7
$\begingroup$

Given a random walk $x_t$,

$$x_t=x_{t-1}+\varepsilon_t,$$

consider estimating the slope coefficient $\beta$ in

$$x_t=\beta x_{t-1}+\varepsilon_t$$

by OLS. This question and the following answer noted that $\hat \beta^{OLS}$ has a skewed distribution. My question is,

Are any of OLS assumptions violated in the model $x_t=\beta x_{t-1}+\varepsilon_t$ given that $x_t$ is a random walk? If so, what are the violations?

$\endgroup$

2 Answers 2

7
$\begingroup$

It is generally assumed that the explanatory variables have finite moments at least up to second order. In this case, as the explanatory variable is a random walk, its variance is not finite. This makes the matrix $Q=\hbox{plim } X′X/n$ not finite, with the consequences discussed below.

The explanatory variable $x_{t-1}$ is not fixed (it is stochastic as it depends on $\epsilon$) and is not independent of the error term $\epsilon_t$. This makes OLS in general biased and inference is not valid in small samples.

The explanatory variable and $\epsilon_t$ are not independent of each other but they are contemporaneously uncorrelated, $E(x_t, u_t) = 0 \forall t$. In the classical regression model this will open the possibility for the the OLS estimator to be consistent in large samples.

If the matrix $Q = \hbox{plim } X′X/n$ were finite and positive definite matrix, then the F-test statistic will follow asymptotically follows the $\chi^2$ distribution. As pointed by @ChristophHanck this matrix is not finite in this context. Hence, the Mann and Wald theorem is not applicable and inference based on OLS will not be reliable even in large samples.

You may be interested in this answer, which discusses similar issues in the context of a stationary AR(q) process.

$\endgroup$
3
  • $\begingroup$ I think "$Q=plim X'X/n$ not finite" is kind of what I was looking for. I just figured that out independently, and now I see it in your answer. $\endgroup$ Commented Apr 11, 2015 at 16:25
  • $\begingroup$ I am little confused here. In the case of a random walk where we try to estimate the equation $X_{t+1} = \beta X_t + \epsilon$, due to the cointegrating realtionsip $X_{t+1} - X_t$ we should converge at the superconsistent rate to $\beta =1$. Why do you think we are not going to have any reliable results? $\endgroup$ Commented May 18, 2017 at 11:29
  • $\begingroup$ @CagdasOzgenc you are right, OLS will in general be consistent. (Phillips (1987) Theorem 3.1). I have edited the answer referring only to problems with inference. $\endgroup$
    – javlacalle
    Commented May 19, 2017 at 18:40
6
$\begingroup$

One of the key assumptions I would list among standard OLS assumptions is that there is no weak LLN for the "average of the $X'X$-matrix" $1/T\sum_tx_{t-1}^2$. Instead, we have weak convergence to a functional of Brownian motion provided we scale by $T^2$, viz.

$$ T^{-2}\sum_{t=1}^Tx^2_{t-1}\Rightarrow\sigma^2\int_0^1W(r)^2d r $$

I would, btw, not quite agree with @Alecos statement in the link you posted that there is no analytical solution to the distribution of the OLSE - we know the asymptotic distribution of the OLSE, when scaled with the suitable superconsistent rate $T$, to be

\begin{eqnarray*} T\left(\hat{\beta}^{OLS}-1\right)&=&T\frac{\sum_{t=1}^Tx_{t-1}\epsilon_{t}}{\sum_{t=1}^Tx_{t-1}^2}\\ &=&\frac{T^{-1}\sum_{t=1}^Tx_{t-1}\epsilon_{t}}{T^{-2}\sum_{t=1}^Tx_{t-1}^2}\\ &\Rightarrow&\frac{\sigma^2/2\{W(1)^2-1\}}{\sigma^2\int_0^1W(r)^2d r}\\ &=&\frac{W(1)^2-1}{2\int_0^1W(r)^2d r}, \end{eqnarray*} the "Dickey-Fuller-distribution" (JASA 1979).

$\endgroup$
4
  • $\begingroup$ Your answer is useful in that it helps to understand what is going wrong, but it does not directly answer the question. What about pinning down a specific assumption that is being violated? $\endgroup$ Commented Apr 11, 2015 at 15:55
  • $\begingroup$ It would be easier to answer if we knew precisely what the OLS assumptions are to you. Take eg assumption 8.3(c) in Hamilton, which requires (translated to our notation) $E(x_{t-1}^2)$ to exist and $1/T\sum_tx_{t-1}^2$ to converge to it. Neither is the case here. $\endgroup$ Commented Apr 11, 2015 at 16:23
  • $\begingroup$ Fair remark. I should have spelled out the assumptions in the OP. But this comment as well as the answer by @javlacalle partly does the job for me and are quite helpful. I was picking more on your formulation, not on the content. $\endgroup$ Commented Apr 11, 2015 at 16:27
  • 1
    $\begingroup$ That was a helpful answer (+1), thank you! (My earlier comments may or may have not come across right.) $\endgroup$ Commented Feb 6, 2023 at 13:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.