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arxiv: 1608.03643 · v2 · pith:Q7WVHRSNnew · submitted 2016-08-12 · 💻 cs.LG · cs.DS· stat.ML

Chi-squared Amplification: Identifying Hidden Hubs

classification 💻 cs.LG cs.DSstat.ML
keywords sigmahubsentrieshiddenalgorithmdeltamodelsqrt
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We consider the following general hidden hubs model: an $n \times n$ random matrix $A$ with a subset $S$ of $k$ special rows (hubs): entries in rows outside $S$ are generated from the probability distribution $p_0 \sim N(0,\sigma_0^2)$; for each row in $S$, some $k$ of its entries are generated from $p_1 \sim N(0,\sigma_1^2)$, $\sigma_1>\sigma_0$, and the rest of the entries from $p_0$. The problem is to identify the high-degree hubs efficiently. This model includes and significantly generalizes the planted Gaussian Submatrix Model, where the special entries are all in a $k \times k$ submatrix. There are two well-known barriers: if $k\geq c\sqrt{n\ln n}$, just the row sums are sufficient to find $S$ in the general model. For the submatrix problem, this can be improved by a $\sqrt{\ln n}$ factor to $k \ge c\sqrt{n}$ by spectral methods or combinatorial methods. In the variant with $p_0=\pm 1$ (with probability $1/2$ each) and $p_1\equiv 1$, neither barrier has been broken. We give a polynomial-time algorithm to identify all the hidden hubs with high probability for $k \ge n^{0.5-\delta}$ for some $\delta >0$, when $\sigma_1^2>2\sigma_0^2$. The algorithm extends to the setting where planted entries might have different variances each at least as large as $\sigma_1^2$. We also show a nearly matching lower bound: for $\sigma_1^2 \le 2\sigma_0^2$, there is no polynomial-time Statistical Query algorithm for distinguishing between a matrix whose entries are all from $N(0,\sigma_0^2)$ and a matrix with $k=n^{0.5-\delta}$ hidden hubs for any $\delta >0$. The lower bound as well as the algorithm are related to whether the chi-squared distance of the two distributions diverges. At the critical value $\sigma_1^2=2\sigma_0^2$, we show that the general hidden hubs problem can be solved for $k\geq c\sqrt n(\ln n)^{1/4}$, improving on the naive row sum-based method.

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