Let X_1,\ldots,X_n be independent Gaussian tensors in \mathbb{R}^{d_1}\otimes\cdots\otimes\mathbb{R}^{d_k} whose covariance is a Kronecker product of k unknown positive-definite factors, and put D=\prod_{a=1}^k d_a and d_{\max}=\max_a d_a. A recent result of Franks et al. (2026) established condition-number-free nonasymptotic guarantees for the tensor-normal maximum likelihood estimator under the sample threshold nD\gtrsim k^2 d_{\max}^3. They asked whether the cubic dependence on d_{\max} could be replaced by the operator-norm scale d_{\max}^2. We answer this question affirmatively. We prove that, for t\geq 1, the maximum likelihood estimator exists uniquely with high probability whenever nD\geq Ck^2 d_{\max}^2 t^2, and satisfies d_{FR}(\widehatΘ,Θ)\leq Ct\sqrt{k},d_{\max}/\sqrt{n} and d_{FR}(\widehatΘ_a,Θ_a)\leq Ct\sqrt{k d_a},d_{\max}/\sqrt{nD}. For every mode of largest dimension, we also obtain the sharp Thompson bound d_{op}(\widehatΘ_a,Θ_a)\leq Ct,d_{\max}/\sqrt{nD}. No sparsity, condition-number bound or warm start is assumed. For fixed k, the threshold has the information-theoretically optimal dependence on d_{\max}, and the displayed rates for the full precision and the largest factor match Gaussian minimax lower bounds up to a factor \sqrt{k}. The proof extends a random Gram bound for local group-orbit directions to the full local Lie algebra, transports it to a fixed Thompson ball by exact conjugation, and combines sensitivity of a constrained maximum likelihood estimator with an equivariant Kirszbraun extension and Gaussian concentration. This removes the Frobenius-to-operator loss responsible for the previous extra factor d_{\max} and resolves the explicit open problem posed in the earlier work.
Tensor-normal maximum likelihood estimation at the operator-norm sample threshold
Let $X_1,\ldots,X_n$ be independent Gaussian tensors in $\mathbb{R}^{d_1}\otimes\cdots\otimes\mathbb{R}^{d_k}$ whose covariance is a Kronecker product of $k$ unknown positive-definite factors, and put $D=\prod_{a=1}^k d_a$ and $d_{\max}=\max_a d_a$.
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