[proofplan] We first formalize the problem as a sequence of null and planted laws on the sample matrix, together with the empirical covariance statistic. The subcritical statement is exactly the spherical rank-one spiked Wishart contiguity theorem, after checking that the present normalization has spike strength $\theta$ and aspect ratio $p_n/n \to \gamma$. The supercritical statement follows from the BBP largest-eigenvalue limit: the null top eigenvalue converges to the Marchenko--Pastur edge, while the planted top eigenvalue converges to a strictly larger outlier, so a fixed threshold between the two limits gives vanishing type-I and type-II errors. [/proofplan]

[step:Define the null and planted laws and the sample covariance statistic] For each $n$, let $p_n \in \mathbb{N}$ denote the ambient dimension and assume $p_n/n \to \gamma$. Let $\mathcal{X}_n := \mathbb{R}^{p_n \times n}$ be the sample space for the data matrix $X_n := (X_1,\dots,X_n)$ whose columns are the observations. Let $\mathbb{P}_{0,n}$ denote the law of $X_n$ under $H_0$, and let $\mathbb{P}_{1,n}$ denote the mixture law under $H_1$ after integrating over the uniformly spherical spike direction $v_n \in S^{p_n-1}$. Define the sample covariance map $S_n: \mathcal{X}_n \to \mathbb{R}^{p_n \times p_n}$ by \begin{align*} S_n(X_n) := \frac{1}{n}\sum_{i=1}^{n} X_i X_i^\top. \end{align*} Let $\operatorname{Sym}_{p_n}(\mathbb{R}) := \{A \in \mathbb{R}^{p_n \times p_n} : A^\top = A\}$ denote the real [vector space](/page/Vector%20Space) of symmetric $p_n \times p_n$ matrices. Let $\lambda_{\max}: \operatorname{Sym}_{p_n}(\mathbb{R}) \to \mathbb{R}$ denote the largest eigenvalue map. The statistic used in the supercritical part is $\lambda_{\max}(S_n)$. [/step]

[step:Apply the subcritical contiguity theorem for the spherical spiked Wishart model]Assume $0<\theta<\sqrt{\gamma}$. We apply the [spherical rank-one spiked Wishart contiguity theorem](/page/Spherical%20Spiked%20Wishart%20Contiguity) in its real Gaussian covariance form. Its hypotheses are satisfied here: the observations are real vectors in $\mathbb{R}^{p_n}$, the null covariance is $I_{p_n}$, the spike prior is the uniform probability measure on $S^{p_n-1}$, the planted covariance conditional on $v_n$ is $I_{p_n}+\theta v_n v_n^\top$, the spike strength $\theta$ is fixed in $n$, and the proportional asymptotic regime is $p_n/n \to \gamma \in (0,\infty)$. The theorem's subcritical hypothesis is $\theta^2<\gamma$, which is equivalent to $\theta<\sqrt{\gamma}$ because $\theta>0$. The theorem concludes that $\mathbb{P}_{1,n}$ is [contiguous](/page/Contiguity) with respect to $\mathbb{P}_{0,n}$, meaning that for every sequence of measurable events $A_n \subset \mathcal{X}_n$, \begin{align*} \mathbb{P}_{0,n}(A_n) \to 0 \implies \mathbb{P}_{1,n}(A_n) \to 0. \end{align*} This is exactly the first assertion.[/step]

[guided]We now use the random matrix input that handles the hard information-theoretic part of the theorem, namely the [spherical rank-one spiked Wishart contiguity theorem](/page/Spherical%20Spiked%20Wishart%20Contiguity) in its real Gaussian covariance normalization. We check its hypotheses explicitly. The sample space is $\mathcal{X}_n=\mathbb{R}^{p_n\times n}$, the null law $\mathbb{P}_{0,n}$ is the law of $n$ independent centered real Gaussian vectors with covariance $I_{p_n}$, and the planted law $\mathbb{P}_{1,n}$ is the mixture obtained by first drawing $v_n$ uniformly from $S^{p_n-1}$ and then drawing the columns independently with covariance $I_{p_n}+\theta v_n v_n^\top$. The theorem requires a proportional limit $p_n/n\to\gamma$ with $\gamma\in(0,\infty)$, which is one of the hypotheses of the statement. It requires a fixed spike strength, and here $\theta$ does not depend on $n$. It requires the spherical rank-one prior, which is exactly the uniform distribution of $v_n$ on $S^{p_n-1}$. It also requires the real Gaussian Wishart normalization in which the sample covariance is $n^{-1}\sum_{i=1}^n X_iX_i^\top$, which is precisely the map $S_n$ defined above. Finally, its subcritical inequality is $\theta^2<\gamma$. Since $\theta>0$, this condition is equivalent to $\theta<\sqrt{\gamma}$. After these hypotheses are verified, the theorem gives [contiguity](/page/Contiguity) of the planted laws with respect to the null laws. In concrete event language, for every sequence of measurable events $A_n\subset\mathcal{X}_n$, \begin{align*} \mathbb{P}_{0,n}(A_n) \to 0 \implies \mathbb{P}_{1,n}(A_n) \to 0. \end{align*} This is exactly what the theorem statement calls the planted distribution being contiguous with respect to the null distribution. The point is that below the threshold, no event that is asymptotically negligible under pure noise can have non-negligible probability under the planted mixture.[/guided]

