A Upper Bound

That is, every remaining interval ("in-between" two heavy elements's (which must then all be consecutive, and at least 7 are "heavy", and the rightmost one is either heavy or light), we are done, since then's, as otherwise those are consecutive, and thus Proof of Lemma 2. Reduction from Lemma 7. 15 A.2 Proof of Interval Mass EstimateAlgorithm 2 Interval-Mass-Estimate Require: m i.i.d. Equation (3). it holds |p( I) ˆ φ(I)| null p(I)/b 1 2 p( I), This then implies that p( I) ˆ φ( I) 2 . Condition on that, we easily have, for any sub-interval J I, p(J) ˆ p(J) = p(I) ˆ p(I) max(2, 8 n/b) since both p and ˆ p are uniform within I . The first point is easy to see since otherwise Line 7 would reject. Lemma 3, will be useful in both aspects of the analysis.