1. Reconstruction on trees and Phylogeny 1
Elchanan Mossel,
U.C. Berkeley
Supported by Microsoft Research and the Miller Institute
9/18/2018

2. General plan Study stochastic process on bounded degree trees.
Vertices of tree T are labeled by random variables.
Interested in asymptotic problems where |T| ! 1. + + + + - + + + - - + - + +
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3. The reconstruction problem
We discuss two related problems.
In both, want to reconstruct/estimate unknown parameters from observations.
The first is the “reconstruction problem”.
Here we are given the tree and
the values of the random variables at the leaves.
Want to reconstruct the value of the random variable at a specific vertex (“root”).
Algorithmically “easy” – but when does it “work”? ??
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4. Phylogeny Here the tree is unknown.
Given a sequence of collections of random variables at the leaves (“species”).
Collections are i.i.d.!
Want to reconstruct the tree (un-rooted).
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5. Phylogeny
Algorithmically “hard”.
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6. Lecture plan Talk 1 [GW 19th cent.; M-Steel,2003]
Introduction.
The “random cluster” model – reconstruction.
The random cluster model – phylogeny.
Talk 2 [Hi77, EKPS2000, M1998,MSW2003]
The Ising = CFN model – reconstruction.
Talk 3 [M2003]
The Ising = CFN model – phylogeny.
Talk 4 [M2003, ¸ 2004]
General Markov model.
Open problems etc. + + + + - + + + - - + - + +
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7. Trees (3-)regular trees.
Binary -- All internal degrees are 3 (bifurcating speciation; results valid if degrees are ¸ 3, or ¸ b+1).
General trees. + + + + - + + + - - + - + +
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8. Trees In biology, all internal degrees ¸ 3.
Given a set of species (labeled vertices) X, an X-tree is a tree which has X as the set of leaves.
Two X-trees T1 and T2 are identical if there’s a graph isomorphism between T1 and T2 that is the identity map on X. u u
Me’ v
Me’ Me’’
Me’’ w w d a c b d a b c c a b d
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9. The “random cluster” model
Infinite set A of colors.
“real life” – large |A|; e.g. gene order.
Defined on an un-rooted tree T=(V,E).
Edge e has (non-mutation) probability (e).
Character: Perform percolation – edge e open with probability (e).
All the vertices v in the same open-cluster have the same color v. Different clusters get different colors. This is the “random cluster” model.
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10. Galton-Watson
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11. Galton-Watson Theorem
For the random cluster model on a rooted binary tree.
If (e) > ½ +  for all e, then
for all v 2 T,
with probability at least s() = 2  / (½ + )2,
there exists u 2 T (below v), with (v) = (u).
If (e) < ½ -  for all e, then
the probability that such u 2 T exists is
at most 3 (1 – 2 )d(v, T)
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12. Reconstruction on random clusters
For the random cluster model on a rooted binary tree.
If (e) > ½ +  for all e, then for all v 2 T, we may reconstruct (v) with probability ¸ (½ + )2s2(e) from T (below v).
Proof: v
If (e) < ½ -  for all e, then the probability of reconstructing (v) is · 3 (1 – 2 )d(v, T).
Proof: True even given more info (open/closed edges).
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13. Phylogeny from log characters for R.C.
Th1[M-Steel,2003]: Suppose that T is an X-tree on n leaves and for all e, ½ +  < (e)< 1 - .
Then k = (2 log n – log )/165 = O(log n - log ) characters suffice to reconstruct the topology with probability ¸ 1-.
Colors of leaves
Definition: A cherry is a pair of leaves at distance 2.
Fact: Every X-tree has at least one cherry.
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14. Testing cherries
If x,y is a cherry then there exist no characters  and leaves x’,y’ 2 T - {x,y} s.t. (x) = (x’)  (y) = (y’). x’ x y’ y
If x,y is a not a cherry then for each character ,
the probability that 9 x’,y’ 2 T - {x,y} s.t (x) = (x’)  (y) = (y’) is at least
r =  s2 /16, where s() = 2  / (½+)2. x
Repeating for k characters, we may find all cherries with error probability bounded by
n2 (1-r)k . y’ x’ y
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15. From cherries to trees
We wish to continue by replacing each cherry (u,v) by replacing the vertex w at distance 1 from v and u.
Problem: We may not know what the color of w is.
But: for each character , with probability at least (½ + )2s2(e) we can reconstruct (w) .
Now we can repeat. u x w v y
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16. Poly. lower bound for R.C. Phylogeny
Th1[M-Steel,2003]: Suppose that n=3 £ 2q and
T is a uniformly chosen (q+1)-level 3-regular X-tree.
For all e, (e)< , and < 1/2.
Then in order to reconstruct the tree with probability ¸ 0.1, the number of characters must satisfy
k ¸ (2)–q+1/100 = (n-log2()+1).
Proof: Suffices to prove the same bound given the topology of the bottom q-L levels and the status of the edges there.
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17. Poly. lower bound for R.C. Phylogeny
Proof:
X=T ? ? L
* k
Known
Known
q-L
* k
If for all k characters random cluster “dies” in bottom q-L levels, then X is independent of the data.
This happens with probability ¸ 1 –k 2L (2 )q-L.
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18. Phylogeny: Conjectures and results
Statistical physics
Phylogeny
Binary tree in ordered phase
conj
k = O(log n)
Binary tree unordered
conj
k = poly(n)
Percolation
critical  = 1/2
Random Cluster
M-Steel2003
CFN
M-2003
Ising model
critical : 22 = 1
Sub-critical
representation
High mutation
M-2003
Problems: How general?
What is the critical point? (extremality vs. spectral)
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