Chapter 4 Limit Theorem

1. Chapter 4 Limit Theorem

2. § 4.1Law of large number1. Convergence in probability Suppose that {Xn} is a sequence of r.v.s, if for any >0, we have it is said that {Xn} convergence to X in probability and denoted it by

3. Means when the probability that the value of Xn Remark fall in interval is increased to 1 a means when

4. 2. Law of Large Numbers (LLN) 1. Chebyshev’s LLN Suppose that {Xk,k=1,2,...} are i.i.d r.v.s with mean  and variance 2>0，then i.e. for any give>0, we have

5. Proof Chebyshev’s inequality, we have where thus

6. 2.Bernoulli’s LLN Set records the numbers of outcomes of A in Bernoulli experiment, , then for any ，we have 3. Khinchine’s LLN Suppose that {Xk,k=1.2,...} are i.i.d sequence with EXk= <, k=1, 2, … then

7. RemarkSuppose that{Xi,i=1.2,...} are i.i.d. r.v.s with E(X1k) <=, then This remark is very important for moment estimation for parameters to be discussed in Chapter 6.

8. § 4.2. Central Limit Theorems1. Convergence in distribution Suppose that {Xn} are i.i.d. r.v.s with d.f. Fn(x), X is a r.v. with F(x), if for all continuous points of F(x) we have It is said that {Xn} convergence to X in distribution and denoted it by

9. 2. Central Limit Theorems (CLT) Levy-Lindeberg’s CLT Suppose that {Xn} are i.i.d. r.v.s wIth mean < and variance 2 <，k=1, 2, …, then {Xn} follows the CLT, which also means that

10. De Moivre-Laplace’s CLT Suppose that n(n=1, 2, ...) follow binomial distribution with parameters n, p(0<p<1), then Example 2 A life risk company has received 10000 policies, assume each policy with premium 12 dollarsand mortality rate 0.6%，the company has to paid 1000 dollars when a claim arrived, try to determine： (1) the probability that the company could be deficit？ (2)to make sure that the profit of the company is not less than 60000 dollars with probability 0.9, try to determine the most payment of each claim.

11. Let X denote the death of one year, then, X~B(n, p), where n= 10000，p=0.6%，Let Y represent the profit of the company, then, Y=1000012-1000X. By CLT, we have (1)P{Y<0}=P{1000012-1000X<0}=1P{X120} 1 (7.75)=0. (2) Assume that the payment is a dollars, then P{Y>60000}=P{1000012-X>60000}=P{X60000/a}0.9. By CLT, it is equal to

12. Convergence in probability Convergence in distribution Chebyshev’s LLN Levy-Lindeberg CLT Bernoulli LLN De Moivre-Laplace’s CLT Khinchine’s LLN