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Assouad's lemma turns many neighbouring testing problems into a lower bound for minimax risk. One direct form estimates an unknown bit vector $\omega \in \{0,1{\}}^k$ from data with law $P_{\omega }$, using Hamming distance as loss. If every neighbouring pair satisfies $\mathrm{TV}(P_{\omega },P_{{\omega }^{\prime }})\le \eta$, where $\omega ,{\omega }^{\prime }$ differ in one bit, thenTo prove it, average the risk over all vertices. For a fixed coordinate, pair vertices differing only in that bit. The sum of the two bit-error probabilities of any hypothesis test is at least $1-\mathrm{TV}$, so the averaged error in that coordinate is at least $(1-\eta )/2$. Sum over coordinates and use that a maximum dominates an average.

An Assouad hypercube embeds this construction in a larger estimation problem. The embedding must also relate the original loss to bit errors; it is that geometric separation which transfers the testing difficulty into a numerical estimation lower bound.

The total variation distance and Kullback-Leibler divergence arewith the divergence $+\infty$ when $P$ is not absolutely continuous with respect to $Q$.

For $P=\mathrm{Laplace}(0)$ and $Q=\mathrm{Laplace}(\mu )$,If $X\sim P$, direct integration on the intervals cut by $0$ and $\mu$ givesConsequently

One squared-loss form of Assouad's lemma is as follows. Suppose $(P_{\omega },{\theta }_{\omega }{)}_{\omega \in \{0,1{\}}^k}$ is an Assouad hypercube such thatand every pair of neighboring vertices satisfies $\mathrm{TV}(P_{\omega },P_{{\omega }^{\prime }})\le \eta$. ThenTo prove it, decode from $\hat{\theta }$ a nearest hypercube vertex. The separation condition converts estimation loss into a constant multiple of its Hamming error. Average this error under the uniform prior on $\omega$ and sum coordinatewise. For each coordinate, the two equally weighted mixtures with that bit equal to zero or one have total variation at most $\eta$ by convexity. The minimum average error of any binary test is $(1-\mathrm{TV})/2$, which gives the displayed bound after the nearest-vertex factor.

We now build such a hypercube inside the monotone cone. Let $k=\lfloor c_0{n}^{1/3}\rfloor$ and $m=\lfloor n/k\rfloor$. Divide the first $km$ coordinates into $k$ consecutive blocks. For $\omega \in \{0,1{\}}^k$, set on block $j$and extend the remaining coordinates at the last level. If $c_0,c_1$ are sufficiently small universal constants, every signal is nondecreasing and belongs to $[0,1{]}^n$.

Neighboring vertices differ by $\delta$ on one block, so their squared Euclidean separation is $m{\delta }^2$. Their product experiments differ only on that block and haveBy Pinsker's inequality, choosing $c_1$ small makes every neighboring total variation distance at most, say, $1/4$.

Apply Assouad's lemma with normalized squared loss $d^2(\theta ,{\theta }^{\prime })=n^{-1}\Vert \theta -{\theta }^{\prime }{\Vert }^2$. Here $a=m{\delta }^2/n=c_1^2/n$, and henceThis proves the lower half of the minimax rate for isotonic sequence estimation.