Lecture Three - STAT 212a

Lecture Seven - STAT 212a

Lecture Eight - STAT 212a

1 Sudakov Minoration Suppose that (T, ρ) is a finite metric space and let {X t , t ∈ T } be a stochastic process indexed by T satisfying

Lecture Six - STAT 212a

Conversely, for every -packing subset t1, . . . , tn of T , the closed balls B(ti, /2), i = 1, . . . , n are disjoing and hence every /2-cover of T must have one point in each of the balls B(ti, /2). As a result, an /2-cover of T must have at least n points. This implies that M( /2, T ) ≥ N( , T ). Lemma 1.2 (Volumetric Argument). Let T = X denote the ball in R of radius Γ centered at the origi...

Lecture Two - STAT 212a

1 f-divergences Recall from last lecture: Here is the definition of an f-divergence: Let f : (0, ∞) → R be a convex function with f (1) = 0. By virtue of convexity, both the limits f (0) := lim x↓0 f (x) and f (∞) := lim x↑∞ f (x)/x exist, although they may be equal to +∞. For two probability measures P and Q, the f-divergence D f (P ||Q) is defined by D f (P ||Q) := {q > 0} f p q dQ + f (∞)P {...

Lecture Five - STAT 212a

1 Varshamov-Gilbert Lemma Lemma 1.1. Fix k ≥ 1. There exists a subset W of {0, 1} with |W | ≥ exp(k/8) such that the Hamming distance, ∆(τ, τ ′) := ∑k i=1{τi 6= τ ′ i} > k/4 for all τ, τ ′ ∈W with τ 6= τ ′. Proof. Consider a maximal subset W of {0, 1} that satisfies: ∆(τ, τ ′) ≥ k/4 for all τ, τ ′ ∈W with τ 6= τ ′. (1) The meaning of maximal here is that if one tries to expand W by adding one m...