Outline. Martingales. Piotr Wojciechowski 1. 1 Lane Department of Computer Science and Electrical Engineering West Virginia University.

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1 Outline Piotr 1 1 Lane Department of Computer Science and Electrical Engineering West Virginia University 8 April, 01

2 Outline Outline 1 Tail Inequalities

3 Outline Outline 1 Tail Inequalities General

4 Outline Tail Inequalities 1 Tail Inequalities General

5 Tail Inequalities We can apply a Chernoff-like bound to Martingale even when the variables are not independent. This bound is referred to as the Azuma-Hoeffding Inequality.

6 Tail Inequalities We can apply a Chernoff-like bound to Martingale even when the variables are not independent. This bound is referred to as the Azuma-Hoeffding Inequality. Azuma-Hoeffding Inequality Theorem If X 0,..., X n is a martingale such that X k X k 1 c k then, for all t > 0 and λ > 0 P( X t X 0 λ) e λ Pt c k=1 k.

7 Tail Inequalities Proof. Let Y i = X i X i 1 for i = 1,..., t. Thus Y i c i.

8 Tail Inequalities Proof. Let Y i = X i X i 1 for i = 1,..., t. Thus Y i c i. Since X 0,..., X n is a martingale we have that E[Y i X 0,..., X i 1 ] = E[X i X i 1 X 0,..., X i 1 ] = E[X i X 0,..., X i 1 ] X i 1 = 0.

9 Tail Inequalities Proof. Let Y i = X i X i 1 for i = 1,..., t. Thus Y i c i. Since X 0,..., X n is a martingale we have that E[Y i X 0,..., X i 1 ] = E[X i X i 1 X 0,..., X i 1 ] = E[X i X 0,..., X i 1 ] X i 1 = 0. We have that Y i = c i 1 Y i c i + c i 1 + Y i c i.

10 Tail Inequalities Proof. Let Y i = X i X i 1 for i = 1,..., t. Thus Y i c i. Since X 0,..., X n is a martingale we have that E[Y i X 0,..., X i 1 ] = E[X i X i 1 X 0,..., X i 1 ] = E[X i X 0,..., X i 1 ] X i 1 = 0. We have that Y i = c i 1 Y i c i Because the function e α x is convex it follows that + c i 1 + Y i c i. e α Y i 1 Y i c i e α c i Y i c i e α c i = eα c i + e α c i + Y i c i (e α c i e α c i ).

11 Tail Inequalities Proof. As shown previously E[Y i X 0,..., X i 1 ] = 0, thus» E[e α Y i e α c i + e α c i X 0,..., x i 1 ] = E = eα c i + e α c i + Y i (e α c i e α c i ) X 0,.., X i 1 c i e (α c i ).

12 Tail Inequalities Proof. As shown previously E[Y i X 0,..., X i 1 ] = 0, thus» E[e α Y i e α c i + e α c i X 0,..., x i 1 ] = E We have that h i E e α (X t X 0 ) = E = eα c i + e α c i t 1 Y 4 e α Y i 3 t Y 5 = E 4 + Y i (e α c i e α c i ) X 0,.., X i 1 c i e (α c i ). e α Y i i=1 i=1 3 t Y E 4 e α Y i 5 e (α c i ) α Pt c k=1 i e. i=1 3 i 5 E he α Y i 1 X 0,..., x i 1

13 Tail Inequalities Proof. Therefore, we have that P(X t X 0 λ) = P(e α (X t X 0 ) e α λ ) E ˆe α (X t X 0 ) e α λ e α P tk=1 c i α λ.

14 Tail Inequalities Proof. Therefore, we have that P(X t X 0 λ) = P(e α (X t X 0 ) e α λ ) E ˆe α (X t X 0 ) By letting α = λ P tk=1 c i we get that e α λ e α P tk=1 c i α λ. P(X t X 0 λ) e λ Pt c k=1 k.

15 Tail Inequalities Proof. Therefore, we have that P(X t X 0 λ) = P(e α (X t X 0 ) e α λ ) E ˆe α (X t X 0 ) By letting α = λ P tk=1 c i we get that e α λ e α P tk=1 c i α λ. P(X t X 0 λ) e λ Pt c k=1 k. We can similarly construct the same bound on P(X t X 0 λ).

16 Tail Inequalities Corollary If X 0,..., X n is a martingale such that X k X k 1 c then, for all t 1 and λ > 0 P( X t X 0 λ c t) e λ.

17 Tail Inequalities Corollary If X 0,..., X n is a martingale such that X k X k 1 c then, for all t 1 and λ > 0 P( X t X 0 λ c t) e λ. Azuma-Hoeffding Inequality Theorem If X 0,..., X n is a martingale such that B k X k X k 1 B k + d k for some constants d k and random variables B k then, for all t 0 and λ > 0 P( X t X 0 λ) e λ P tk=1 d k.

18 Outline Tail Inequalities General 1 Tail Inequalities General

19 Tail Inequalities General Lipschitz condition A function f( X) = f(x 1, X,..., X n) satisfies the Lipschitz condition with bound c if for any i and any x 1,..., x n and y i, f(x 1,..., x i,..., x n) f(x 1,..., x i 1, y i, x i+1,..., x n) c.

20 Tail Inequalities General Lipschitz condition A function f( X) = f(x 1, X,..., X n) satisfies the Lipschitz condition with bound c if for any i and any x 1,..., x n and y i, f(x 1,..., x i,..., x n) f(x 1,..., x i 1, y i, x i+1,..., x n) c. Theorem Let f be a functions satisfying the Lipschitz condition with bound c and let Z 0,... be the Doob martingale defined by Z 0 = E[f(X 1,..., X n)] and Z k = E[f(X 1,..., X n) X 1,..., X k ]. We have that for each k there exists a random variable B k depending on Z 0,..., Z k 1 such that B k Z k Z k 1 B k + c.

