How Can I calculate Power function of Feng and Sun test?

I'm wondering how I can calculate the power function of the test statistic

$$T_n=\frac{1}{n_1(n_1-1)}\frac{1}{n_2(n_2-1)}\sum_{k=1}^p \underset{i\not=j}{\sum^{n_1}\sum^{n_1}} \underset{s\not=t}{\sum^{n_2}\sum^{n_2}}\frac{(X_{1ik}-X_{2sk})(X_{1jk}-X_{2tk})}{\hat{\sigma}^2_{1k(i,j)}+\gamma\hat{\sigma}^2_{2k(s,t)}}$$

where $\gamma=n_1/n_2$, $\hat{\sigma}^2_{1k(i,j)}$ is the sample variance of $\{X_{1lk}\}_{l=1}^{n_1}$ excluding $X_{1ik}$ and $X_{1jk}$

which is given in Feng and Sun (2015) [1], page 31, from theorem 1 (same page):

Under Conditions (C1)--(C3), as $p,n \to \infty$, $\frac{T_n-E(T_n)}{\sqrt{\mathrm {var}(T_n)}}\overset{\mathcal{L}}{\longrightarrow} N(0,1).$

Any help will be appreciated.

[1]: L.Feng and F.Sun (2015),
"A note on high-dimensional two-sample test",
Statistics and Probability Letters, 105, 29–36
http://www.sciencedirect.com/science/article/pii/S0167715215001789
[arXiv version arXiv:1502.05455 [stat.ME] here]

• I have added some brief quotes of the precise things referred to though additional context would help Commented Apr 4, 2017 at 22:53
• I know my question was bad! You edit is even more appreciated than your answer, thanks @Glen_b Commented Apr 5, 2017 at 21:32

The short answer is you can't compute power from theorem 1. That's not what it's about. Theorem 1 establishes the asymptotic distribution of the test statistic under the null hypothesis, as Feng and Sun state quite directly, just above the statement of the theorem:

The following theorem establishes the asymptotic null distribution of $T_n$.

(emphasis mine)

The null distribution doesn't tell us about power (which is the rejection rate under some alternative).

The authors do subsequently perform some power calculations, which they describe in a later section, and which they obtain via simulation.

Power is calculated for a sequence of alternatives by specifying the exact situation at each specific alternative, simulating samples (from whichever high-dimensional distribution they chose to look at) many times, performing their test each time and calculating the proportion of rejections at that specific alternative, and repeating the process at every alternative they want to give the power for. It is possible to compute power curves and power surfaces in this fashion, given sufficient computational effort.

[If such an exercise in power simulation is not something you've done before, don't start with this test - start with something much simpler, such as an investigation of power for the Wilcoxon-Mann-Whitney univariate two sample test. You'll have enough hurdles to overcome doing that, but once you've done generated a few power curves for simpler cases, much about this more complex situation will become considerably clearer.]