Discrete transport problems and the concavity of entropy

Transcription

1 Discrete transport problems and the concavity of entropy Bristol Probability and Statistics Seminar, March 2014 Funded by EPSRC Information Geometry of Graphs EP/I009450/1 Paper arxiv:

2 Motivating Problem I Suppose we have a pile of soil we need to move somewhere (say along R, or along Z). Each spadeful moved from point x to point y costs us something. Fix cost function e.g. c(t) = t p for p 1. Moving one spadeful from x to y costs c(y x) = y x p. Can we find a moving strategy that minimises the total cost...?

3 ... Yes We Can! Source and destination piles need to have same size. Suppose piles have the same shape. Intuitive solution: just translate (move everything same distance).

4 More general case What if piles not the same shape? For many cost functions c, intuitive non-crossing principle.

5 Non-crossing principle w x y z Suppose spadefuls of soil at w and x to move to y and z. Take cost c(t) = t 2 for simplicity. Strategy 1: w z, x y. Cost c 1 = (z w) 2 + (y x) 2. Strategy 2: w y, x z. Cost c 2 = (y w) 2 + (z x) 2. c 1 c 2 = 2(x w)(z y) 0. Prefer Strategy 2: not to let soil cross over. Similar argument for any convex cost function.

6 Transport of probability measures Can rephrase problem more mathematically. Transport probability density function (or mass function) f 0 to f 1. Equivalently think in terms of distribution functions F 0 and F 1. Write Γ(F 0, F 1 ) for the set of joint probability distributions with marginals F 0 and F 1 (couplings). Joint density f (x, y) codes the amount of mass to be moved from x to y for particular strategy.

7 Transport of probability measures on {0, 1} Example Consider marginals f 0 = (3/4, 1/4) and f 1 = (1/4, 3/4). Could define f (x, y) as follows: f (x, y) y = 0 y = 1 x = 0 1/4 1/2 f 0 (0) = 3/4 x = 1 0 1/4 f 0 (1) = 1/4 f 1 (0) = 1/4 f 1 (1) = 3/4 Cost of strategy f is f (x, y)c(y x) = x,y x,y f (x, y) y x p.

8 Distance between probability measures This gives us a way to measure how similar F 0 and F 1 are measure cost to move one distribution to the other under optimal strategy.

9 Distance between probability measures Definition Given F 0 and F 1 and cost function c(t) = t p, write ( W p (F 0, F 1 ) = inf F Γ(F 0,F 1 ) y x p df (x, y)) 1/p. Using non-crossing principle, optimal strategy gives ( 1 1/p W p (F 0, F 1 ) = F0 1 (t) F1 1 (t) dt) p. 0 This is the Wasserstein distance... (or Vasershtein)... (or earth mover s)... (or Mallows)... (or Kantorovich)... (or Kantorovich-Rubinstein)... (or Monge-Kantorovich)... (or Tanaka)... (or transport)... (or transportation)... SEO hell!

10 Motivating problem II Suppose want to run from x = 0 to x = D in T units of time. Suppose to maintain a speed of v costs us v 2 in energy. What is correct speed to run to minimise total energy use? Represent trajectory in terms of a function x(t), with x(0) = 0 and x(t ) = D. Wish to minimise T 0 x (t) 2 dt.

11 Constant speed paths Obvious strategy: x(t) = td/t. Gives T 0 x (t) 2 dt = T (D/T ) 2 = D 2 /T. Now by Cauchy-Schwarz: ( T 0 ) ( T ) 1dt x (t) 2 dt 0 ( T 0 ) 2 x (t)dt = D 2 That is T T 0 x (t) 2 dt D 2. Obvious strategy (constant speed path) is optimal.

12 What does this tell us about the Wasserstein distance? We saw how to move probability density f 0 to f 1 on R. Can think of this as taking 1 unit of time. Now suppose that we interrupt the process at time t. Where would we have got to? Can use ideas from fluid dynamics. Benamou Brenier proved variational characterization of W 2. Works for R, R d, Riemannian manifolds... but not e.g. Z.

13 Benamou Brenier formula Given distribution functions F 0 and F 1, write P R (F 0, F 1 ) for the set of densities f t (x) such that F 0 (x) = x f 0(y)dy and F 1 (x) = x f 1(y)dy. Given a sequence of densities, define velocity field v t (x) by Theorem (Benamou Brenier) t f t(x) = x (v t(x)f t (x)). The quadratic Wasserstein distance on R is given by W 2 (F 0, F 1 ) = ( inf f t P R (F 0,F 1 ) 1 0 ( ) 1/2 f t (y)v t (y) 2 dy dt).

14 Benamou Brenier geodesics If f t P R (F 0, F 1 ) achieves infimum in Benamou Brenier, call it a geodesic. Geodesics have nice properties. Theorem Geodesics satisfy fixed speed property: W 2 (F s, F t ) = t s W 2 (F 0, F 1 ), for all s and t. Say that W 2 induces a length space. Fits with idea that geodesics are straight lines.

