Notes 12 : Kesten-Stigum bound

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1 Notes : Kesten-Stigum bound MATH Fall 0 Lecturer: Sebastien Roc References: [EKPS00, Mos0, MP03, BCMR06]. Kesten-Stigum bound Te previous teorem was proved by sowing tat majority is a good root estimator up to p = p. Here we sow tat tis result is best possible. Of course, majority is not te best root estimator: in general its error probability can be iger tan maximum likeliood. (See Figure 3 in [EKPS00] for an insigtful example were majority and maximum likeliood differ.) However, it turns out tat te critical tresold for majority, called te Kesten-Stigum bound, coincides wit te critical tresold of maximum likeliood in te CFN model. Note tat te latter is not true for more general models [Mos0]. THM. (Tigtness of Kesten-Stigum Bound) Let θ = p = /. Ten wen p p te ancestral reconstrution problem is not solvable. Along eac pat from te root, information is lost troug mutation at exponential rate maeasured by θ = p. Meanwile, te tree is growing exponentially and information is duplicated measured by te brancing ratio b =. Tese two forces balance eac oter out wen bθ =, te critical tresold in te teorem. To prove Teorem. we analyze te maximum likeliood estimator. Let µ (s 0 s ) be te conditional probability of te root state s 0 given te states s at level. It will be more convenient to work wit te following related quantity Z = µ (+ σ ) µ ( σ ) = µ (σ ) [µ+ (σ ) µ (σ )] = µ (+ σ ), wic, as a function of σ, is a random variable. Note tat E[Z ] = 0 by symmetry. It is enoug to prove a bound on te variance of Z. LEM. It olds tat µ + µ E[Z ].

2 Notes : Kesten-Stigum bound Proof: By Bayes rule and Caucy-Scwarz µ + (s ) µ (s ) = µ (s ) µ (+ s ) µ ( s ) s s = E Z E[Z ]. Let z = E[Z ]. Te proof of Teorem. will follow from lim z = 0. We apply te same type of recursive argument we used for te analysis of majority: we condition on te root to exploit conditional independence; we apply te Markov cannel on te top edge. Distributional recursion We first derive a recursion for Z. Let σ be te states at level below te first cild of te root and let µ be te distribution of σ. Define Ż = µ (+ σ ) µ ( σ ), were µ (s 0 ṡ ) is te conditional probability tat te root is s 0 given tat σ = ṡ. Similarly, denote wit a double dot te same quantities wit respect to te subtree below te second cild of te root. LEM.3 It olds pointwise tat Z = Ż + Z + Ż Z.

3 Notes : Kesten-Stigum bound 3 Proof: Using µ + (s ) = µ + (ṡ ) µ + ( s ), note tat Z = = = γ µγ (σ ) µ (σ ) µ ( σ ) µ ( σ ) µ (σ ) µ ( σ ) µ ( σ ) µ (σ ) = µ ( σ ) µ ( σ ) (Ż + µ (σ ) Z ), γ µγ ( σ ) µ γ ( σ ) µ ( σ ) µ ( σ ) ( ) γ + γż ( + γ Z ) were µ (σ ) µ ( σ ) µ ( σ ) = µ γ (σ ) µ ( σ ) µ ( σ ) = µ γ ( σ ) µ γ ( σ ) µ ( σ ) µ ( σ ) = ( ) + γż ( + γ Z ) = + Ż Z. Define Ż = µ (+ σ ) µ ( σ ), were µ (s 0 σ ) is te condition probability tat te first cild of te root is s 0 given tat te states at level below te first cild are σ. Similarly, LEM.4 It olds pointwise tat Ż = θż. Proof: Te proof is similar to tat of te previous lemma and is left as an exercise.

4 Notes : Kesten-Stigum bound 4 3 Moment recursion We now take expectations in te previous recursion for Z. Note tat we need to compute te second moment. However, an important simplification arises from te following observation: E + [Z ] = s µ + (s )Z (s ) = s µ (s ) µ+ (s ) µ (s ) Z (s ) = s µ (s )( + Z (s ))Z (s ) = E[( + Z )Z ] = E[Z ], so it suffices to compute te (conditioned) first moment. Proof:(of Teorem.) Using te expansion we ave tat + r = r + r + r, Z = θ(ż + Z ) θ 3 (Ż + Z )Ż Z + θ 4 Ż Z Z θ(ż + Z ) θ 3 (Ż + Z )Ż Z + θ 4 Ż Z, () were we used Z. To take expectations, we need te following lemma. LEM.5 We ave E + [Ż ] = θe + [Ż ], and E + [Ż ] = ( θ)e[ż ] + θe + [Ż ] = E[Ż ] = E + [Ż ]. Proof: For te first equality, note tat by symmetry E + [Ż ] = ( p)e + [Ż ] + pe [Ż ] = ( p)e + [Ż ].

5 Notes : Kesten-Stigum bound 5 Te second equality is proved similarly and is left as an exercise. Taking expectations in (), using conditional independence and symmetry z θ z θ 4 z + θ 4 z = θ z θ 4 z. References [BCMR06] Cristian Borgs, Jennifer T. Cayes, Elcanan Mossel, and Sébastien Roc. Te Kesten-Stigum reconstruction bound is tigt for rougly symmetric binary cannels. In FOCS, pages , 006. [EKPS00] [Mos0] W. S. Evans, C. Kenyon, Y. Peres, and L. J. Sculman. Broadcasting on trees and te Ising model. Ann. Appl. Probab., 0():40 433, 000. E. Mossel. Reconstruction on trees: beating te second eigenvalue. Ann. Appl. Probab., ():85 300, 00. [MP03] E. Mossel and Y. Peres. Information flow on trees. Ann. Appl. Probab., 3(3):87 844, 003.