# 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

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.
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.' – kasa Feb 4 '18 at 12:05