# Local Asymptotics: The Simplest Possible Example

If you study enough econometrics, you will eventually come across an asymptotic argument in which some parameter is assumed to change with sample size. This peculiar notion goes by a variety of names including “Pitman drift,” a “sequence of local alternatives,” and “local mis-specification,” and crops up in a wide range of problems from weak instruments, to model selection, to power analysis.1 Whatever you choose to call it, the idea of a parameter that changes with sample size is bizarre, and I remember spending weeks trying to understand it when I was a graduate student. How could parameters, fixed quantities that we’re trying to estimate, possibly know anything about our sample size?

Do we expect parameters to be smaller when we have more data? Do we expect them to be larger when we have less data? The answer to both questions is a resounding NO. Like all asymptotics, what I will call local asymptotics are nothing more than a thought experiment that we set up for mathematical convenience. Ideally we would derive finite sample results for every problem of interest, but this is rarely possible in practice. For this reason we turn to asymptotic results, such as the central limit theorem. Sometimes this works out OK, and sometimes it’s a disaster. The goal of local asymptotics is to derive results that more closely approximate the finite sample behavior that we can understand from simple examples, in the hope that this will lead to better approximations in more complicated problems. In this post, I’ll illustrate the usefulness of local asymptotics in the simplest example I could think of: a one-sided test for the mean of a normal distribution with known variance. No advanced statistics or econometrics are used below, so even if you found the preceding paragraph off-putting give the rest a go: you may be pleasantly surprised!

Suppose that we observe $X_1, X_2, \dots, X_{n} \overset{iid}{\sim}N(\mu, 1)$ and want to test $$H_0\colon \mu = 0$$ against the one-sided alternative $$H_1\colon \mu >0$$. In this admittedly very simple example, the Econometrics 101 test statistic is $T_{n} = \sqrt{n} \bar{X}_{n} \sim N\left(\mu \sqrt{n}, 1\right)$ where $$\bar{X}_{n}$$ is the sample mean. We reject when $$\sqrt{n} \bar{X}_{n}>z_{1-\alpha}$$ where $$z_{1-\alpha}$$ is the $$1-\alpha$$ quantile of a standard normal distribution. Let’s calculate the power of this test: the probability of rejecting the null hypothesis given that it is false. We find that $\begin{eqnarray*} \mbox{Power}(T_{n}) &=& P\left(\sqrt{n} \bar{X}_{n}>z_{1-\alpha}\right) = P\left(Z + \mu\sqrt{n} >z_{1-\alpha}\right)\\ &=&P\left(Z >z_{1-\alpha} - \mu\sqrt{n}\right) = 1 - \Phi\left(z_{1-\alpha} - \mu\sqrt{n}\right) \end{eqnarray*}$ where $$Z$$ is a standard normal random variable and $$\Phi$$ is the standard normal CDF.

Now suppose we decided to do something completely crazy: throw away half our sample. Let $$\bar{X}_{n/2}$$ denote the sample mean based on observations $$1, 2, \dots, \lfloor N/2 \rfloor$$ only, where $$\lfloor x \rfloor$$ denotes the floor function, i.e. the greatest integer less than or equal to $$x$$.2 We can still construct a perfectly valid test with size $$\alpha$$ as follows. Define $T_{n/2} = \sqrt{\lfloor n/2 \rfloor } \bar{X}_{n/2} \sim N\left(\mu \sqrt{\lfloor n/2 \rfloor }, 1\right)$ and reject if $$\sqrt{n} \bar{X}_n > z_{1-\alpha}$$. But there’s an obvious problem here: there must be a cost for throwing away perfectly good data. Indeed, if we calculate the power for this crazy test, we’ll find that it’s strictly lower than that of the sensible test based on the full sample. In particular, $\mbox{Power}(T_{n/2}) = 1 - \Phi\left(z_{1-\alpha} - \mu\sqrt{\lfloor n/2 \rfloor }\right)$ using the same argument as above with $$\lfloor N/2 \rfloor$$ in place of $$n$$. Unsurprisingly, it turns out to be a bad idea to throw away half of your data!

