Repeated observations shrink the noise side of the gap

Suppose all noise observations are independent and centered, with moment-generating bound \(\mathbb E e^{\lambda\eta}\le e^{\sigma^2\lambda^2/2}\). Averaging \(R\) observations at each of \(n+1\) sample sites replaces \(\sigma\) by \(\sigma/\sqrt R\). A sufficient recovery condition is

The probability of simultaneous recovery is then at least \(1-\delta\), subject to the same separated-jump sampling model.

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Take \(n=256\), \(M=1/2\), \(\sigma=1/50\), \(\delta=1/20\), and \(a_{\min}=2/25\). Since \(\log(10240)<10<(16/5)^2\), choosing \(R=16\) makes the right-hand side strictly less than

Thus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a claim of optimal sample complexity. The exact checker certifies \(e^{10}>10240\) using a rational Taylor partial sum, so the displayed logarithm bound is not based on rounded numerical logarithms.

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Neighboring differences share one sample. The proof does not require those differences to be independent: it bounds each tail and applies a union bound across all cells. Independence of the underlying sample noises supplies the stated subGaussian difference proxy. For a fixed separated signal model, a noise scale satisfying \(\sigma_n\sqrt{\log n}\to0\) gives the stated recovery consistency in probability. A fixed positive noise scale with one observation per site does not satisfy that condition.

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