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One of my textbooks on time-series analysis claims that

Dependency in the second moments of the residuals contradicts the assumption of a constant, time-invariant variance. Thus [the residual] is not pure white noise.

This confuses me. The most popular definition of a white noise process $u_t$ seems to be $$ u_t \sim (\,0, \, \sigma^2 \in \mathbb{R} \,) \text{ for all }t, \quad \mathrm{Cov}(u_t, u_s) = 0 \text{ for all } t \neq s,$$ where $\sigma^2$ is some constant. But the definiton does not imply anything, in my opinion, when it comes to the correlation (or lack thereof) between $ u_t^2$ and $u_s^2$.

Could you help me understand, why the dependency between the second-order moments deny the assumption of a white noise process?

Thank you for reading this, and stay healthy!

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Suppose (something which is not always given) your textbook distinguishes between errors $u_t$ - the difference between the dependent variable and the systematic part of the model - and residuals $\hat u_t$ - the difference between the dependent variable and the fitted values of the fitted model.

Then, we know that the vector of residuals when fitting via least squares is given by, using well known notation $M=I-X(X'X)^{-1}X'$. $\hat u=My$. Hence, even if errors $u$ are spherical in the sense that $Cov(u)=\sigma^2I$, $$ Cov(\hat u)=Cov(My)=MCov(y)M'=MCov(y)M=M\sigma^2IM=\sigma^2M, $$ which, in general, is not a diagonal matrix anymore. (Here, I assumed a model $y=X\beta+u$ and $X$ to be constant for simplicity, else, the argument would be conditional on $X$.) The result uses that $M$ is symmetric and idempotent, see e.g. here.

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  • $\begingroup$ I understand what you are trying to say: uncorrelated errors do not imply uncorrelated residuals. I sincerely thank you for your effort, however I must explain that my question does not concern the claim "uncorrelated errors imply uncorrelated residuals." In the textbook, the author is trying to motivate the ARCH model by observing some data have "volatility clusters" (large (small) variance is followed by large (small) variance in the next time index), and therefore the model should not assume white noise "errors" (following your choice of terminology.) Again, thank you for your help! $\endgroup$ – Ko Byeongmin Jun 16 at 10:28
  • $\begingroup$ I see. Can you give the precise reference of your textbook? Maybe the author defines white noise as independent errors, which are indeed not given when there are (G)ARCH effects. $\endgroup$ – Christoph Hanck Jun 16 at 11:02

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