15 Entropy Definition Recall we measure randomness of probability density f by entropy H(f ) = f (x) log f (x)dx. R Interested in how entropy varies along paths f t. In particular, what is behaviour along geodesics?

16 Behaviour of entropy along paths Definition Given a path f t (x), introduce functions g t and h t such that f t (x) t = g t(x), x 2 f t (x) t 2 = 2 h t (x) x 2. Theorem Writing H(t) = H(f t ) for the entropy along the path, under integrability conditions: H (t) = R R (h t (x) g t(x) 2 ) 2 ( f t (x) x f t (x) ( gt (x) f t (x) x 2 (log f t(x)) dx )) 2 dx.

17 Behaviour of entropy along paths Proof. H (t) = = H (t) = R R 2 f t (x) t 2 2 h t (x) x 2 f t (x) t log f t (x)dx R 1 log f t (x)dx f t (x) R log f t (x)dx Integration by parts deals with these terms. Key is an explicit expression for 2 x 2 log f t (x). R 1 f t (x) ( ) ft (x) 2 dx t ( ) gt (x) 2 dx. x

18 Behaviour of entropy along geodesics Along BB geodesics turns out g t (x) = v t (x)f t (x) and h t (x) = v t (x) 2 f t (x). In above theorem ( ) H vt (x) 2 (t) = f t (x) dx 0. R x This concavity used in information geometry. Properties of W 2 are key. Special case of Sturm Lott Villani theory. For example: Theorem For a Riemannian manifold (M, d) concavity of entropy along every geodesic is equivalent to positivity of the Ricci curvature tensor.

19 Discrete random variables Situation less clear for random variables supported on discrete sets. Will consider random variables supported on Z or in fact {0, 1,..., n}.

20 For discrete problems, W 2 is not a length space Example Consider marginals f 0 = (3/4, 1/4) and f 1 = (1/4, 3/4) Obvious (and optimal) strategy f t = (3/4 t/2, 1/4 + t/2). Could define f t (x, y) as follows: f t (x, y) y = 0 y = 1 x = 0 3/4 - t/2 t/2 f 0 (0) = 3/4 x = 1 0 1/4 f 0 (1) = 1/4 f t (0) = 3/4 t/2 f t (1) = 1/4 + t/2 Cost of f t is W 2 2 (F 0, F t ) = x,y f t(x, y) y x p = t/2. Hence W 2 (F 0, F t ) = tw 2 (F 0, F 1 ) not a length space.

21 Concavity of entropy: Shepp Olkin conjecture Consider n independent Bernoulli random variables, with parameters p = (p 1,... p n ). Their sum has mass function f p (k) for k = 0, 1,..., n. Consider the entropy of f p, defined by H(p) := n f p (k) log f p (k). k=0 Conjecture (Shepp Olkin (1981)) H(p) is a concave function of p. Sufficient to consider concavity for affine t, i.e. take p i (t) = p i (0)(1 t) + p i (1)t.

22 Known cases Folklore: n = 1. Shepp Olkin (1981): n = 2, n = 3 (claim with no proof, in paper). Shepp Olkin (1981): for all i, p i (t) = t (binomial case). Yu Johnson (2009): for all i, either p i (0) = 0 or p i (1) = 0. Hillion (2012): for all i, either p i (t) = t or p i (t) constant (binomial translation case).

23 Motivating example: binomial case Example Write spatial derivative 1 f (k) = f (k) f (k 1). For 0 p < q 1, define p(t) = p(1 t) + qt. Write Bin n,p (k) := ( n k) p k (1 p) n k. Write f t (k) = Bin n,p(t) (k). Simple calculation (e.g. Mateev, Shepp Olkin) shows: ) f t (k) = 1 (n(q p)bin t n 1,p(t) (k).

24 Motivating example: binomial case (cont.) Example We rewrite this using an idea of Yu: ( (k + 1) Bin n 1,p (k) = Bin n,p (k + 1) + 1 k ) Bin n,p (k). n n Suggests we introduce mixtures of mass functions: for f t (k) ( ) = 1 vg (α) t (k), t g (α) t (k) = α t (k + 1)f t (k + 1) + (1 α t (k))f t (k) Here α t (k) = k/n for all k and t and v = n(q p). Remember continuous equation t f t(x) = x (v t(x)f t (x)).

25 Discrete Benamou Brenier formula Definition Write P Z (f 0, f 1 ) for the set of probability mass functions f t (k), given end constraints f t (k) t=0 = f 0 (k) and f t (k) t=1 = f 1 (k). Write A for the set of α(k) with α t (0) 0, α t (n) 1 and with 0 α t (k) 1 for all k.

