# Why is the dickey fuller test different from a simple t-test

I am trying to understand why should there be different distribution for t-statistic, in case of AR model, Dickey-Fuller test

For e.g. Say, the model is $Y_t = \beta_lY_{t-1} + \varepsilon_{t}$.

Why should I not use Simple linear regression model like $y_i = \beta_0 + \beta_1x_i+\epsilon_i$, where $x_i = Y_{t-1}$ and $y_i = Y_t$, and get the coefficient estimate as $$\hat\beta_1=\frac{\sum_ix_iy_i-n\bar x\bar y}{n\bar x^2-\sum_ix_i^2}$$ and its standard error estimate as $$s_{\hat\beta_1}=\sqrt{\frac{\sum_i\hat\epsilon_i^2}{(n-2)\sum_i(x_i-\bar x)^2}}.$$

Once we get the coefficient estimate and its standard error estimate, why can not we say the t-stat ($\frac{{\hat\beta_1}}{s_{\hat\beta_1}}$) follows a t-distribution, just like how we do in the case of simple linear regression. Are we violating any particular assumption in doing so?

• One remark next to the answer: in the t-ratio, you would need to subtract the value under the null from the point estimate. In this formulation of the test regression, you would test the null that $\beta=1$. – Christoph Hanck Feb 4 '18 at 11:59

## 1 Answer

You are right that the test statistic is just a standard t-statistic.

It, however, follows a different null distribution, i.e., using critical values from the t or normal distribution would lead to tests that would not reject in $\alpha$% of the cases when the null is true.

See Estimation of unit-root AR(1) model with OLS for an assumption that is violated and How is the augmented Dickey–Fuller test (ADF) table of critical values calculated? for some information on the asymptotic null distribution.

From the first link, we note that $$T^{-1}\sum_{t=1}^Tx_{t-1}\epsilon_{t}\Rightarrow\sigma^2/2\{W(1)^2-1\}.$$ In particular, $W(1)^2-1$ is a demeaned $\chi^2_1$ random variable (as the Wiener process has $W(s)\sim N(0,s)$), which has probability 0.682 of being less than zero, leading to the skew in the distribution of the DF statistic.

• Brilliant answers, sir! So, I just wanted to confirm my understanding: 'the non-finiteness of $\frac{X'X}{n}$ is the reason behind the skewness of the distribution in the non-stationary case.' – honeybadger Feb 4 '18 at 12:05
• Thanks - in fact, I would argue that the numerator is resonsible for the skew, see my edit. – Christoph Hanck Feb 4 '18 at 17:16