I'm working on some exercises for my econometrics class and I'm a little confused. I'm meant to consider a model
$$Y=\beta_0 +\beta_1X+u$$
and propose a test (test statistic and critical value) of $H_0:\beta_1 =0$ against $H_1:\beta_1 \ne 0$ such that the Type 1 error goes to 5% and the power of the test goes to 1 as the sample size goes to infinity.
Pretty standard I think, I'm pretty sure that we just want to use the test statistic $\tau=| \frac {\hat{\beta_1}}{\hat{SE}(\beta_1)}|$ with a critical value of 1.96. By the central limit theorem, under the null hypothesis, we have
$\frac {\hat{\beta_1}}{\hat{SE}(\beta_1)}\longrightarrow_d N(0,1)$
and so we can show the type 1 error in the limit pretty easy.
If anything is wrong so far, please let me know.
The problem I'm having is showing that the power of the test approaches 1. Under $H_1$, for large sample sizes
$\frac {\hat{\beta_1}}{\hat{SE}(\beta_1)}\approx \frac {\hat{\beta_1}}{{SE}(\beta_1)}=\frac{\hat{\beta_1}}{\sqrt{\frac{\sigma^2}{nVar(X)}}}\longrightarrow_d ???????$
Now this is where I'm confused... Does that term converge to something? Am I meant to make an argument about convergence in distribution?
The one thing I was thinking is that we can argue that
$$\frac{\hat{\beta_1}}{\sqrt{\frac{\sigma^2}{nVar(X)}}}\longrightarrow_p \frac{\beta_1}{\sqrt{\frac{\sigma^2}{nVar(X)}}}$$
And that this number is clearly increasing with n, so the test statistic goes to infinity in the limit. Is this a correct argument though? My problem with this is that, why don't we then say that that implies that the test statistic goes to 0 in the limit under $H_0$? We don't say that, we say that it goes to a distribution, not a number, but