1. From Branching processes to Phylogeny: A history of reproduction
Elchanan Mossel, U.C. Berkeley
9/22/2018

2. Trees in genetics Darwin + Today - tree of species.
Today: tree of fathers.
Today: tree of mothers.
Today: Descendants for a few generations if there are no crossovers (graph has no cycles).
Noah
Shem
Japheth
Ham
Put
Cush
Kannan
Mizraim
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3. Information on the tree
No brain,
Can’t move
Information on the tree
Tree of species: characteristics. Won’t discuss.
Tree of mothers: Mitochondria.
Tree of fathers: Y chromosome.
No crossover: Full DNA.
Stupid
Flies
Stupid
Walks
Stupid
Swims
Too smart
Barely moves
acctga
Noah
Shem
Japheth
Ham
acctaa
acctga
acctga
Put
Cush
Kannan
Mizraim
acctga
9/22/2018
acctga
agctga
acctga

4. Genetic mutation models
How does the D.N.A and other genetic data mutates along each edge of the tree?
Actual mutation is complicated (for Y: “indels, snips, micro satellites, mini satellites”).
A simplified model: Split sequence to subsequences. All subsequences have same mutation rate.
That’s the standard model in theoretical phylogeny.
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5. Formal model – finite state space
Finite set A of information values.
Tree T=(V,E) rooted at r.
Vertex v in V, has information σv in A.
Edge e=(v, u), where v is the parent of u, has a mutation matrix Me of size |A| by|A|:
A sample of the process is (v)v in T.
For each sample, we know v, for v in B(T),
where B(T) is the boundary of the tree; |B(T)| = n.
We are given k independent samples 1B(T),…, kB(T).
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6. Formal model – infinite state space
Infinite set A – assuming no mutations back – “Homoplasy-free models”
Defined on an un-rooted tree T=(V,E).
Edge e has (non-mutation) parameter (e).
Sample: 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.
We are given k independent samples 1B(T),…, kB(T).
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7. Reconstructing the topology
Given 1B(T),…, kB(T), want to reconstruct the topology, i.e., the structure of the underlying tree on the set of labeled leaves.
Formally, we want to find for all u and v in B(T) their graph-metric distance d(u, v).
Assume all internal degrees are at least 3.
Assume B(T) consists only of leaves – vertices of degree 1. u u
Me’ v
Me’ Me’’
Me’’ w w
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8. Summary: 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
Homoplasy free
Ising model
CFN
critical : 22 = 1
Sub-critical
random cluster
High mutation
Problems: How general?
What is the critical point? (extremality vs. spectral)
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9. Homoplasy free models and percolation
Th1[M2002]: Suppose that n=3*2q and
T is a uniformly chosen (q+1)-level 3-regular tree.
For all e, (e)< , and < 1/2.
Then in order to reconstruct the topology with probability  = 0.1, at least k = (–q) = nΩ(-log ) samples are needed.
Th2[M2002]: Suppose that T has n leaves and
For all e, ½ +  < (e)< 1 - .
Then k = O(log n – log ) samples suffice to reconstruct the topology with probability 1-.
9/22/2018

