The power of chi-square test 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!!
 A: 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 $\mu^\top\mu$.
