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This question comes after reading: https://spuriousregression.com/the-dangers-of-time-series-unit-roots/

Consider the humble one-dimensional random walk process: $$ X_t = X_{t-1} + \epsilon_t $$ Where $\epsilon_t$ is drawn i.i.d. from some distribution with zero mean and finite, constant variance. This process is well known to be non-stationary - the variance is proportional to time. I have the following questions:

  1. If we consider the regression $X_t = \beta_1X_{t-1} + \epsilon_t$, and the OLS estimator $\hat{\beta_1}$, which assumption of the Gauss-Markov theorem is violated? I assume it is strict endogeneity $(\mathbb{E}[\epsilon_s | X_1, \cdots, X_T] = 0)$ which is violated, but I am not sure how to prove it. Is the only consequence of this a biased-estimate of $\beta_1$ ? (E.g. do we keep things like consistency?)
  2. Is there a reason that such a regression would give us spurious results (Inflated T-statistics and R-Square)? I have of course seen some works metion Granger's work on this, and how the statistics converge to functionals of Brownian motion in the limit. As a follow-up to this, if we consider the cousin of the random walk, the $AR(1)$ process: $$ X_t = \rho X_{t-1} + \epsilon_t $$ with $|\rho| < 1$, then this process is well-behaved (stationary) and in particular, the estimator is consistent. Does the AR(1) process violate "fewer" OLS assumptions than the random walk as a result of its stationarity? Which assumptions does it keep intact that the random walk does not?

Thank you for all of your help!

rubikscube09
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  • Most explicitly: the variance of $X$ **at time *t*** is proportional to time. In the limit of time series length, its variance is not finite. – Alexis May 12 '21 at 16:45
  • @Alexis, this applies to $X_t$, not $\varepsilon_t$. Do we have an assumption of constant conditional variance of $X_t$? – Richard Hardy May 12 '21 at 16:47
  • @RichardHardy [I.i.d.](https://en.wikipedia.org/wiki/Independent_and_identically_distributed_random_variables)? – Alexis May 12 '21 at 16:50
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    @RichardHardy - thank you for the link - it's helpful! I think this question can be closed. – rubikscube09 May 12 '21 at 16:51
  • @Alexis, I do not think so. The relevant assumption is spelled out in the answers of the linked thread. – Richard Hardy May 12 '21 at 17:08
  • @RichardHardy Oh, whoops! I missed the "conditional" in "conditional variance". Thank you for the suggested duplicate link. :) – Alexis May 12 '21 at 19:00

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