t)}$, where $\Gamma$ is the variance-covariance matrix made from an ARMA(p,q) model, then $\Gamma^{-\frac{1}{2}}y=\Gamma^{-\frac{1}{2}} Z \beta+ \Gamma^{-\frac{1}{2}}x$, so that we can write the model as

$$y^\star=Z^\star \beta+ \delta,$$

where $y^\star=\Gamma^{-\frac{1}{2}}y,Z^\star=\Gamma^{-\frac{1}{2}} Z$, and $\delta=\Gamma^{-\frac{1}{2}}x$. Consequently, the covariance matrix of $\delta$ is the identity and the model is in the classical linear model form. It follows that the weighted estimate of $\beta$ is $\hat{\beta_w}=(Z^{' \star} Z^\star)^{-1} Z^{' \star} y^\star =(Z' \Gamma^{-1} Z)^{-1} Z' \Gamma^{-1}y$, and the variance-covariance matrix of the estimator is $var(\hat{\beta_w}) = (Z' \Gamma^{-1} Z)^{-1}$. If $x_t$ is white noise, then $\Gamma = \sigma^2 I$ and these results reduce to the usual least squares results."

During this process, I try to take the covariance of $\delta$ then I should get the following, $$cov(\delta)=cov(\Gamma^{-\frac{1}{2}}x) =\sigma^2 \Gamma^{-\frac{1}{2}} \Gamma \Gamma^{-\frac{1}{2}}=\sigma^2 I.$$ This is derived from the fact that $cov(x)=\sigma^2 \Gamma$ and using the property that $cov(Ax)=A cov(x) A'$ along with the property that $(\Gamma^{-\frac{1}{2}})'=\Gamma^{-\frac{1}{2}}$ since $\Gamma^{-\frac{1}{2}}$ is a symmetric matrix due to properties of covariance matrices. Additionally since $\Gamma^{-1}$ is formed from an ARMA(p,q) covariance matrix, the matrix has a $\sigma^2$ multiplier in it, so it is possible to factor it out from $\Gamma$ which provides the final result of $cov(\delta)=\sigma^2 I$.

So I have three questions that I am trying to understand,

  1. Is my thought process correct in understanding this?
  2. Does this mean that $\hat{\beta_w}$ doesn't depend on $\sigma^2$
  3. Would this mean that the new transformed errors, $\delta$, are now iid such that $\delta \sim N(0,\sigma^2)$?

Thank you kindly

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1 Answer 1

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  1. You have a minor error in your expression for $cov(\delta)$; the exponent on $\Gamma$ shouldn't be $-1$.

  2. Yes, in OLS the coefficient estimates don't depend on $\sigma^2$. The variance-covariance matrix of the estimates does, but the estimates themselves do not.

  3. There is no assumption of Normality required, just that the errors have a finite variance. If we added the assumption that $x$ was Normally distributed, with covariance matrix $\Gamma$, then, yes, $\delta = \Gamma^{-1/2}x \sim \mathrm{N}(0, \sigma^2\mathrm{I})$.

Updated in response to comments:

Note that it is unnecessary to make any distributional assumptions about $x$ (other than finite variance and independence from $Z$) for the asymptotic normality of $\hat{\beta}$ to hold; we do need some straightforward conditions on $Z$, e.g., $Z$ bounded, $\lim_{T \to \infty} Z'Z/T$ is finite and nonsingular.

The distribution of $\delta$ will depend on the distribution of $x$. For example, if $x \sim \mathrm{MVt}(\nu, \Gamma)$ (a multivariate $t$ distribution) then $\delta$ will have an $\mathrm{MVt}(\nu, I)$ distribution, i.e., each element of $\delta \sim t(\nu)$. Other cases may not work out so nicely, however.

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  • $\begingroup$ Thank you for the quick reply, I went ahead and fixed the error in 1. For question 2, yes I can see that when we have to take the variance of $\hat{\beta_w}$. Question 3, is still a bit confusing for me. If we don't assume that $x$ was normally distributed then what kind of distribution would $\delta$ follow? Thanks $\endgroup$ Jan 18, 2023 at 18:23

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