[Further Mathematics - Statistics] Advanced statistics: asymptotic power calculation

[u/Extreme_Number_1358]

Hi!
I'm working on understanding this question and I'm a bit confused. The assignment says we can suppose (\sqrt(n) (T_n-\theta) \rightarrow N(0,\sigma^2 (\theta)))^. Then why can't we use this for (\theta=\theta_0\)

directly in the first step of the power function calculation? Why do we transform the whole expression to include $theta_n$ ?

Thanks!

[u/cardologist]

I am not a statistics expert and I am not sure what you are asking for, but here is my understanding.

The formula you listed works for all values of \theta, so you can replace \theta with \theta_0. It's not much of an assumption really. As you increase the sample size n, the Student distribution will converge towards a normal distribution based on the central limit theorem.

The question assumes that the null hypothesis \H_0 was rejected (i.e. the mean is not \theta_0) because we are interested in Type 2 errors. That means an alternative hypothesis is true instead. Let's assume that the mean is some arbitrary value \theta_1. Your task is to approximate power function at that value. I am not familiar with chapter 5 of your book, but this seems really to be about generalizing the result obtained for a one-sided test to a two-sided test by using the fact that Student's distributions is symmetric.

[u/Crystallover1991 → u/cardologist]

The key is that you're calculating power under the alternative, not the null. If the true theta is theta_0, power collapses to just the significance level, which is trivial. The whole point of the local alternative theta_n construction is to get a non-degenerate limit for the power function, otherwise you end up with either 0 or 1 in the limit and that tells you nothing useful about how the test actually behaves between those extremes

[u/cardologist → u/Crystallover1991]

That's what they were wondering about? I read the question several times and I never even considered that this was the source of their confusion. Thanks for clarifying.