# STAT 830 Convergence in Distribution

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1 STAT 830 Convergence in Distribution Richard Lockhart Simon Fraser University STAT 830 Fall 2011 Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

2 Purposes of These Notes Define convergence in distribution State central limit theorem Discuss Edgeworth expansions Discuss extensions of the central limit theorem Discuss Slutsky s theorem and the δ method. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

3 Convergence in Distribution p 72 Undergraduate version of central limit theorem: Theorem If X 1,...,X n are iid from a population with mean µ and standard deviation σ then n 1/2 ( X µ)/σ has approximately a normal distribution. Also Binomial(n, p) random variable has approximately a N(np, np(1 p)) distribution. Precise meaning of statements like X and Y have approximately the same distribution? Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

4 Towards precision Desired meaning: X and Y have nearly the same cdf. But care needed. Q1: If n is a large number is the N(0,1/n) distribution close to the distribution of X 0? Q2: Is N(0, 1/n) close to the N(1/n, 1/n) distribution? Q3: Is N(0,1/n) close to N(1/ n,1/n) distribution? Q4: If X n 2 n is the distribution of X n close to that of X 0? Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

5 Some numerical examples? Answers depend on how close close needs to be so it s a matter of definition. In practice the usual sort of approximation we want to make is to say that some random variable X, say, has nearly some continuous distribution, like N(0, 1). So: want to know probabilities like P(X > x) are nearly P(N(0,1) > x). Real difficulty: case of discrete random variables or infinite dimensions: not done in this course. Mathematicians meaning of close: Either they can provide an upper bound on the distance between the two things or they are talking about taking a limit. In this course we take limits. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

6 The definition p 75 Def n: A sequence of random variables X n converges in distribution to a random variable X if E(g(X n )) E(g(X)) for every bounded continuous function g. Theorem The following are equivalent: 1 X n converges in distribution to X. 2 P(X n x) P(X x) for each x such that P(X = x) = 0. 3 The limit of the characteristic functions of X n is the characteristic function of X: for every real t E(e itxn ) E(e itx ). These are all implied by M Xn (t) M X (t) < for all t ǫ for some positive ǫ. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

7 Answering the questions X n N(0,1/n) and X = 0. Then 1 x > 0 P(X n x) 0 x < 0 1/2 x = 0 Now the limit is the cdf of X = 0 except for x = 0 and the cdf of X is not continuous at x = 0 so yes, X n converges to X in distribution. I asked if X n N(1/n,1/n) had a distribution close to that of Y n N(0,1/n). The definition I gave really requires me to answer by finding a limit X and proving that both X n and Y n converge to X in distribution. Take X = 0. Then and E(e txn ) = e t/n+t2 /(2n) 1 = E(e tx ) E(e tyn ) = e t2 /(2n) 1 so that both X n and Y n have the same limit in distribution. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

8 First graph N(0,1/n) vs X=0; n= X=0 N(0,1/n) N(0,1/n) vs X=0; n= X=0 N(0,1/n) Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

9 Second graph N(1/n,1/n) vs N(0,1/n); n= N(0,1/n) N(1/n,1/n) N(1/n,1/n) vs N(0,1/n); n= N(0,1/n) N(1/n,1/n) Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

10 Scaling matters Multiply both X n and Y n by n 1/2 and let X N(0,1). Then nxn N(n 1/2,1) and ny n N(0,1). Use characteristic functions to prove that both nx n and ny n converge to N(0, 1) in distribution. If you now let X n N(n 1/2,1/n) and Y n N(0,1/n) then again both X n and Y n converge to 0 in distribution. If you multiply X n and Y n in the previous point by n 1/2 then n 1/2 X n N(1,1) and n 1/2 Y n N(0,1) so that n 1/2 X n and n 1/2 Y n are not close together in distribution. You can check that 2 n 0 in distribution. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

11 Third graph N(1/sqrt(n),1/n) vs N(0,1/n); n= N(0,1/n) N(1/sqrt(n),1/n) N(1/sqrt(n),1/n) vs N(0,1/n); n= N(0,1/n) N(1/sqrt(n),1/n) Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

12 Summary To derive approximate distributions: Show sequence of rvs X n converges to some X. The limit distribution (i.e. dstbn of X) should be non-trivial, like say N(0,1). Don t say: X n is approximately N(1/n,1/n). Do say: n 1/2 (X n 1/n) converges to N(0,1) in distribution. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

