1. Reconstruction on trees and Phylogeny 2
Elchanan Mossel,
U.C. Berkeley
Supported by Microsoft Research and the Miller Institute
11/17/2018

2. Reconstruction on Ising-CFN model
We study the reconstruction problem for the Ising-CFN model on regular trees. + + + + - + + - + - + - + +
11/17/2018

3. Markov models on trees Finite set A of information values.
Tree T=(V,E) rooted at r.
Vertex v 2 V, has information σv 2 A.
Edge e=(v, u), where v is the parent of u, has a mutation matrix Me of size |A| £ |A|:
Mi,j (v,u) = P[u = j | v = i]
For each character , we are given T = (v)v 2 T,
where T is the boundary of the tree.
We will focus on the Ising-CFN model:
11/17/2018

4. Statistical physics
Statistical physics is a sub-field of mathematical physics where we study complex systems with simple microscopic interactions.
The Ising model on a graph is a probability measure (“Gibbs distribution”) on the space of configurations σ from vertices to {-1,1} such that
P[σ] ~ exp(Σ(v, w) ε E σ(v)σ(w)/T).
Traditionally studied on cubes in Zd.
The Ising model on
200 x 200 grid
11/17/2018

5. Statistical physics on trees
The Ising model on the binary tree can be defined:
Set σr, the root spin, to be +/- with probability ½.
For all pairs of (parent, child) = (v, w), set σw = σv, with probability , otherwise σw = +/- with probability ½.
This is exactly the CFN model.
Studied in statistical physics [Spitzer 75, Higuchi 77, Bleher-Ruiz-Zagrebnov 95, Evans-Kenyon-Peres-Schulman 2000, Ioffe 99, M 98, Haggstrom-M 2000, Kenyon-M-Peres 2001, Martinelli-Sinclair Weitz 2003, Martine 2003] + + + + - + +
11/17/2018 - + - + - + +

6. Reconstruction solvability
Let T be an infinite rooted tree and Tn denote the first n levels of T.
We say that the reconstruction problem is solvable if one of the following equivalent conditions hold:
9  s.t. (8 non-degenerate ) limn ! 1 I(X0,Xn) > 0, where I(X0,Xn) = H(X0) + H(Xn) – H(X0,Xn); H is the entropy operator, H(X) = -x P[X = x] log2 P[X = x].
9 i,j s.t. limn ! 1 | Pni - Pnj | > 0, where Pnj denotes the distribution of Xn conditional on X0 = j.
If X0 has the uniform distribution then, liminfn ! 1 n > 1/m, where n is the probability of correct reconstruction of X0 given Xn.
9  (8 non-degenerate ) liminfn ! 1 Var[E[X0|Xn]] > 0.
11/17/2018

7. The Ising model on the 3-regular tree
mutual information:
H(σ∂) + H(σr)) - H(σr,σ∂)
Temp 
σr | σ∂≡ 1
Uniq
I(σr,σ∂)
Free measure
high
< 1/2
unbiased V
→ 0
extremal
med.
(1/2,1/√2)
biased X
low
> 1/√2
Inf > 0
Non-ext
11/17/2018

8. Reconstruction for the CFN model
Thm: The reconstruction problem for the Ising model on the (b+1)-regular tree is solvable if and only if b 2 > 1.
“Easy direction” [Higuchi 77]: prove that a certain reconstruction algorithm works when b 2 > 1.
Higuchi argument extends to general chains and general trees.
Will also show an argument from [M98] useful for phylogeny.
“Hard direction” [¸ 95]: Non-reconstruction?
6 different proofs!
All involve a magic.
None extends to other markov models.
Will follow a coupling proof [Martinelli-Siclair-Weitz]
11/17/2018

9. Non-reconstruction - Coupling down
Copying rule. For i =+,-:
P[i ! i] = .
P[i ! Uniform] = 1 – .
Continuing down the tree, non-coupled elements form a branching process with parameter .
+ / -
+ / - = =
+ / - = = = = = = = = = =
If b  · 1, branching process dies ) coupling.
More generally, at level n, the expected number of uncoupled sites is bnn.
(Doesn’t work all the way to b 2 · 1).
11/17/2018