10. Cavendar-Farris-Neyman model:
The CFN model
Cavendar-Farris-Neyman model:
2 data types: 1 and –1.
Mutation along edge e: with probability (e) copy data from parent. Otherwise, choose 1/-1 with probability ½ independently of everything else.
Thm[CFN] Suppose that for all e, 1 -  > (e) >  > 0.
Then given k samples of the process at the n leaves,
It is possible to reconstruct the underlying topology with probability 1 - , if k = nΩ(-log ).
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11. Phase transition for the CFN model
Th1[M2002]: Suppose that n=3*2q and
T is a uniformly chosen (q+1)-level 3-regular tree.
For all e, (e)< , and 22 < 1.
Then in order to reconstruct the topology with probability  > 0.1, at least k = (2-q-2q) = nΩ(-log ) samples are needed.
Th2[M2002]: Suppose that T is a balanced tree on n leaves and
For all e, min < (e)< max and 22min > 1, max < 1.
Then k = O(log n – log ) samples suffice to reconstruct the topology with probability 1-.
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12. Metric spaces on trees Let D be a positive function on the edges E.
Define
Claim: Given D(v,u) for all v and u in B(T), it is possible to reconstruct the topology of T.
Proof: Suffices to find d(u, v) for all u, v B(T), where d is the graph metric distance.
d(u1,u2) = 2 iff for all w1 and w2 it holds that
D’(u1,u2,w1,w2) = D(u1,w1)+D(u2,w2) –D(u1,u2)–D(w1,w2) is non-negative (“Four point condition”). w1 u1 w1 u1 w2 u2 w2 u2
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13. Metric spaces on trees
Similarly, d(u1,u2) = 4 iff for all w1 and w2 s.t. for i=1,2 and j=1,2, d(ui,wj) = 4, it holds that
D’(u1,u2,w1,w2) = 0.
Remark 1: If D’(u1,u2,w1,w2) > 0, then
| D’(u1,u2,w1,w2) |  2 min_e D(e). Therefore, it suffices to know D with accuracy min_e D(e)/4.
Remark 2: For a balanced tree, in order to reconstruct the underlying topology up to l levels from the boundary, it suffices to know D(u,v) for all u and v s.t. d(u,v)  2 l + 2 (with accuracy min_e D(e)/4).
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14. Proof of CFN theorem Define D(e) = - log (e).
D(u,v) = -log(Cov(v,u)), where Cov(v, u) = E[vu].
Estimate Cov(v, u) by Cor(v, u) where
Need D with accuracy m = min D(e)/4 = c, or
Cor = (1  c)Cov.
Cor(v, u) is a sum of k i.i.d.  1 variables with expected value Cov(v, u).
Cov(v, u) may be a small as  2 depth(T) = n-O(-log ).
Given k = nΩ(-log ) samples, it is possible to estimate D and therefore reconstruct T with high probability.
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15. Extensions
Steel 94: Trick to extend to general Me provided that that det(Me) [-1,-1+]  [- , ]  [1 - , 1],
using D(e) = -log( |det(Me)| ) (more or less).
ESSW: If tree has small depth then k is smaller.
For Balanced trees: In order to reconstruct up to l levels from the boundary, suffices to have k = Θ(log(n)).
Proof: Cov(v, u). for u and v with d(u,v)  2 l + 2 is at
least 2l + 2.
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16. What’s next? It’s important to minimize the number of samples.
Do we need k = nΩ(1), or
do we need k = polylog(n)?
Since the number of trees on n leaves is exponential in n log n, and each sample consists of n bits, we need at least Ω(log(n)) samples.
9/22/2018

17. The Ising model on the binary tree
The (Free)-Ising-Gibbs measure on the binary tree:
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.
Different Perspective: Topology is known and looking at a single sample. + + + + - + + - + - + - + + +
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18. The Ising model on the binary tree
Studied in statistical physics [Spitzer 75, Higuchi 77, Bleher-Ruiz-Zagrebnov 95, Evans-Kenyon-Peres-Schulman 2000, M 98]
Interesting phenomena: double phase transition (different from Ising model in Zd).
When 2   1, unique Gibbs measure.
When 2  2  1, free measure is extremal.
In other words,
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19. The Ising model on the binary tree
From BRZ or EKPS:
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
Remark: 2  2 = 1 phase transition also transition for mixing time of Glauber dynamics for Ising model on tree (with Kenyon and Peres)
9/22/2018

