Why information criterion (not adjusted $R^2$) are used to select appropriate lag order in time series model? In time series models, like ARMA-GARCH, to select appropriate lag or order of the model different information criterion, like AIC, BIC, SIC etc, are used. 
My question is very simple, why donot we use adjusted $R^2$ to choose appropriate model? We can select model which lead to higher value of adjusted $R^2$. Because both adjusted $R^2$ and information criterion penalize for additional number of regressors in the model, where former penalize $R^2$ and later penalize likelihood value.  
 A: The penalty in $R^2_{adj}$ does not yield the nice properties in terms of model selection as posessed by the AIC or BIC. The penalty in $R^2_{adj}$ is enough to make $R^2_{adj}$ an unbiased estimator of the population $R^2$ when none of the regressors actually belongs to the model (as per Dave Giles' blog posts "In What Sense is the "Adjusted" R-Squared Unbiased?" and "More on the Properties of the "Adjusted" Coefficient of Determination"); however, $R^2_{adj}$ is not an optimal model selector. 
(There could be a proof by contradiction: if AIC is optimal in one sense and BIC is optimal in another, and $R^2_{adj}$ is not equivalent to either of them, then $R^2_{adj}$ is not optimal in either of these two senses.)
A: I would argue that at least when discussing linear models (like AR models), adjusted $R^2$ and AIC are not that different. (This discussion is based on Hansen's Econometrics textbook, if I remember correctly.)
Consider the question of whether $X_2$ should be included in
$$
y=\underset{(n\times K_1)}{X_1}\beta_1+\underset{(n\times K_2)}{X_2}\beta_2+\epsilon
$$
This is equivalent to comparing the models
\begin{eqnarray*}
\mathcal{M}_1&:&y=X_1\beta_1+u\\
\mathcal{M}_2&:&y=X_1\beta_1+X_2\beta_2+u,
\end{eqnarray*}
where $E(u|X_1,X_2)=0$. We say that $\mathcal{M}_2$ is the true model if $\beta_2\neq0$.
Notice that $\mathcal{M}_1\subset\mathcal{M}_2$. The models are thus nested.
A model selection procedure $\widehat{\mathcal{M}}$ is a data-dependent rule that selects the most plausible of several models.
We say
$\widehat{\mathcal{M}}$ is consistent if
\begin{eqnarray*}
\lim_{n\rightarrow\infty}P\bigl(\widehat{\mathcal{M}}=\mathcal{M}_1|\mathcal{M}_1\bigr)&=&1\\
\lim_{n\rightarrow\infty}P\bigl(\widehat{\mathcal{M}}=\mathcal{M}_2|\mathcal{M}_2\bigr)&=&1
\end{eqnarray*}
Consider adjusted $R^2$. That is, choose $\mathcal{M}_1$ if $\bar{R}^2_1>\bar{R}^2_2$. As $\bar{R}^2$ is monotonically decreasing in $s^2$, this procedure is equivalent to minimizing $s^2$. In turn, this is equivalent to minimizing $\log(s^2)$. For sufficiently large $n$, the latter can be written as
\begin{eqnarray*}
\log(s^2)&=&\log\left(\widehat{\sigma}^2\frac{n}{n-K}\right) \\
&=&\log(\widehat{\sigma}^2)+\log\left(1+\frac{K}{n-K}\right) \\
&\approx&\log(\widehat{\sigma}^2)+\frac{K}{n-K} \\
&\approx&\log(\widehat{\sigma}^2)+\frac{K}{n},
\end{eqnarray*}
where $\widehat{\sigma}^2$ is the ML estimator of the error variance. Model selection based on $\bar{R}^2$ is therefore asymptotically equivalent to choosing the model with the smallest
$\log(\widehat{\sigma}^2)+K/n$.
This procedure is inconsistent.
Proposition:
$$\lim_{n\rightarrow\infty}P\bigl(\bar{R}^2_1>\bar{R}^2_2|\mathcal{M}_1\bigr)<1$$
Proof:
\begin{eqnarray*}
P\bigl(\bar{R}^2_1>\bar{R}^2_2|\mathcal{M}_1\bigr)&\approx&P\bigl(\log(s^2_1)<\log(s^2_2)|\mathcal{M}_1\bigr) \\
&=&P\bigl(n\log(s^2_1)<n\log(s^2_2)|\mathcal{M}_1\bigr) \\
&\approx&P(n\log(\widehat{\sigma}^2_1)+K_1<n\log(\widehat{\sigma}^2_2)+K_1+K_2|\mathcal{M}_1) \\
&=&P(n[\log(\widehat{\sigma}^2_1)-\log(\widehat{\sigma}^2_2)]<K_2|\mathcal{M}_1) \\
&\rightarrow&P(\chi^2_{K_2}<K_2) \\
&<&1,
\end{eqnarray*}
where the 2nd-to-last line follows because the statistic is the LR statistic in the linear regression case that follows an asymptotic $\chi^2_{K_2}$ null distribution.
QED
Now consider Akaike's criterion,
$$
AIC=\log(\widehat{\sigma}^2)+2\frac{K}{n}
$$
Thus, the AIC also trades off the reduction of the SSR implied by additional regressors against the "penalty term," which points in the opposite direction. Thus, choose $\mathcal{M}_1$ if
$AIC_1<AIC_2$, else select $\mathcal{M}_2$.
It can be seen that the $AIC$ is also inconsistent by continuing the above proof in line three with $P(n\log(\widehat{\sigma}^2_1)+2K_1<n\log(\widehat{\sigma}^2_2)+2(K_1+K_2)|\mathcal{M}_1)$. The adjusted $R^2$ and the $AIC$ thus choose the "large" model $\mathcal{M}_2$ with positive probability, even if $\mathcal{M}_1$ is the true model.
As the penalty for complexity in AIC is a little larger than for adjusted $R^2$, it may be less prone to overselect, though. And it has other nice properties (minimizing the KL divergence to the true model if that is not in the set of models considered) that are not addressed in my post.
