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Suppose we have an estimator $\hat\mu$ of population parameter $\mu$ and we know that

$$\sqrt{N}(\hat\mu-\mu)\overset{d}{\to}N(0,1).$$

We are interested in the following hypothesis scheme:

$$H_0: \mu=0$$ $$H_1: \mu\ne0$$

Suppose that $\mu=\delta$ for some arbitrarily small $\delta>0$. I need to show that the probability of rejecting $H_0$ tends to 1 as the sample size goes to $\infty$. Why is this like so? I think that it has to do with the fact that the convergence in distribution implies (not completely positive of that) that $\hat\mu\overset{p}{\to}\mu>0$ and the probability of our statistic $\hat\mu$ being exactly $\mu$ is zero thus with a large sample any value of $\hat\mu$ that is slightly different from $\mu$ will lead to the rejection of $H_0$. Is this reasoning correct?

Any help is appreciated.

Thanks.

manifold
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1 Answers1

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Sort of, but not quite: $\hat\mu$ being exactly zero isn't needed and you've left out some important information.

Consider the distributions. If $\mu=\delta$, then $$\sqrt{N}(\hat\mu-\delta)\stackrel{d}{\to}N(0,1)$$ but if $\mu=0$ then $$\sqrt{N}(\hat\mu-0)\stackrel{d}{\to}N(0,1).$$ So under $H_0$, $\hat \mu$ is close to 0 in large samples and otherwise $\hat\mu$ is close to $\delta$ in large samples.

Now, one reasonable kind of test would be the test that rejects if $|\hat\mu_N|>C$ for some $C>0$. If $0<C<\delta$, then $P(|\hat\mu|<C)\to 1$ under $H_0$ and $P(\hat\mu>C)\to 1$ under $H_1$. So for any test of this sort the probability of rejecting $H_0$ at any level $\alpha$ goes to 1.

We can then argue that since this test has power converging to 1 no-one would use a test that doesn't have power converging to 1, and so we are done.

We do need an argument of this kind, because you didn't specify what test you wanted to use and there are bad tests out there. Suppose we took a test that rejected $H_0$ if $\hat\mu\neq 0$. For this test, the probability of rejection is zero for all $N$. It's an unbiased test, but it's a bad test. Or we could reject $H_0$ if $\hat\mu\in [-0.42,\,0.69]$. That's a bad test and whether the probability of rejection goes to 1 depends on $\delta$. Or we could generate an independent $U$ from a uniform distribution on $[0,1]$ and reject if $U<\alpha$ for some specified level $\alpha$. That's an unbiased exact test and extremely bad, and the probability of rejection doesn't go to 1.

Thomas Lumley
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