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!!



The power of chi-square test
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](https://en.wikipedia.org/wiki/Noncentral_chi-squared_distribution#Derivation_of_the_pdf) 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 $\mu^\top\mu$.