20. Lower bound on number of samples
Th1[M2002]: Suppose that n=3*2q and
T is a uniformly chosen (q+1)-level 3-regular tree.
For all e, (e)< , and 22 < 1.
Then in order to reconstruct the topology with probability  > 0.1, at least k = (2-q-2q) = nΩ(-log ) samples are needed.
Proof of lower bound uses information inequalities.
Lemma 1[EKPS]: For the single sample process on the binary tree of q levels, I(σr,σ∂)  (2  2) q.
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21. Lower bound on number of samples
Lemma 2[Fano’s inequality]: X  Y
Want to reconstruct the random variable X given the random variable Y,
X takes m values.
Let pe be the error probability. Then
1 + pe log2(m) > H(X|Y).
Lemma 3[“Data Processing Lemma”]: X  Y  Z
If X and Z are independent given Y, then
I(X, Z)  min { I(X, Y) , I(Y, Z) }.
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22. Lower bound on number of samples
Assume that topology of bottom q - l levels is known, i.e., we know d(u, v) if d(u, v)  2(q – l).
Then the the conditional distribution of the topology T of the first l levels is uniform.
Let σlt for t=1,…,k be the data at level l for sample t. Recall that we are given the Data = (σ∂t).
By “Data Processing”, l X ? ? Y
* k
q - l
Known
Known Z
* k
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23. Lower bound on number of samples
By independence,
After some manipulations and by EKPS Lemma So
Let Since T
is chosen uniformly,
Now
By Fano’s Lemma, the probability of error in reconstructing the topology, pe, satisfies
In order to get pe < 1 - , need
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24. Upper bound on number of samples
Th2[M2002]: Suppose that T is a balanced tree on n leaves and
For all e, min < (e)< max and 22min > 1, max < 1.
Then k = O(log n – log ) samples suffice to reconstruct the topology with probability 1-.
Proves a conjecture of M. Steel.
9/22/2018

25. Algorithmic aspects of phase transition
Higuchi [77]: For Ising model, when 2  2 > 1, and all q level binary trees E[σr Maj(σ∂)] >  > 0, where Maj is the majority function ( is independent of q).
Looks good for phylogeny because can apply Maj even when do not know the topology.
But, doesn’t work when  is non-constant.
All edges on blue area have  1
All edges on black area have  2
 1 <  2 is close to 1.
Maj(σ∂) is very close to Maj of black tree.
Maj of black tree very close to σv .
σv and σr are weakly correlated. r v
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26. Algorithmic aspects of phase transition
Mossel [98]: For Ising model, when 2  2 > 1, and all q level binary trees E[σr Rec-Majl(σ∂)] >  > 0, where Rec-Majl is the recursive majority function of l levels ( is independent of q; l depends on ).
Rec-Majl for l=1
Looks bad for phylogeny, as we need to know the tree topology.
But main lemma of Mossel[98] is extendable to non-constant .
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27. Main Lemma [M2001]
Lemma: Suppose that 2  min2 > 1, then there exists an l, and  > 0 such that the CFN model on the binary tree of l levels with
(e)   min, for all e not adjacent to ∂T.
(e)    min , for all e adjacent to ∂T.
satisfies E[σr Maj(σ∂)]  .
Roughly, given data of “quality  ”, we can reconstruct the root with “quality  ”.
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28. Reconstructing the topology [M2001]
The algorithm: Repeat the following:
Reconstruct the topology up to l levels from the boundary using 4-points method.
For each sample, reconstruct the data l levels from the boundary using majority algorithm.
Recall: reconstruction near the boundary take O(log n) samples.
By main lemma quality stays above .
Remark: The same algorithm gives (almost) tight upper bounds also when 2  min2 < 1.
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29. Remarks and open problems
For CFN model, algorithm is very nice:
Polynomial time.
Adaptive (don’t need to know  min and  max in advance).
Nearly optimal.
Main problem: extending main lemma to non-balanced trees and other mutation models (reconstructing local topology still works).
Secondary problem: extending lower bounds to other models.
9/22/2018

30. Proving main Lemma
Need to estimate E[σr Maj(σ∂)]. Estimate has two parts:
Case 1: For all e adjacent to ∂T, (e) is small. Here we use a perturbation argument, i.e. estimate partial derivatives of E[σr Maj(σ∂)] with respect to various variables (using something like Russo formula).
Case 2: Some e adjacent to ∂T has large (e). Use percolation theory arguments.
Both cases uses isoperimetric estimates for the discrete cube.
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