26 Discrete Benamou Brenier formula Definition For f t (k) P Z (f 0, f 1 ) and α A, define probability mass function g (α) t (k), velocity field v α,t (k) and distance V n by (k) = α t (k + 1)f t (k + 1) + (1 α t (k))f t (k) f ( ) t t (k) = 1 v α,t (k)g (α) t (k) ( 1 n 1 ) V n (f 0, f 1 ) = inf g (α) t (k)v α,t (k) 2 dt g (α) t f t P Z (f 0,f 1 ), α t(k) A 0 k=0 Refer to any path achieving the infimum as a geodesic. 1/2.

27 Discrete Benamou Brenier formula Definition Example: binomial path is geodesic with v α,t (k) n(q p). Call path with v α,t (k) fixed in k and t a constant speed path. Proposition V n is a metric for probability measures on {0,... n}. V n defines a length space: for any geodesic f, distance V n (f s, f t ) = t s V n (f 0, f 1 ). If there exists a constant speed path then f 0 and f 1 are stochastically ordered. Wasserstein distance W 1 and V n coincide.

28 Framework for concavity of entropy Want conditions under which entropy is concave. Give conditions in terms of α t (k) to generalize binomial case. Recall that in that case, α t (k) k/n.

29 k-monotonicity condition Condition (k-mon) Given t, we say that the α t (k) are k-monotone at t if α t (k) α t (k + 1) for all k = 0,..., n 1.

30 t-monotonicity condition Condition (t-mon) Given t, we say that the α t (k) are t-monotone at t if α t (k) t 0 for all k = 0,..., n. Given a constant speed path f t (k) ( = v 1 t introduce h(k) such that g (α) t ) (k), 2 f t (k) t 2 = v (h(k)). t-mon condition provides an upper bound on h(k).

31 GLC condition Condition (GLC) We say f t (k) is α-generalized log-concave at t, if for all k = 0,..., n 2, GLC(α t )(k) := α t (k + 1)(1 α t (k + 1))f t (k + 1) 2 0. α t (k + 2)(1 α t (k))f t (k)f t (k + 2)

32 Theorem (Hillion Johnson 2014) Consider constant speed path f t (k) and associated optimal α(t). If Conditions k-mon, t-mon and GLC hold at given t = t, the entropy H(f t ) is concave in t at t = t.

33 Proof Dealing with logarithm remains key but harder. k-mon and GLC together imply that f t (k)g t (k + 1) f t (k + 1)g t (k) 1 and f t (k + 2)g t (k) f t (k + 1)g t (k + 1) 1. Also log v θ(v) = 1/(2v) v/2, for v 1. Hence ( ) ft (k)f t (k + 2) log f t (k + 1) ( 2 ) ft (k)g t (k + 1) = log f t (k + 1)g t (k) ( ) ft (k)g t (k + 1) θ + θ f t (k + 1)g t (k) ( ft (k + 2)g t (k) log f t (k + 1)g t (k + 1) ) ( ft (k + 2)g t (k) f t (k + 1)g t (k + 1) )

34 Proof (cont.) H (t) = n k=0 2 f t (k) t 2 k=0 log f t (k) n k=0 ( ) 1 ft (k) 2 f t (k) t n ( 1 (vg t (k))) 2 n = v (h t (k)) log f t (k) f t (k) k=0 k=0 n ( ( )) = v 2 ft (k)f t (k + 2) n ( 1 (vg t (k))) 2 h t (k) log f t (k + 1) 2 f t (k) k=0 k=0 n ( ( ) ( )) v 2 ft (k)g t (k + 1) ft (k + 2)g t (k) h t (k) θ + θ f t (k + 1)g t (k) f t (k + 1)g t (k + 1) n ( 1 (vg t (k))) 2 k=0 f t (k)

35 Proof (cont.) n 2 k=0 f t (k)f t (k + 1)f t (k + 2) 2g t (k)g t (k + 1)... and then, as if by magic, this becomes minus a perfect square!! Details best left to Mathematica... H (t) becomes v 2 times... ( g t (k) 2 f t (k)f t (k + 1) g t (k + 1) 2 f t (k + 1)f t (k + 2) Would like to know how to interpret this cf (above) ( ) H vt (x) 2 (t) = f t (x) dx 0. x R ) 2

36 Relating this to Shepp Olkin Proposition For Shepp Olkin interpolations, if all p i have the same sign ( monotone case ): We have a constant speed path k-mon condition holds. GLC condition holds. However, t-mon condition fails for some Shepp Olkin paths. Entropy remains concave if replace by t-mon by weaker Condition 4. Condition 4 holds for Shepp Olkin paths.

37 Main result of our paper Theorem (Hillion Johnson 2014) If all p i have the same sign, H(p) is a concave function of p. Call this monotone Shepp Olkin theorem. General case remains open (not constant speed path).