Prohorov s theorem. Bengt Ringnér. October 26, 2008

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1 Prohorov s theorem Bengt Ringnér October 26, The theorem Definition 1 A set Π of probability measures defined on the Borel sets of a topological space is called tight if, for each ε > 0, there is a compact set such that P () > 1 ε for all P Π Theorem 1 A tight set, Π, of probability measures on the Borel sets of a metric topological space, X, is relatively compact in the sense that for each sequence, P 1,P 2, in Π there exists a subsequence that converges to a probability measure P, not necessarily in Π, in the sense that g dp nj g dp for all bounded continuous integrands Conversely, if the metric space is separable and complete, then each relatively compact set is tight This is a generalisation of the Helly selection and the Helly-Bray theorems in which X is the real line If F 1,F 2, is a sequence of right-continuous cumulative distribution functions, F n (x) = P(X n x), then there is a subsequence and a limiting F such that, as j, F nj (x) F(x) if F is continuous at x Centre for Mathematical Sciences, Lund University, Lund, Sweden Homepage: bengtr 1

2 This is the same as, for all g continuous with compact support, g(x)df nj (x) g(x)df(x), in other words, g dp nj g dp where P n (A) = P(X n A) for Borel sets A The limiting F can be chosen as right continuous and nondecreasing, but is not neccessarily a distribution function, since one has lim x F(x) 1 and lim x F(x) 0, but not necessarily equality Equality is equivalent to the original sequence being tight In this case pointwise convergence at continuity points is the same as g dfnj converges for all bounded continuous integrands Note also that working with continuous functions with compact support is only possible in locally compact spaces, a concept which is too restrictive in probability theory Here X may contain trajectories of random processes, and one might be interested in P n ({x : sup t 2 Proof of the direct part x(t) M}) = P(supX n (t) M) t The idea the same as on the real line First prove convergence on a denumerable dense subset, then use equicontinuity To get a dense subset of integrands, we need a lemma Lemma 1 Let be a compact set in a space X equipped with a metric topology Then the space of bounded real valued functions on X is separable in the sense that there exist a countable subset, ϕ 1,ϕ 2, such that, for any bounded continuous g, sup g(x) ϕ k (x) can be made arbitrarily small, and sup x X g(x) ϕ k (x) 3sup g(x) x X Proof Being compact in a metric space, has a countable dense subset, x 1,x 2, From the (proof of the) Stone-Weierstrass approximation theorem it follows that any continuous function f can be approximated in the sup-norm on with functions of the form ϕ = min 0<j j 0 max 0<k k 0 ψ j k 2

3 where the ψ are of the form ψ(x) = a bρ(x,x i ) and ρ is any metric generating the topology Since is bounded, we may let a and b be rational Finally, make the ϕ bounded by replacing them with min(max(ϕ, inf ϕ),sup ϕ) To prove the theorem, choose compact 1, 2, such that P ( m ) < 1 1/m for all P Π For each such m the lemma gives a dense subset of functions Let ϕ 1,ϕ 2, be an enumeration of all these For a given bounded continuous g and ε > 0, choose m > 1/ε and ϕ k with g ϕ k < ε on m Then g ϕ k dp = g ϕ k dp + g ϕ m }{{} k m c }{{} <ε 3sup f Therefore, for any given g and ε > 0 there is a ϕ k such that g ϕ k dp < ε sup P Π dp (1 + 3sup g )ε X By the Bolzano-Weierstrass theorem combined with Cantor s diagonal method, there is a subsequence such that, for each k, ϕ k dp nj converges as j This gives lim sup g dp nj lim inf g dp nj < 2ε Since this holds for arbitrarily small ε > 0, the limit exists, so we can define a functional I by I(g) = lim g dp nj Clearly, I is a linear functional and I(g) 0 if g 0 To prove that it can be represented with a probability measure, we use the Stone-Daniell representation theorem Let g k ց 0 pointwise For a given ε > 0 choose a compact such that P ( c ) < ε for all P in Π Then g k dp nj g k dp nj + c g k dp nj sup g k (x) + εsupg 1 (x), x 3

4 so I(g k ) sup g k (x) + εsup g 1 (x), x By Dini s theorem, g k 0 uniformly on, so the first term tends to zero as k, giving lim sup k I(g k ) εsup x g 1 (x) for all ε > 0, which gives I(g k ) 0 Since g and min(1,g) are bounded continuous if g is so, I is a Daniell integral on a Stone lattice Therefore, and since I(1) = 1, it can be represented as an integral with respect to a probability measure, so lim g dp nj = I(g) = which proves the first part of the theorem 3 Proof of the converse All sets of the form = i=1 k j i=1 B(x i,1/j) g dp in which B(x i,1/j) = {x : ρ(x,x i ) < 1/j} are compact, since X is complete and is closed and totally bounded In order to make large enough, we use separability and choose x 1, as a dense subset We shall use the fact that, if P k P is the sense of the theorem, then 1 lim inf k P k(u) P (U) for all open sets U Let ε > 0 There is a with P () > 1 ε for all P Π if we can prove that for each j there is a k j such that for all P in Π P ( k j i=1 B(x i,1/j)) > 1 ε/2 j 1 This is a part of the so called Portemanteau Theorem It follows by considering g n(x) = min(1, nρ(x, U c )) Then g n increases towards the indicator function of U, and P k (U) g n dp k g n dp, k, which gives lim inf P k(u) k g n dp P (U), n 4

5 If this does not hold, there is a j 0 such that, for each k there is a P k with P k ( k i=1 B(x i,1/j 0 )) 1 ε/2 j 0 Of course, this also holds with k i=1 if k k By assumption, there is a converging subsequence, so P ( k i=1 B(x i,1/j 0 )) lim inf n P k n ( k i=1 B(x i,1/j 0 )) 1 ε/2 j 0 But this would give 1 = P (X) = lim k P ( k i=1 B(x i,1/j 0 )) 1 ε/2 j 0 Therefore holds and the proof is complete 5