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I was wondering if someone could help me clarify something from my lecture notes. It concerns the last step. I was wondering why we test if

$\frac{\hat{\beta}}{\textrm{SE}(\hat{\beta}}<0$

and what the rejection rule is in this case. I am quite confused by this.

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My own thoughts were to test $H_0:\beta=0$ against $H_1:\beta \neq 0$ using the $t$-statistic $\frac{\hat{\beta}}{\textrm{SE}(\hat{\beta}}$ (we compare this to the critical values of a $t$ distribution with $N-K$ degrees of freedom where $N$ is the number of observations and $K$ the number of parameters; here $K=3$ since we estimate $\beta,\tilde{\alpha}$ and $\tilde{\gamma}$). Then, if we do not reject $H_0$, this thus means that the model is the same as the pure linear trend model, which corresponds to the special case of $\rho=0$. Hence, if we do not reject $H_0:\beta=0$ then we do reject $H_0:\rho=1$.

user133993
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1 Answers1

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(1) If $\beta=0$ then the specification you were testing simplifies to

$$x_t=x_{t-1}+\tilde\alpha+\tilde\gamma t+\epsilon_t,$$

which is the equation at the top for $\rho=1$, correct?

How is that a pure linear trend model?

(2) One tests against $\rho<1$ (and hence $\beta<0$) because roots greater than unity are typically ruled out.

(3) One beef I have with the lecture notes is that one tests coefficient values, not values of estimates.

(4) Correcting for degrees of freedom and using t-distributions only makes sense if you're willing to assume that your data are drawn from a normal distribution, the $\epsilon_t$'s are iid, and your sample size is small.

JPi
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  • Hello, thank you very much for your answer. I think for (1) in your answer, if $\beta=0$, then it does reduce to $x_t=\tilde{\alpha}+\tilde{\gamma}t + \epsilon_t$. which, according to the notes, is a pure linear trend model. – user133993 Apr 09 '14 at 17:10
  • With regards to what I wrote under the lecture notes, would you say that there are huge mistakes in that thinking? And how does $\rho<1$ lead to testing $\beta<0$.I am sorry if these questions are bit silly but I am trying to wrap my head around it to have it make sense because I find it a bit confusing ... @peter – user133993 Apr 09 '14 at 17:15
  • No, it reduces to $x_t=x_{t-1}+\tilde\alpha+\tilde\gamma t +\epsilon_t$. As a consequence, what you wrote at the bottom is incorrect. – JPi Apr 10 '14 at 00:00