3 min read

Which infinite sequence?

I’m at ENAR, and there are talks with asymptotic theory1. One thing that caught my attention is problems with two different sample sizes, eg, a main sample and a validation sample. Call the two sample sizes \(m\) and \(n\). Theorems are then proved under the assumption that \(m/n\) converges to finite non-zero constant \(C\). What is the statistical content of this assumption?

In an application, we have one data set, with one particular value of \(m\) and \(n\). The assumption \(m/n\to C\) is about which infinite sequence we want to think of the data as an element of2 If we embedded the actual \(m\) and \(n\) in an infinite sequence where we used \(m/n=2\) on Wednesdays and \(m/n=0.5\) on Fridays, would it matter?

This is where it’s useful to extract subsequences. Suppose \(m/n\) is bounded above and bounded away from zero. Suppose the two samples are each iid so that \(m\) and \(n\) are just measuring the amount of information. Any subsequence \((m_k,n_k)\) of our infinite sequence has a subsubsequence \((m_{k_j},n_{k_j})\)that converges to some finite, non-zero \(C\). We could just choose that subsubsequence as our infinite sequence and get convergence of \(m/n\) for free. We’d get the same asymptotic approximation to the finite sample distribution, so the assumption \(m/n\to C\) is not a loss of generality3 in statistical terms.

Another way to write this is to standardise the limit distribution. Suppose \[\sqrt{n}(\hat\theta-\theta)\stackrel{d}{\to}N(0,\sigma^2(C))\] Instead, we can do4 \[\sqrt{\frac{n}{\sigma^2(m/n)}}(\hat\theta-\theta)\stackrel{d}{\to}N(0,1).\] Any subsequence has a subsubsequence where \(m/n\to C\), and along this subsubsequence the asymptotic distribution is \(N(0,1)\). That implies the asymptotic distribution along the whole sequence is \(N(0,1)\). So, again, it’s not a loss of generality from a statistical point of view to assume the limit is \(N(0,1)\) for the whole sequence and use \(N(0,\sigma^2(m/n))\) as the asymptotic approximation to the data you actually have, with \((m,n)\).

When will this break down? The standardised version indicates one potential problem. Suppose the value of \(m/n\) affects the asymptotic mean as well as the asymptotic variance. For example, in perinatal data, \(n\) might be the number of deliveries and \(m\) the number of babies. The ratio \(m/n\) tells us about the fraction of multiple births, which could easily be informative. As another example, suppose \(n\) is the number of households sampled and \(m\) the number of people; \(m/n\) is then the mean household size, and could matter. In both cases it could still be that there’s no problem, but more argument is needed; it’s not immediate.

  1. And in some cases it’s relevant

  2. Never use a preposition to end a sentence.

  3. and so doesn’t have any real content

  4. yes, I’m assuming something about \(\sigma^2(\cdot)\)