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Under the null hypothesis, if we have $$\sqrt{n} \vec{x} \, \rightarrow_d \, N(0, I_p),$$ the test statistic can be construct as: $$\hat{\Psi} = n \vec{x}^{\top} \vec{x} \, \rightarrow_d \,\chi^2_p.$$ And we reject the null hypothesis if $\hat{\Psi} > \chi^2_{p, 1 - \alpha}$ under level $\alpha$.

Now, if under the alternative hypothesis $H_1$, $$\sqrt{n} \left( \vec{x} - \vec{\mu} \right)\, \rightarrow_d \, N(0, \Sigma),$$ I want to know the power of using test statistic $\hat{\Psi}$, i.e. $$\mathrm{P} \left( \left. \hat{\Psi} > \chi^2_{p, 1 - \alpha} \right| H_1 \right) = ?$$

I know under $H_1$, $\hat{\Psi}$ can be decomposed as: $$\hat{\Psi} \, \rightarrow_d \, \sum_{j = 1}^p \xi_j^2, \quad \left( \xi_1, \cdots, \xi_p \right)^{\top} \, \sim \, N(\vec{\mu}, \Sigma),$$ but I don't know how to deal with this square sum. Can anyone help me? Thanks a lot!!

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  • $\begingroup$ It seems we cannot use the above method to calculate the power of this test. Since $n^{-1} \hat{\Psi} \, \rightarrow_p \, \vec{\mu}^{\top}\vec{\mu}$, I guess, under $H_1$, $n^{-1/2} \left( \hat{\Psi} - \vec{\mu}^{\top}\vec{\mu} \right)$ is asymptotically normal? So the power goes to 1 when $n \rightarrow \infty$, it that right? $\endgroup$
    – 香结丁
    Jul 29, 2020 at 1:29

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If $\Sigma=I_p$, then the distribution of $\sum_{j=1}^p\xi_j^2$ for $(\xi_1,\cdots,\xi_p)^\top\sim N(\vec{\mu},\Sigma)$ is the non-central chi-square distribution with $p$ degrees of freedom and non-centrality parameter $\vec{\mu}^\top\vec{\mu}$.

If $\Sigma\ne I_p$, then the distribution of $\sum_{j=1}^p\xi_j^2$ for $(\xi_1,\cdots,\xi_p)^\top\sim N(\vec{\mu},\Sigma)$ has no name or closed-form expression of its pdf or cdf -- even when $\vec{\mu}=\vec0$.


Responding to a comment by the OP: Let $x_n:=\vec x$ and $\mu:=\vec\mu$. If $$Z_n:=\sqrt n(x_n-\mu) \to_d Z\sim N(0,\Sigma), \tag{1}$$ then, for any fixed $\mu\ne0:=\vec0$, $$nx_n^\top x_n=(\sqrt n\mu+Z_n)^\top(\sqrt n\mu+Z_n) =n\mu_n^\top\mu_n+O_P(\sqrt n),$$ so that $nx_n^\top x_n$ converges in probability to $\infty$, rather than to a finite random variable.

To get a finite random variable in the limit, you need to consider alternative hypotheses close enough to the null one; that is, in this case, you need to consider nonzero alternative asymptotic mean vectors close enough to $0$. In particular, it will make sense to fix some $\mu\ne0$ and consider the alternative values $\mu/\sqrt n$ of the asymptotic mean.

So, now instead of (1) we are assuming that $$Z_n:=\sqrt n x_n-\mu=\sqrt n(x_n-\mu/\sqrt n) \to_d Z\sim N(0,\Sigma).$$ Then $$nx_n^\top x_n=(\mu+Z_n)^\top(\mu+Z_n)\to_d (\mu+Z)^\top(\mu+Z),$$ and $\mu+Z\sim N(\mu,\Sigma)$, so that the limit random variable $(\mu+Z)^\top(\mu+Z)$ will have the non-central chi-square distribution with $p$ degrees of freedom and non-centrality parameter $\mu^\top\mu$.

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  • $\begingroup$ Thanks so much. But is there a feasible method to tackle $$\mathrm{P} \left( \sum_{j = 1}^p \xi_j^2 > \chi_{p, 1 - \alpha}^2\right)$$ when $\Sigma \neq I_p$? $\endgroup$
    – 香结丁
    Jul 29, 2020 at 0:38
  • $\begingroup$ @香结丁 : As I said, if $\Sigma\ne I$, then little can be said even for $\vec\mu=\vec0$; see e.g. link.springer.com/article/10.1007%2Fs00440-003-0262-6 for some inequalities. $\endgroup$ Jul 29, 2020 at 1:09
  • $\begingroup$ @ Iosif Appreciate your helping! It seems we cannot use this method to calculate the power. Since $n^{-1} \hat{\Psi} \, \rightarrow_p \, \vec{\mu}^{\top}\vec{\mu}$, I guess, under $H_1$, $n^{-1/2} \left( \hat{\Psi} - \vec{\mu}^{\top}\vec{\mu} \right)$ is asymptotically normal? It that right? $\endgroup$
    – 香结丁
    Jul 29, 2020 at 1:24
  • $\begingroup$ Thanks for your generous help! Now I know what I should do! $\endgroup$
    – 香结丁
    Jul 30, 2020 at 1:40
  • $\begingroup$ @香结丁 : As a relatively new user, you may want to look at these guidelines $\endgroup$ Jul 30, 2020 at 15:30

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