Erratum to “Lower bounds for trace reconstruction”

Abstract

Lemma 3.1 asserts that Eyn[Z(̃yn)] − Exn[Z(̃xn)] = (n−1/2) and Eyn[Z(̃yn)] > Exn[Z(̃xn)] for all sufficiently large n. Our proof was not correct: As Benjamin Gunby and Xiaoyu He pointed out to us, we missed four terms in the computation of equation (3.3). Those terms contribute a negative amount, so the proof is more delicate. Here is a correct proof. The intuition behind the result is that a string with a defect of the type we consider, namely, a 10 in a string of 01’s, is likely to cause more 11’s in the trace than a string without the defect. Since the defect in yn is shifted to the right as compared to the defect in xn, the defect of yn is slightly more likely to “fall into” the test window { 2np + 1 , . . . , 2np + √npq } of the trace than is the defect of xn. More precisely, the difference in probability is of order n−1/2. In the proof below, we make this intuition rigorous. PROOF. We assume throughout the proof that k ∈ { 2np + 1 , . . . , 2np + √npq }. Let E(m,k) denote the event that bit m in the input string is copied to position k in the trace. First observe that