upper bound on power of neyman-pearson hypothesis test Let $H_0$ and $H_1$ be two distributions.  The Neyman-Pearson lemma says that of all rejection regions $R$ with fixed probability $\alpha$ under $H_0$, the one with maximal probability under $H_1$ is the set of the form $R = \{x: \frac{p_1(x)}{p_0(x)} \ge c\}$ with $c$ chosen such that  $\mathbb{P}_{x \sim H_0}(x \in R) = \alpha$.
The power of the test is then $\mathbb{P}_{x \sim H_1}(x \in R)$.
In my case, $H_0 = \mathcal{N}(0, \tau^2 I_d)$ and $H_1 = \mathcal{N}(\mu, \sigma^2 I_d)$ with $\tau > \sigma$.  The dimension is high (hundreds of thousands).
For this choice of $H_0$ and $H_1$, computing the rejection region in closed form does not appear to be possible, so I'd like to compute an upper bound on the power of the NP test at level $\alpha$ without actually computing the rejection region.
Are there any generic methods to compute an upper bound on the power of a Neyman-Pearson test at level $\alpha$?  I'm looking for an exact bound, not an approximate bound based on e.g. the CLT.
In textbooks and papers, I've seen many ways to upper-bound the power as a function of $c$ (the likelihood ratio threshold), but none to upper-bound the power as a function of $\alpha$.  That said, if I had a right tail bound on $\frac{p_1(x)}{p_0(x)}$ under $H_1$ and a left tail bound on $\frac{p_1(x)}{p_0(x)}$ under $H_0$, I could combine those to get an upper bound on the power as a function of $\alpha$.
 A: $\renewcommand{\P}{\operatorname{\mathsf P}}
\newcommand{\al}{\alpha}
\newcommand{\be}{\beta}
\newcommand{\la}{\lambda}
\newcommand{\D}{\overset{\text{D}}=}
\newcommand{\eD}{\overset{\text{D}}\to} 
\newcommand{\si}{\sigma}
\newcommand{\tZ}{\tilde Z}$
Let $(X_1,\dots,X_d)$ be the observed normal random vector. Then the Neyman--Pearson test will reject $H_0$ if $S:=\sum_1^d(X_i-k\mu_i)^2<c$, where $\mu=:(\mu_1,\dots,\mu_d)$ and  $c$ is a critical value and 
\begin{equation*}
 k:=\frac{\tau^2}{\tau^2-\sigma^2}. \tag{0.1}
\end{equation*}
With $\al\in(0,1)$ fixed, for large $d$ the hypotheses $H_0$ and $H_1$ will be too easy to distinguish from each other (so that the power be close to $1$) unless $H_0$ and $H_1$ are close enough to each other. The following rather natural conditions will provide for that: 
\begin{equation*}
 d\to\infty,\quad\tau^2\to\tau_0^2\in(0,\infty),\quad \frac{\tau^2-\si^2}{\si^2}=\frac{\tau^2}{\si^2}-1\sim\frac{w^2}{\sqrt d},\quad \frac{\|\mu\|^2}{\si^2}\sim b^2 \tag{0.2}
\end{equation*}
for some fixed $w,b$ in $(0,\infty)$, so that 
\begin{equation*}
 \tau\sim\si,\quad k\sim\frac{\sqrt d}{w^2}\to\infty. \tag{0.3}
\end{equation*}
Let $Z,Z_1,Z_2,\dots$ be iid standard normal random variables. 
Under $H_0$, by the spherical symmetry, 
\begin{equation*}
 S\D\sum_1^d(\tau Z_i-k\mu_i)^2
 \D\tau^2((a_0+Z_1)^2+Z_2^2+\dots+Z_d^2), 
% \eD\tau^2(a_0^2+d+Z\sqrt{2d}), 
\end{equation*}
where $\D$ means the equality in distribution and 
\begin{equation*}
 a_0^2:=(k/\tau)^2\|\mu\|^2\sim k^2b^2\sim\frac{b^2}{w^2}\,d\to\infty. \tag{1}
\end{equation*} 
By the central limit theorem, 
\begin{equation*}
 \frac{Z_2^2+\dots+Z_d^2-d}{\sqrt d}\eD Z\sqrt2,
\end{equation*}
where $\eD$ means the convergence in distribution. Also, by (1), 
\begin{equation*}
 \frac{(a_0+Z_1)^2-a_0^2}{\sqrt d}=\frac{2a_0Z_1+Z_1^2}{\sqrt d}\to\frac{2b}w\,Z_1  
\end{equation*}
pointwise and hence in distribution. Thus, under $H_0$,
\begin{equation*}
(H_0):\quad \frac{S/\tau^2-(a_0^2+d)}{\sqrt d}\eD \frac{2b}w\,Z_1+Z\sqrt2\D\la Z,\quad \la:=\sqrt{\frac{4b^2}{w^2}+2}. \tag{2}
\end{equation*}
Similarly, under $H_1$, 
\begin{equation*}
 S\D\sum_1^d(\si Z_i-(k-1)\mu_i)^2
 \D\si^2((a_1+Z_1)^2+Z_2^2+\dots+Z_d^2), 
% \eD\si^2(a_1^2+d+Z\sqrt{2d}), 
\end{equation*}
where 
\begin{equation*}
 a_1^2:=((k-1)/\si)^2\|\mu\|^2\sim a_0^2\sim\frac{b^2}{w^2}\,d\to\infty, \tag{3}
\end{equation*}
and hence, under $H_1$,
\begin{equation*}
(H_1):\quad \tZ:=\frac{S/\si^2-(a_1^2+d)}{\la \sqrt d}\eD Z; \tag{3.1}
\end{equation*}
cf. (2). 
Choosing now the critical value 
\begin{equation*}
 c:=\tau^2(a_0^2+d+z_\al\la\sqrt{d}), \tag{3.2}
\end{equation*}
where $z_\al:=\Phi^{-1}(\al)$ and $\Phi$ is the standard normal cdf, we see that the size of the test is 
\begin{equation*}
 \P_{H_0}(S<c)=\P_{H_0}\Big(\frac{S-\tau^2(a_0^2+d)}{\la\sqrt d}<z_\al\Big)\to\P(Z<z_\al)=\al, 
\end{equation*}
as desired. 
Recalling (3.1), (3.2), (1), (3), (0.3), (0.2), the relation $k^2-(k-1)^2\sim2k$, and the definition of $\la$ in (2), for the power of the test we now have 
\begin{align*}
 \P_{H_1}(S<c)&=\P_{H_1}\Big(\tZ<z_\al\frac{\tau^2}{\si^2}
 +\frac{\tau^2-\si^2}{\la\si^2}\,\sqrt d
 +\frac{[k^2-(k-1)^2]\|\mu\|^2}{\la\si^2\sqrt{d}}\Big) \\ 
& \to\Phi\Big(z_\al+(w^2+2b^2/w^2)\Big/\sqrt{\frac{4b^2}{w^2}+2}\Big),  
\end{align*}
which latter is $>\al$, and it is close to $1$ if either $w$ or $b$ is large, which seems to agree with the intuition.