Now, for an example this simple we’d never resort to asymptotics, but suppose we did. How do these two tests compare as the sample size goes to infinity? The asymptotic size in this example is the same as the finite-sample size since we know the exact sampling distribution of the test statistics under the null and neither depends on sample size. But what about the power? We have, $\begin{eqnarray*} \lim_{n\rightarrow \infty} \mbox{Power}(T_{n}) &=& \lim_{n\rightarrow \infty}\left[1 - \Phi\left(z_{1-\alpha} - \mu\sqrt{n}\right) \right] = 1\\ \lim_{n\rightarrow \infty} \mbox{Power}(T_{n/2}) &=& \lim_{n\rightarrow \infty}\left[1 - \Phi\left(z_{1-\alpha} - \mu\sqrt{\lfloor n/2 \rfloor }\right) \right] = 1 \end{eqnarray*}$ In other words, both of these tests are consistent: as the sample size goes to infinity, the power goes to one. Think about this for a moment: we know that for any fixed sample size a test based on the full sample is strictly more powerful but in the limit this difference disappears. This strongly suggests that something is wrong with comparing two tests on the basis of their asymptotic power. Clearly the second test is worse than the first, but the asymptotics obscure this.

You might object that I’ve cooked up a particularly perverse example, but it turns out that this phenomenon is quite general. It’s easy to find consistent tests, in fact it’s difficult to find tests that aren’t consistent. But we know from simulation studies that not all consistent tests are created equal: some have much better finite sample power than others and it’s ultimately finite sample performance that we care about. One way around this problem would be to only compare the finite-sample properties of different tests and never use asymptotics. But we almost never know the exact sampling distribution of our test statistics.3

This is where local alternatives come in. Rather than evaluating our tests against a fixed alternative $$\mu$$, suppose we were to evaluate it against a sequence of local alternatives that drift towards the null at rate $$n^{-1/2}$$. In other words, our alternative becomes $$H_{1,n} \colon \mu = \delta / \sqrt{n}$$ where, for this one-sided test, $$\delta > 0$$. If we substitute $$\delta/\sqrt{n}$$ for $$\mu$$ and take the limit as $$n\rightarrow \infty$$, we find $\begin{eqnarray*} \lim_{n\rightarrow \infty} \mbox{Power}(T_{n}) &=& \lim_{n\rightarrow \infty}\left[1 - \Phi\left(z_{1-\alpha} - \frac{\delta}{\sqrt{n}}\sqrt{n}\right) \right]\\ &=& 1 - \Phi\left(z_{1-\alpha} - \delta \right) \end{eqnarray*}$ and similarly $\begin{eqnarray*} \lim_{n\rightarrow \infty} \mbox{Power}(T_{n/2}) &=& \lim_{n\rightarrow \infty}\left[1 - \Phi\left(z_{1-\alpha} - \frac{\delta}{\sqrt{n}}\sqrt{\lfloor n/2 \rfloor }\right) \right]\\ &=& 1 - \Phi\left(z_{1-\alpha} - \frac{\delta}{\sqrt{2}} \right) \end{eqnarray*}$ Wow! Our problem has disappeared! The asymptotic power of the two tests now differs in essentially the same way as the finite sample power. Also note that the power no longer converges to one. Intuitively, this is because the drifting sequence of alternatives $$\delta/\sqrt{n}$$ makes it “harder and harder” to reject the null as the sample size grows by shrinking just fast enough but not so fast that the power goes to zero. This type of calculation is called a local power analysis. A test that has asymptotic power greater than zero in such a setting is said to have “power against local alternatives.”

This example was a bit silly since we already knew the answer. But this is precisely what made it so obvious that local asymptotics make more sense in this setting than fixed-parameter asymptotics. Now that you understand this basic intuition, I hope you’ll feel more confident tackling examples of local asymptotics that come up in the econometrics literature.

1. I’ve used this idea in some of my own work on moment selection and model selection.↩︎

2. If you don’t like this, ignore it and pretend that $$n$$ is an even number!↩︎

3. This blog post is a very special exception created for pedagogical purposes :)↩︎