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

  • $\begingroup$ What is your noise model? Presumably your $X$ values are fixed and known, rather than random. If the noise is Gaussian, e.g., you know the distribution of the test statistic exactly under the null hypothesis and for each fixed value of the parameter under the alternative. Everything should be able to be written pretty explicitly under such assumptions. $\endgroup$
    – cardinal
    Commented Oct 10, 2012 at 23:34

1 Answer 1


You are almost there, but you need to make your arguments more formal.

Rewrite the null and alternative hypothesis more generally so that they read $$ \begin{align} \mathfrak{h}_0{}:{}\beta_1 &= \beta^0_1\\ \mathfrak{h}_a{}:{}\beta_1&=\beta_1^a \end{align} $$

Then, using standard machinery, and under the linear regression model, we can show that under the null: $$ \begin{align} \dfrac{\sqrt{n}\left(\widehat{\beta}_1- \beta_1^0\right)}{\widehat{\text{a.var.}}(\widehat{\beta}_1)} &\rightarrow^d X\\ &{\sim}\text{t}(n-K) \end{align} $$ where $t(n-K)$ is the t-distribution with $n-K$ degrees of freedom, and $K$ is equal to the total number of regressors ($K=2$). Now recall that the definition of convergence in probability is that the distribution functions of the sequence of statistics converges to the distribution function of the limiting random variable at each continuity point of the limiting distribution. In this case,

$$ \scriptstyle \begin{align} \lim_{n\to\infty}\mathbb{P}\left[\left|\frac{\sqrt{n}\left(\widehat{\beta}_1- \beta_1^0\right)}{\widehat{\text{a.var.}}(\widehat{\beta}_1)} \right| > t_{1-\tfrac{\alpha}{2}}(n-K)\right] &= \lim_{n\to \infty}\mathbb{P}\left[\frac{\sqrt{n}\left(\widehat{\beta}_1- \beta_1^0\right)}{\widehat{\text{a.var.}}(\widehat{\beta}_1)} > t_{1-\tfrac{\alpha}{2}}(n-K)\right] \\ &\quad+ \lim_{n\to \infty}\mathbb{P}\left[\frac{\sqrt{n}\left(\widehat{\beta}_1- \beta_1^0\right)}{\widehat{\text{a.var.}}(\widehat{\beta}_1)} \leq -t_{1-\tfrac{\alpha}{2}}(n-K)\right] \\ &= \mathbb{P}\left[X > t_{1-\tfrac{\alpha}{2}}(n-K) \right] + \mathbb{P}\left[X \leq -t_{1-\tfrac{\alpha}{2}(n-K)} \right] \\ %&= 1-\Phi(z_{1-\tfrac{\alpha}{2}}) + \Phi(z_{\tfrac{\alpha}{2}}) \\ &= \alpha \end{align} $$ where $\alpha$ is the nominal size of the test, and $t_\alpha(d)$ is the $\alpha$-th quantile of the t-distribution with $d$ degrees of freedom. This proves that the t-test has the correct size asymptotically.

Now, under the alternative, we have that $$ \begin{align} \dfrac{\sqrt{n}\left(\widehat{\beta}_1- \beta_1^a\right)}{\widehat{\text{a.var.}}(\widehat{\beta}_1)} &\rightarrow^d X\\ &{\sim}\text{t}(n-K) \end{align} $$

This implies that $$ \begin{align} \dfrac{\sqrt{n}\left(\widehat{\beta}_1- \beta_1^0\right)}{\widehat{\text{a.var.}}(\widehat{\beta}_1)} &=\dfrac{\sqrt{n}\left(\widehat{\beta}_1- \beta_1^a\right)}{\widehat{\text{a.var.}}(\widehat{\beta}_1)}+\dfrac{\sqrt{n}\left(\beta^a_1- \beta_1^0\right)}{\widehat{\text{a.var.}}(\widehat{\beta}_1)} \\ &= \mathcal{O}_p(1)+ \mathcal{O}_p(\sqrt{n}) \\ &= \mathcal{O}_p(\sqrt{n}) \end{align} $$ By definition of orders of magnitude, this means that w.p.a. 1 under the alternative, the t-statistic is greater than any fixed (finite) critical value.


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