[step:Separate the null edge from the planted outlier above the BBP threshold] Assume $\theta>\sqrt{\gamma}$. By the null Marchenko--Pastur upper-edge conclusion in the real Wishart version of the [BBP transition for spiked covariance matrices](/page/BBP%20Transition), equivalently the zero-spike case of that theorem, the hypotheses are satisfied under $\mathbb{P}_{0,n}$ because the columns are independent real Gaussian vectors with covariance $I_{p_n}$ and $p_n/n\to\gamma\in(0,\infty)$. This formulation covers every positive aspect ratio, including $\gamma>1$, where $S_n$ may have zero eigenvalues because $p_n>n$ asymptotically; the upper spectral edge is still $(1+\sqrt{\gamma})^2$. Hence, under $\mathbb{P}_{0,n}$, \begin{align*} \lambda_{\max}(S_n) \to (1+\sqrt{\gamma})^2 \end{align*} in probability. The supercritical rank-one outlier conclusion in the same real Wishart [BBP transition for spiked covariance matrices](/page/BBP%20Transition) applies conditionally on $v_n$ because the covariance matrix $I_{p_n}+\theta v_nv_n^\top$ has one population eigenvalue $1+\theta$ in the direction $v_n$ and all remaining population eigenvalues equal to $1$; the spike strength $\theta$ is fixed, the spike rank is one, the aspect ratio satisfies $p_n/n\to\gamma\in(0,\infty)$, and the supercritical hypothesis is $\theta>\sqrt{\gamma}$. Let $O(p_n):=\{Q\in\mathbb{R}^{p_n\times p_n}:Q^\top Q=I_{p_n}\}$ denote the real orthogonal group, let $e_1:=(1,0,\dots,0)\in\mathbb{R}^{p_n}$ denote the first coordinate vector, and choose an orthogonal matrix $Q_n\in O(p_n)$ such that $Q_nv_n=e_1$. The conditional law of the eigenvalues of $S_n$ does not depend on the particular value of $v_n$: under this change of coordinates, $Q_nX_n$ has covariance $I_{p_n}+\theta e_1e_1^\top$ and $S_n(Q_nX_n)=Q_nS_n(X_n)Q_n^\top$, which has the same eigenvalues as $S_n(X_n)$. Hence the conditional convergence holds with the same deterministic limit for every $v_n$, and integrating over the spherical prior transfers the convergence to the mixture law $\mathbb{P}_{1,n}$. Therefore, under $\mathbb{P}_{1,n}$, \begin{align*} \lambda_{\max}(S_n) \to (1+\theta)\left(1+\frac{\gamma}{\theta}\right) \end{align*} in probability. The two deterministic limits are strictly separated. Indeed, \begin{align*} (1+\theta)\left(1+\frac{\gamma}{\theta}\right)-(1+\sqrt{\gamma})^2 = \frac{(\theta-\sqrt{\gamma})^2}{\theta}>0. \end{align*} Choose a threshold $\tau$ satisfying \begin{align*} (1+\sqrt{\gamma})^2 < \tau < (1+\theta)\left(1+\frac{\gamma}{\theta}\right). \end{align*} [/step]

[step:Use the separated eigenvalue limits to build a consistent largest-eigenvalue test] Define the rejection event $B_n\subset\mathcal{X}_n$ by \begin{align*} B_n:=\{X_n\in\mathcal{X}_n:\lambda_{\max}(S_n(X_n))>\tau\}. \end{align*} Let $\mathbb{1}_{B_n}:\mathcal{X}_n\to\{0,1\}$ denote the indicator function of $B_n$, so $\mathbb{1}_{B_n}(X_n)=1$ for $X_n\in B_n$ and $\mathbb{1}_{B_n}(X_n)=0$ for $X_n\notin B_n$. Define the [test function](/page/Test%20Function) $\varphi_n: \mathcal{X}_n \to \{0,1\}$ by \begin{align*} \varphi_n(X_n) := \mathbb{1}_{B_n}(X_n). \end{align*} Here $\varphi_n=1$ means that the test rejects $H_0$ in favor of $H_1$. Since $\lambda_{\max}(S_n) \to (1+\sqrt{\gamma})^2$ in $\mathbb{P}_{0,n}$-probability and $\tau$ is strictly larger than this limit, \begin{align*} \mathbb{P}_{0,n}(\varphi_n=1)=\mathbb{P}_{0,n}(\lambda_{\max}(S_n)>\tau) \to 0. \end{align*} Since $\lambda_{\max}(S_n) \to (1+\theta)(1+\gamma/\theta)$ in $\mathbb{P}_{1,n}$-probability and $\tau$ is strictly smaller than this limit, \begin{align*} \mathbb{P}_{1,n}(\varphi_n=0)=\mathbb{P}_{1,n}(\lambda_{\max}(S_n)\le \tau) \to 0. \end{align*} Therefore the total error of the largest-eigenvalue test satisfies \begin{align*} \mathbb{P}_{0,n}(\varphi_n=1)+\mathbb{P}_{1,n}(\varphi_n=0) \to 0. \end{align*} This proves the supercritical detection assertion and completes the proof. [/step]