13 The Central Limit Theorem pp Theorem If X 1,X 2, are iid with mean 0 and variance 1 then n 1/2 X converges in distribution to N(0, 1). That is, P(n 1/2 X x) 1 2π x e y2 /2 dy. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

14 Proof of CLT As before E(e itn1/2 X) e t2 /2. This is the characteristic function of N(0,1) so we are done by our theorem. This is the worst sort of mathematics much beloved of statisticians reduce proof of one theorem to proof of much harder theorem. Then let someone else prove that. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

15 Edgeworth expansions In fact if γ = E(X 3 ) then φ(t) 1 t 2 /2 iγt 3 /6+ keeping one more term. Then log(φ(t)) = log(1+u) where u = t 2 /2 iγt 3 /6+. Use log(1+u) = u u 2 /2+ to get log(φ(t)) [ t 2 /2 iγt 3 /6+ ] [ ] 2 /2+ which rearranged is log(φ(t)) t 2 /2 iγt 3 /6+. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

16 Edgeworth Expansions Now apply this calculation to log(φ T (t)) t 2 /2 ie(t 3 )t 3 /6+. Remember E(T 3 ) = γ/ n and exponentiate to get φ T (t) e t2 /2 exp{ iγt 3 /(6 n)+ }. You can do a Taylor expansion of the second exponential around 0 because of the square root of n and get φ T (t) e t2 /2 (1 iγt 3 /(6 n)) neglecting higher order terms. This approximation to the characteristic function of T can be inverted to get an Edgeworth approximation to the density (or distribution) of T which looks like f T (x) 1 2π e x2 /2 [1 γ(x 3 3x)/(6 n)+ ]. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

17 Remarks The error using the central limit theorem to approximate a density or a probability is proportional to n 1/2. This is improved to n 1 for symmetric densities for which γ = 0. These expansions are asymptotic. This means that the series indicated by usually does not converge. When n = 25 it may help to take the second term but get worse if you include the third or fourth or more. You can integrate the expansion above for the density to get an approximation for the cdf. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

18 Multivariate convergence in distribution Def n: X n R p converges in distribution to X R p if E(g(X n )) E(g(X)) for each bounded continuous real valued function g on R p. This is equivalent to either of Cramér Wold Device: a t X n converges in distribution to a t X for each a R p. or Convergence of characteristic functions: for each a R p. E(e iat X n ) E(e iatx ) Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

19 Extensions of the CLT 1 Y 1,Y 2, iid in R p, mean µ, variance covariance Σ then n 1/2 (Ȳ µ) converges in distribution to MVN(0,Σ). 2 Lyapunov CLT: for each n X n1,...,x nn independent rvs with E(X ni ) = 0 Var( i X ni ) = 1 E( Xni 3 ) 0 then i X ni converges to N(0,1). 3 Lindeberg CLT: 1st two conds of Lyapunov and E(X 2 ni 1( X ni > ǫ)) 0 each ǫ > 0. Then i X ni converges in distribution to N(0,1). (Lyapunov s condition implies Lindeberg s.) 4 Non-independent rvs: m-dependent CLT, martingale CLT, CLT for mixing processes. 5 Not sums: Slutsky s theorem, δ method. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

20 Slutsky s Theorem p 75 Theorem If X n converges in distribution to X and Y n converges in distribution (or in probability) to c, a constant, then X n +Y n converges in distribution to X +c. More generally, if f(x,y) is continuous then f(x n,y n ) f(x,c). Warning: the hypothesis that the limit of Y n be constant is essential. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

21 The delta method pp 79-80, Theorem Suppose: Sequence Y n of rvs converges to some y, a constant. X n = a n (Y n y) then X n converges in distribution to some random variable X. f is differentiable ftn on range of Y n. Then a n (f(y n ) f(y)) converges in distribution to f (y)x. If X n R p and f : R p R q then f is q p matrix of first derivatives of components of f. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

22 Example Suppose X 1,...,X n are a sample from a population with mean µ, variance σ 2, and third and fourth central moments µ 3 and µ 4. Then n 1/2 (s 2 σ 2 ) N(0,µ 4 σ 4 ) where is notation for convergence in distribution. For simplicity I define s 2 = X 2 X 2. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