21 Tail Inequalities General Example Suppose that we are throwing m balls independently and uniformly at random into n bins. Let X i denote the bin into which the i th ball falls and let F denote the number of empty bins after all m balls are thrown.

22 Tail Inequalities General Example Suppose that we are throwing m balls independently and uniformly at random into n bins. Let X i denote the bin into which the i th ball falls and let F denote the number of empty bins after all m balls are thrown. We will use the Azuma-Hoeffding Inequality to determine the tightness of the distribution of F.

23 Tail Inequalities General Example Suppose that we are throwing m balls independently and uniformly at random into n bins. Let X i denote the bin into which the i th ball falls and let F denote the number of empty bins after all m balls are thrown. We will use the Azuma-Hoeffding Inequality to determine the tightness of the distribution of F. We have that the sequence Z i = E[F X 1,..., X i ] is Doob martingale.

24 Tail Inequalities General Example Suppose that we are throwing m balls independently and uniformly at random into n bins. Let X i denote the bin into which the i th ball falls and let F denote the number of empty bins after all m balls are thrown. We will use the Azuma-Hoeffding Inequality to determine the tightness of the distribution of F. We have that the sequence Z i = E[F X 1,..., X i ] is Doob martingale. Since F depends on X 1,..., X m, there is a function f such that F = f(x 1,..., X m).

25 Tail Inequalities General Example Suppose that we are throwing m balls independently and uniformly at random into n bins. Let X i denote the bin into which the i th ball falls and let F denote the number of empty bins after all m balls are thrown. We will use the Azuma-Hoeffding Inequality to determine the tightness of the distribution of F. We have that the sequence Z i = E[F X 1,..., X i ] is Doob martingale. Since F depends on X 1,..., X m, there is a function f such that F = f(x 1,..., X m). Because changing which bin a single ball lands in changes F by at most 1, we have that f satisfies the Lipschitz condition with bound 1.

26 Tail Inequalities General Example Suppose that we are throwing m balls independently and uniformly at random into n bins. Let X i denote the bin into which the i th ball falls and let F denote the number of empty bins after all m balls are thrown. We will use the Azuma-Hoeffding Inequality to determine the tightness of the distribution of F. We have that the sequence Z i = E[F X 1,..., X i ] is Doob martingale. Since F depends on X 1,..., X m, there is a function f such that F = f(x 1,..., X m). Because changing which bin a single ball lands in changes F by at most 1, we have that f satisfies the Lipschitz condition with bound 1. Thus applying the second Azuma-Hoeffding Inequality we get that P( F E[F] ǫ) = P( Z m Z 0 ǫ) e ǫ P mk=1 c = e ǫ m.

27 Tail Inequalities General Example Let G be a random graph in G n,p. The Chromatic number, χ(g), is the minimum number of colors needed to color all the verticies of a graph so that no two adjacent verticies are the same color.

28 Tail Inequalities General Example Let G be a random graph in G n,p. The Chromatic number, χ(g), is the minimum number of colors needed to color all the verticies of a graph so that no two adjacent verticies are the same color. We will use the Azuma-Hoeffding Inequality to determine the tightness of the distribution of χ(g).

29 Tail Inequalities General Example Let G be a random graph in G n,p. The Chromatic number, χ(g), is the minimum number of colors needed to color all the verticies of a graph so that no two adjacent verticies are the same color. We will use the Azuma-Hoeffding Inequality to determine the tightness of the distribution of χ(g). Let Z be the vertex exposure martingale for G. Thus if G i is subgraph of G induced by the verticies 1,..., i, we have that Z 0 = E[χ(G)] and Z i = E[χ(G) G 1,..., G i ].

30 Tail Inequalities General Example Let G be a random graph in G n,p. The Chromatic number, χ(g), is the minimum number of colors needed to color all the verticies of a graph so that no two adjacent verticies are the same color. We will use the Azuma-Hoeffding Inequality to determine the tightness of the distribution of χ(g). Let Z be the vertex exposure martingale for G. Thus if G i is subgraph of G induced by the verticies 1,..., i, we have that Z 0 = E[χ(G)] and Z i = E[χ(G) G 1,..., G i ]. Introducing a vertex into the graph increases he chromatic number by at most 1 so we have that for each i, 0 Z i Z i 1 1.

31 Tail Inequalities General Example Let G be a random graph in G n,p. The Chromatic number, χ(g), is the minimum number of colors needed to color all the verticies of a graph so that no two adjacent verticies are the same color. We will use the Azuma-Hoeffding Inequality to determine the tightness of the distribution of χ(g). Let Z be the vertex exposure martingale for G. Thus if G i is subgraph of G induced by the verticies 1,..., i, we have that Z 0 = E[χ(G)] and Z i = E[χ(G) G 1,..., G i ]. Introducing a vertex into the graph increases he chromatic number by at most 1 so we have that for each i, 0 Z i Z i 1 1. Thus, we can apply the second Azuma-Hoeffding Inequality to get that P( χ(g) E[χ(G)] λ n) e λ.

Randomized Algorithms

Randomized Algorithms 南京大学尹一通 Martingales Definition: A sequence of random variables X 0, X 1,... is a martingale if for all i > 0, E[X i X 0,...,X i1 ] = X i1 x 0, x 1,...,x i1, E[X i X 0 = x 0, X 1