10. Non-reconstruction - Coupling up
We try to couple two configurations which differ at level n so that they agree at the root.
First consider the case where they differ at exactly one site. = =
+ / - u v = = = = = = =
+ / -
Lemma [Mossel-Kenyon-Peres]: Among all boundary conditions , E [u = 1 | v = 1] – E[u = -1 | v = 1] is maximized for the free boundary.
) P[not coupling at u] · .
) P[not coupling at the root] · n.
11/17/2018

11. Coupling up – path coupling
We got that if  and  are two boundary conditions which differ in one position at level n, then
|E[()] – E[()] · 2 n, where  is the root.
) if  and  are two boundary conditions which differ at k sites, then
|E[()] – E[()] · 2 k n.
Pf: If  and  differ at k sites, then we can find a sequence  = (0),(1),…,(k) = , such that i and i+1 differ in exactly one site.
|E[()] – E[()] ·
i=1k |E(i)[()] – E(i-1)[()]| · 2 k n.
11/17/2018

12. Non reconstruction for b 2 < 1
Fix  such that b 2 < 1.
We will show that E+[E[() | +]] – E-[E[() | -] ! 0,
where + = boundary conditions conditioned on () = +.
Let (+,-) be given by the “down coupling”.
Let K(+,-) = number of disagreements between +,-.
E+[E[() | +]] – E- E[() | -]]
= E_{+,-}[E[() | +] - E[() | -]]
· E+,-[2 K(+,-) n] = 2 n E+,-[K(+,-)] (“up coupling”).
= 2 n £ bn n (“down coupling”)
= 2 (b 2)n ! 0 exp. fast in n.
11/17/2018

13. Where we stopped …
Thm: The reconstruction problem for the Ising model on the (b+1)-regular tree is solvable if and only if b 2 > 1.
We showed that if b 2 < 1, it is impossible to reconstruct (“hard” direction).
We now show that if b 2 > 1, we can reconstruct.
11/17/2018

14. Reconstruction via majority
Fix  such that b 2 > 1.
Let X = Xn = #(+) - #(-) at level n.
We claim that Xn is a good estimator of ().
E+[Xn] = bn n ; E-[Xn] = -bn n.
We show that E+/-[Xn2] · c(E+/-[Xn])2 = c b2n 2n.
Let f = fn (g = gn) be the density of the + (-) measure with respect to some reference measure .
2 bn n = E+[X] – E-[X] = s X (f – g) d  =
= s X (f1/2 – g1/2) (f1/2 + g1/2) d 
· (s X2 (f1/2 + g1/2)2 d)1/2 £ (s (f1/2 – g1/2)2 d )1/2
· (4 s X2 f d+ 4 s X2 g d)1/2 £ (s |f – g| d)1/2
· (8 c b2n 2n)1/2 (DTV(+,-))1/2.
11/17/2018

15. Bounds on the second moment
Write Xn = v (v), where the sum is over all v in level n.
E+[Xn2] = v,w E+[(v) (w)].
For each edge with prob.  the two end points are the same and with prob. 1- the two points are independent.
If there is a red edge on the path between v and w, then E+[(v) (w)] = 0. v w v w
Otherwise, (v) = (w).
E+[(v) (w)] = d(v,w).
E+[Xn2] = bn(1 + i=1n (bi – bi-1)2i)
= bn(1 + (b-1) 2 i=0n-1 bi 2i).
= O(b2n 2n) iff b 2 > 1. v 1 2 4
11/17/2018

16. Remarks on the second moment
Kamea/ Higuchi argument is very robust.
Works for general trees when br(T) 2 > 1.
Works for general markov chains, where  = 2nd eigenvalue of M (M-Peres 2002).
Kesten-Stigum (1966!) proved that for all markov chains
if b 2 > 1, then the limiting law of the count depends on the root.
If b 2 < 1, then the limiting law is normal for all root values.
M-Peres (2002) count reconstruction is impossible if b 2 < 1.
11/17/2018

17. Recursive reconstruction for Ising models
An alternative proof for reconstruction for b 2 > 1 [M98]
Advantage: Works also when we have lower bound on . Majority doesn’t.
Blue edges have 1 , black 2, 1 < 2 ~ 1.
Maj(σ∂) ~ Maj of black tree.
Maj of black tree ~ σv .
σv and σ have exp. small correlation.
Phylogeny: reconstruction given bounds. v
Instead we will use recursive-majority.
11/17/2018