Stochastic Models (Lecture #4)

Transcription

1 Stochastic Models (Lecture #4) Thomas Verdebout Université libre de Bruxelles (ULB)

2 Today Today, our goal will be to discuss limits of sequences of rv, and to study famous limiting results.

3 Convergence of sequences of rv Let X 1, X 2,... be i.i.d. rv, that is, rv that are independent and identically distributed. Assume X 1 is square-integrable, and write µ := E[X 1 ] and σ 2 := Var[X 1 ]. Let X n := 1 n n i=1 X i. Then E[X n ] = 1 n n i=1 E[X i] = E[X 1 ] = µ, and Var[X n ] = 1 Var[ n n 2 i=1 X i] = 1 n n 2 i=1 Var[X i] = 1 n Var[X 1] =,which converges to 0 as n. σ 2 n Consequently, we feel intuitively that X n X, where X = µ. How to make this convergence precise?

4 Convergence of sequences of r.v. s Consider a sequence of r.v. s (X n ) and a r.v. X, defined on (Ω,A,P). How to define X n X (as n )? X n a.s. X (almost surely) P[lim sup n X n X = 0] = 1. X n P X (in probability) limn P[ X n X > ε] = 0, ε > 0. X n L r X (in L r, r > 0) E[ X n X r ] 0. X n D X in distribution (or in law) F X n (x) F X (x) for all x at which F X is continuous.

5 Convergence of sequences of r.v. s Consider as example (Ω,A,P) = ([0, 1],B,λ). X n a.s. X : pointwise convergence (modulo null set). X n P X : convergence in measure. X n L r X : 1 0 X n(t) X(t) r dt 0.

6 Convergence of sequences of r.v. s A principal question: What is the relation among those 4 types of convergence?

7 Convergence of sequences of r.v. s Lemma. (Markov inequality). If E Y r <, r > 0, then P[ Y > ε] E Y r ε r. With Y = X E[X] and r = 2 this becomes Chebyshev s inequality: P[ X E[X] > ε] Var(X) ε 2. Proof. Since Y r ε r I{ Y > ε}, it follows that E Y r ε r P[ Y > ε]. An easy consequence is that X n L r X implies X n P X.

8 Convergence of sequences of r.v. s The other direction is not true, we give a counter example. Example 1: Let Y 1, Y 2,... be i.i.d. rv, with common distribution Define X n = n i=1 Y i. distribution of Y i values 0 2 probabilities The distribution of X n is distribution of X n values 0 2 n probabilities n 1 2 n We feel that X n X, where X = 0.

9 Convergence of sequences of rv In probability: For all ε > 0, P[ X n X > ε] = P[X n > ε] P[X n > 0] = 1 2 n 0, as n. X n P X, as n. In L 1 : E[ X n X ] = E[X n ] = 0 ( n ) + 2 n 1 2 n = 1, which does not go to zero, as n. the convergence does not hold in the L 1 sense.

10 Convergence of sequences of rv One can further show that X n a.s. X implies X n P X. Again the other direction is not true: Example 2: (Ω,A,P) = ([0, 1],B((0, 1]), m) (where m is the Lebesgue measure). Further let X 1 (ω) = I (0,1/2] (ω) X 2 (ω) = I (1/2,1] X 3 (ω) = I (0,1/4] X 4 (ω) = I (1/4,2/4] X 7 (ω) = I (0,1/8] X 8 (ω) = I (1/8,2/8] For k {2 n 1, 2 n,...,2 n+1 2} we have P( X k 0 > 0) = 2 n. BUT P(lim sup k X k 0 = 0) = 0.

11 Convergence of sequences of rv Thus we have for the moment X n a.s. X = Xn P X = Xn L r X

12 Convergence of sequences of rv P D Assume X n X. We wish to prove that then Xn X. For the following argument we use (exercise) P(A) P(B) P(A B c ). Set A = [X x ε] and B = [ X n X ε]. Then we have A B c = [X x ε] [ X n X < ε] [X n x]. Thus P[X x ε] P[ X n X ε] P[X n x]. A similar argument (exercise) shows that P[X n x] P[ X n X ε]+p[x x +ε].

13 Convergence of sequences of rv By the previous inequalities we infer that for any ε > 0 F X (x ε) = P[X x ε] lim infp[x n x] n lim supp[x n x] n P[X x +ε] = F X (x +ε). Now let ε tend to zero and use that F X is continuous in x. Hence X n P X implies Xn D X.

14 Convergence of sequences of rv Thus we have X n a.s. X = P = L Xn X r Xn X X n D X

15 Convergence of sequences of rv A useful criterion: Lemma (Borel-Cantelli-Lemma). If k 1 P( X k X > ε) < holds for any ε > 0, then Hence in Example 1, X n a.s. X. X k a.s. X. Also, if P( X n X > ε) 0, then there is a subsequence {n k } such that P( X nk X > ε) <. k 1 Hence X n P X implies Xnk a.s. X along some properly chosen subsequence.

