Dicky Fuller Test - why is it important that the CLT does not apply under assumption of non-stationarity? I am trying to understand the Dicky-Fuller test (simple case, no augmentation or lags). 
I can get this far: 
For $X_t = \alpha + \rho X_{t-1} + \epsilon_t$, 
$\alpha$ is the drift term, and $\epsilon_t$ is i.i.d noise.
$H_0: \rho = 1$, i.e. non-stationary
$H_1: \rho < 1$, i.e. stationary 
This is then rearranged to: 
$\Delta X_t = \alpha + (\rho - 1) X_{t-1} + \epsilon_t$ 
This is where I get stuck. Why can't we just use a standard t-test to test whether $\rho=1$ for $X_t = \alpha + \rho X_{t-1} + \epsilon_t$? 
From my research I've seen the answer that "Because $X_t$ is non-stationary under the null, the Central Limit Theorem doesn't apply and therefore t-test doesn't hold." Can you please explain what this means? 
I've looked at the following resources but still don't understand: 
These slides: https://www.bauer.uh.edu/rsusmel/phd/ec2-5.pdf
Ben Lambert's awesome YouTube series: https://www.youtube.com/watch?v=2GxWgIumPTA 
 A: The t-test is a very specific type of test and to be truthful, you could use the t-test if you restricted the test to 1 degree of freedom.  It would, however, be too conservative.  To see why, look at the t-table with one degree of freedom at .0001%.  The look at the same test for the DF table.  You could use it with one degree of freedom because $\mu$ is no longer a mean and $\sigma$ is no longer a standard deviation.  
To see  this you could take Student's distribution in reduced form to account for the one degree of freedom which is $$\frac{1}{\pi}\frac{\sigma}{\sigma^2+(x-\mu)^2}.$$  Compared to the more general form, this is easy to work with.  Now if you take the expectation of x you will find that $$\int_{-\infty}^\infty\frac{1}{\pi}\frac{x\sigma}{\sigma^2+(x-\mu)^2}\mathrm{d}x$$
You can see from the indefinite integral that the expectation diverges so that no population mean exists at all.  The indefinite integral is $$\frac{\sigma\log(\mu^2-2\mu{x}+\sigma^2+x^2)-2\mu\tan^{-1}\left(\frac{\mu-x}{\sigma}\right)}{2\pi}.$$  If you take $\lim_{x\to\infty}$ you end up with $\infty-\infty$ adjusting out the constants as meaningless as you cannot scale infinity or add or subtract from it.
So $\mu$ is the median of the density and $\sigma$ is the half-width at half-maximum.  It is a scale parameter, but clearly is not a variance as you cannot have a variance about a mean that does not exist.
The reason you could get away with using it is that it is approximately the Bayesian solution for the hypothesis that $\beta\ge{1}$.
The slides do have a mistake in them.  They argue that in the AR(1) equation $$x_{t+1}=\beta{x}_t+\epsilon_{t+1}$$ only has the case where $|\beta|\le{1}$.  This is not remotely true.  Consider the case of GOOG stock.  $\beta>1$ or nobody would invest in it.  Who would purposefully invest, if they could avoid it, if they believed they would lose money at every iteration.
The proper null is $\beta=1$ and the proper alternative is $\beta\ne{1}$.  If $\beta<1$ then it is clearly stationary and $\hat{\beta}-\beta$ will be drawn from a normal distribution.  If $\beta>1$ then no admissible Frequentist solution exits, but an admissible Bayesian solution does exist.  You could also use something such as Theil's regression if you felt you must use a Frequentist method.  It would not meet the standards of admissibility, but it would be an unbiased estimator that is valid for all values of $\beta.$  There would be a monumental loss if efficiency, but it would be a Frequentist solution.
There is a valid, first-principles Bayesian solution to $x_{t+1}=\beta{x}_t+\epsilon_{t+1},\forall\beta\in\mathbb{R}.$
The DF and the ADF test should never be used unless you are testing that a solution is a steady-state solution.  If you believe it is growing then the DF test should never be used.  If you believe it is contracting, it should also never be used.  It also should never be used as a diagnostic.
The problem has to do with the extremely fragile nature of your model at $\beta=1$.  The DF test cannot distinguish between $\beta=1$ and $\beta=1.000001$.  Usually the range of potential parameters is quite wide.  Nonetheless, the mathematical implications of it being anywhere except precisely at unity are enormous.  The likelihood functions are entirely different and the properties are different.  
If $\beta>1$ then you cannot use things like Bellman equations as your equation is a source and not a sink.  Conversely, if $|\beta|<1$ then this implies the system can learn from its mistakes and that mistakes will gradually vanish from the system.
If $\beta\ge{1}$ then no mean based solution can exist for any problem, so there could not possibly be such a thing as mean-variance finance.  If $|\beta|<1$ then mean based estimators will stochastically dominate all others and so you shouldn't be using alternative methods such as Theil's regression, barring other issues.
The central limit theorem requires a second moment exist in the population that the data was drawn from.  Many distributions lack a first moment and a few have a first moment, but no second.