23 How to apply δ method 1 Write statistic as a function of averages: Define W i = [ ] X 2 i. X i See that [ X 2 W n = ] X Define f(x 1,x 2 ) = x 1 x 2 2 See that s 2 = f( W n ). 2 Compute mean of your averages: [ ] [ µ W E( W E(X 2 n ) = i ) µ = 2 +σ 2 E(X i ) µ ]. 3 In δ method theorem take Y n = W n and y = E(Y n ). Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

24 Delta Method Continues 7 Take a n = n 1/2. 8 Use central limit theorem: n 1/2 (Y n y) MVN(0,Σ) where Σ = Var(W i ). 9 To compute Σ take expected value of (W µ W )(W µ W ) t There are 4 entries in this matrix. Top left entry is (X 2 µ 2 σ 2 ) 2 This has expectation: E { (X 2 µ 2 σ 2 ) 2} = E(X 4 ) (µ 2 +σ 2 ) 2. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

25 Delta Method Continues Using binomial expansion: E(X 4 ) = E{(X µ+µ) 4 } = µ 4 +4µµ 3 +6µ 2 σ 2 +4µ 3 E(X µ)+µ 4. So Σ 11 = µ 4 σ 4 +4µµ 3 +4µ 2 σ 2. Top right entry is expectation of which is Similar to 4th moment get (X 2 µ 2 σ 2 )(X µ) E(X 3 ) µe(x 2 ) µ 3 +2µσ 2 Lower right entry is σ 2. So [ µ4 σ Σ = 4 +4µµ 3 +4µ 2 σ 2 µ 3 +2µσ 2 ] µ 3 +2µσ 2 σ 2 Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

26 Delta Method Continues 7 Compute derivative (gradient) of f: has components (1, 2x 2 ). Evaluate at y = (µ 2 +σ 2,µ) to get a t = (1, 2µ). This leads to n 1/2 (s 2 σ 2 ) n 1/2 [1, 2µ] [ X 2 (µ 2 +σ 2 ) X µ ] which converges in distribution to (1, 2µ)MVN(0, Σ). This rv is N(0,a t Σa) = N(0,µ 4 σ 4 ). Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

27 Alternative approach Suppose c is constant. Define Xi = X i c. Sample variance of Xi is same as sample variance of X i. All central moments of Xi same as for X i so no loss in µ = 0. In this case: [ a t µ4 σ = (1,0) Σ = 4 ] µ 3 µ 3 σ 2. Notice that a t Σ = [µ 4 σ 4,µ 3 ] a t Σa = µ 4 σ 4. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

28 Special Case: N(µ,σ 2 ) Then µ 3 = 0 and µ 4 = 3σ 4. Our calculation has n 1/2 (s 2 σ 2 ) N(0,2σ 4 ) You can divide through by σ 2 and get n 1/2 (s 2 /σ 2 1) N(0,2) In fact ns 2 /σ 2 has χ 2 n 1 distribution so usual CLT shows (n 1) 1/2 [ns 2 /σ 2 (n 1)] N(0,2) (using mean of χ 2 1 is 1 and variance is 2). Factor out n to get n n 1 n1/2 (s 2 /σ 2 1)+(n 1) 1/2 N(0,2) which is δ method calculation except for some constants. Difference is unimportant: Slutsky s theorem. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

29 Example median Many, many statistics which are not explicitly functions of averages can be studied using averages. Later we will analyze MLEs and estimating equations this way. Here is an example which is less obvious. Suppose X 1,...,X n are iid cdf F, density f, median m. We study ˆm, the sample median. If n = 2k 1 is odd then ˆm is the kth largest. If n = 2k then there are many potential choices for ˆm between the kth and k +1th largest. I do the case of kth largest. The event ˆm x is the same as the event that the number of X i x is at least k. That is P(ˆm x) = P( i 1(X i x) k) Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

30 The median So P(ˆm x) = P( 1(X i x) k) i ( = P n(ˆfn (x) F(x)) ) n(k/n F(x)). From Central Limit theorem this is approximately ( n(k/n ) F(x)) 1 Φ. F(x)(1 F(x)) Notice k/n 1/2. Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

31 Median If we put x = m+y/ n (where m is true median) we find F(x) F(m) = 1/2. Also n(f(x) 1/2) f(m) where f is density of F (if f exists). So P( n(ˆm m) y) 1 Φ( 2f(m)y) That is, n(ˆm 1/2) N(0,1/(4f 2 (m))). Richard Lockhart (Simon Fraser University) STAT 830 Convergence in Distribution STAT 830 Fall / 31

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