16 Convergence of sequences of rv The previous example shows that arrows can sometimes be reverted. Here some other sufficient conditions. P X n X there exists a subsequence (Xnk ) for which a.s. X nk X. X n P X and the Xn s are uniformly integrable ) X n L 1 X. X n D const Xn P const. ) Definition: the X n s are uniformly integrable lim α sup n { X n α} X n dp = 0. The latter condition implies that sup n E[ X n ] <, thus if this condition is violated, our sequence will not be uniformly integrable. (See Example 1).

17 Limiting theorems Here are the two most famous limiting results in probability and statistics...

18 Limiting theorems The law of large numbers (LLN): Let X 1, X 2,... be i.i.d. integrable rv. Write µ := E[X 1 ]. Then X n := 1 n n i=1 X i a.s. µ. Remark: Interpretation for favorable/fair/defavorable games of chance.

19 Limiting theorems An example... Let X 1, X 2,... be i.i.d., with X i Bern(p). Then µ = E[X i ] = p, so that X n := 1 n n i=1 X i a.s. p. Remark: If Y 1, Y 2,... are i.i.d. and X i = I{Y i x} then X n = F n (x) is called the empirical distribution function. Note that EX i = F Y (x). We see that F Y n (x) a.s. F Y (x). This result holds uniformly in x and is called Fundamental Theorem of Statistics.

20 Limiting theorems Let us prove the following weak law of large numbers: if X 1, X 2, X 3,... are i.i.d. with E[X 1 ] = µ and Var[X n ] = σ 2 <, then X n := 1 n P X i µ. n Proof. We have for any ε > 0: i=1 P[ X n µ > ǫ] Var[X n] ε 2 σ2 nε 2 0 (n ).

21 Limiting theorems The central limit theorem (CLT): Let X 1, X 2,... be i.i.d. square-integrable rv. Write µ := E[X 1 ] and σ 2 := Var[X 1 ]. Then X n µ σ/ n D Z, with Z N(0, 1). Remarks: X n µ σ/ n = X n E[ X n] Var[ X = Sn E[Sn], where S n = X 1 + +X n. n] Var[Sn] It says sth about the speed of convergence in X n a.s. µ. It allows for computing P[ X n B] for large n... It is valid whatever the distribution of the X i s!

22 Limiting theorems Two examples... (A) Let X 1, X 2,... be i.i.d., with X i Bern(p). Then µ = E[X i ] = p and σ 2 = Var[X i ] = p(1 p), so that n( Xn p) p(1 p) D Z, with Z N(0, 1). Application: This allows for getting an approximation of P[ X n B] for large n, by using that n( Xn p) N(0, 1) for large n. p(1 p)

23 Limiting theorems (B) Let us assume that we play 30 games of roulette, each time we bet 1 Euro either on red or black. What is the probability that that in the end we made a gain? Define X n = Y Y n where Y n = 1 if we win, or 1 if we loose. Then X n is the amount of money we won after n games. Here E[X 30 ] = 30 E[Y 1 ] = 30 37, Var[X 30 ] = 30 Var[Y 1 ] = 30(1 (1/37) 2 ). Hence ( P(X 30 > 0) = P X > 30 ) 37 ( = P (X )/ 30(1 (1/37) 2 ) > 30 ) 37 / 30(1 (1/37) 2 ) P(N(0, 1) > 0.15) = 1 Φ(0.15) 0.44.

24 Limiting theorems This should be compared to 0.37 for the exact distribution. The following table shows the normal approximation for the previous problem for sample sizes n = 30 k, k = 1,...,10 and the corresponding exact probabilities. k approx exact [1,] [2,] [3,] [4,] [5,] [6,] [7,] [8,] [9,] [10,]

25 Sketch of the proof Let us assume that X 1, X 2,... are i.i.d. Then we define ϕ X (t) := E exp(itx 1 ). This is the so-called characteristic function of X 1. (By i.i.d. assumption all X i have the same characteristic function.) One can show: For random variables X and Y, if ϕ X (t) = ϕ Y (t) then X D = Y. (Uniqueness theorem) If ϕ Xn (t) ϕ X (t) for all t, then X n D X. (Continuity theorem) If X and Y are independent, then ϕ X+Y (t) = ϕ X (t)ϕ Y (t). ϕ ax (t) = ϕ X (at). ϕ X+µ (t) = e itµ ϕ X (t).

26 Sketch of the proof By assumption X 1, X 2,... are i.i.d., it follows that for Z n = 1 σ n S n, where S n = (X 1 µ)+(x 2 µ)+...+(x n µ). ϕ Zn (t) = ϕ Sn (t/(σ n)) = (ϕ X1 µ(t/(σ n))) n = (Ee it(x 1 µ)/(σ ) n n) ( [ E 1+ it σ n (X 1 µ) t2 ] ) n 2σ 2 n (X 1 µ) 2 }{{} = where Z N(0, 1). ) n (1 t2 2n e t2 /2 = Ee itz, 2 term Taylor expansion