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For $V\sim {\chi }_n^2$ and $x>0$, the upper tail is at most $e^{-x}$ above $n+2\sqrt{nx}+2x$, and the lower tail is at most $e^{-x}$ below $n-2\sqrt{nx}$. Indeed $\log \mathbb{E}{e}^{s(V-n)}=-ns-(n/2)\log (1-2s)\le n{s}^2/(1-2s)$ for $0<s<1/2$. The Chernoff bound with $s=\sqrt{x/n}/(1+2\sqrt{x/n})$ proves the upper estimate. For the lower estimate, $\log \mathbb{E}{e}^{-s(V-n)}\le n{s}^2$ for $s>0$, and optimization gives the result. The lower threshold can be negative; the chi-squared Chernoff lower-tail bound provides an always-positive alternative.

Put $\ell =\log (d/\delta )$ and $r=\sqrt{\ell /n}\le 1$. Use the quadratic scan statistic $\Lambda$ and reject when $\Lambda >1+4r$. Under the null hypothesis, each $nu(T{)}^{\top }\hat{\Sigma }u(T)$ has the chi-squared distribution with $n$ degrees of freedom. The chi-squared concentration inequality gives $P_0^{\otimes n}(\Lambda >1+2r+2r^2)\le d{e}^{-\ell }=\delta$, by the union bound. Since $2r+2r^2\le 4r$, the proposed Type I error is at most $\delta$.

For the true set $S$, $nu(S{)}^{\top }\hat{\Sigma }u(S)/(1+\theta )$ has the same chi-squared distribution. We need a lower-tail bound that remains positive even when $\ell /n$ is close to one. Set $a=e^{-4r}$. The chi-squared Chernoff lower-tail bound gives $P(V\le na)\le \exp [-n(a-1-\log a)/2]\le e^{-n{r}^2}=e^{-\ell }$: indeed $e^{-4r}-1+4r\ge 2r^2$ for $0\le r\le 1$. To verify this last inequality, its derivative is $4(1-r-e^{-4r})$; the expression in parentheses is strictly concave, starts at zero, and changes sign once, so the minimum occurs at an endpoint, where the inequality holds.

The function $g(r)=(1+4r)e^{4r}-1$ is convex with $g(0)=0$, so $g(r)\le rg(1)<300r$ on this interval. Consequently $\theta >300r$ implies $(1+\theta )e^{-4r}>1+4r$. Failure to reject then requires $V\le na$, giving Type II error at most $e^{-\ell }\le \delta$, uniformly in $S$. An explicit, deliberately conservative answer isThe sharp constant is unnecessary; the detection scale is $\sqrt{\log (d/\delta )/n}$.