I understand that using $R^2$ in time series models may not be the best as $R^2$ is non-decreasing.
I also read this post: What is the problem with using R-squared in time series models? on the problems with $R^2$
However, is there a mathematical way of proving why $R^2$ may not be the most useful in time series data?
$$R^2 = 1 – \frac {\frac {1}{T}\sum_{t=1}^T \hat u_i^2}{\frac {1}{T}\sum_{t=1}^T (Y_t-\bar Y)^2}$$
Using this formula, I tried to find the probability limit of $R^2$ for a stationary AR(1) regression and a random walk.
Is it right to say that in both a stationary AR(1) regression and a random walk, $$\frac {1}{T}\sum_{t=1}^T \hat u_i^2$$ converges to 0,
and thus, the probability limit of $R^2$ is equal to 1?
I am not sure if my method is right. Please advice thank you!
Best Answer
Actually, for a random walk, we indeed have that $R^2\to_p1$, but not for the reason you posit.
Whether for a random walk or a stationary AR(1), $$\frac {1}{T}\sum_{t=1}^T \hat u_i^2\to_p\sigma^2,$$ i.e., the numerator of $R^2$ is consistent for the error variances driving the process $y_t$.
But \begin{eqnarray*} R^2&=&1-\frac{\sum\hat{u}_t^2}{\sum (y_t-\overline{y})^2}\\ &=&1-\frac{\frac{1}{T}\sum\hat{u}_t^2}{\frac{1}{T}\sum (y_t-\overline{y})^2}\to_p1 \end{eqnarray*} in the RW case, because it can be shown that (see, e.g., Chapter 17 in Hamilton) $$ \frac{1}{T^2}\sum (y_t-\overline{y})^2\Rightarrow\int_0^1W^\mu(r)^2d r $$ That is to say, when scaled by $T^2$, the totoal sum of squares weakly converges to some functional of Brownian motion - an integral over a squared demeaned BM, to be precise.
However, the more important take-away point here is that this result implies that $$ \frac{1}{T^2}\sum (y_t-\overline{y})^2=O_p(1) $$ That is, $\frac{1}{T^2}\sum (y_t-\overline{y})^2$ is stochastically bounded, which means $\frac{1}{T}\sum (y_t-\overline{y})^2$ diverges at rate $T$. Hence, the denominator in the $R^2$ formula "explodes" when the process is a random walk.