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730 Lecture 7. Today’s lecture: More on CR bound (pp25-29) Rao-Blackwellisation (pp29-33). CR lower bound. If S is an unbiased estimate of q , then. Joint density. CR lower bound. U is the score function I( q ) is the information matrix q is a vector parameter
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730 Lecture 7 Today’s lecture: More on CR bound (pp25-29) Rao-Blackwellisation (pp29-33) 730 Lecture 7
CR lower bound • If S is an unbiased estimate of q, then Joint density 730 Lecture 7
CR lower bound • U is the score function • I(q) is the information matrix • q is a vector parameter • Cov(S) ³ I(q)-1 means that the matrix Cov(S) - I(q)-1 is positive definite. 730 Lecture 7
CR lower bound • Proof follows from the following steps: • Step 1 730 Lecture 7
CR lower bound (cont) • Step 2: This is 0! 730 Lecture 7
Then Cov(T)= is PD Thus is PD CR lower bound (cont) Step 3: Put T=(S,U) Thus Cov(T) -1 is PD • Thus Cov(S)-I(q) -1 is PD 730 Lecture 7
Useful fact: 730 Lecture 7
Examples • Text p 27: Example 14 • Normal distribution • Text p 28: Example 15 • Multinomial distribution • Review 310 stuff as well 730 Lecture 7
p11 p12 p21 p22 Example • Consider 2 x 2 contingency table 730 Lecture 7
Example(cont) Assume a multinomial probability model, with cell probabilities ie an independence model. 730 Lecture 7
Example(cont) Log likelihood is 730 Lecture 7
Example(cont) Score function U is 730 Lecture 7
Example(cont) Information is 730 Lecture 7
Example(cont) Similarly Hence 730 Lecture 7
Example(cont) Consider usual estimate Must be UMVUE!! 730 Lecture 7
CR upper bound (cont) When is the bound attained? Consider If bound is attained, Cov(S)=I(q)-1, so we get 0!!!! 730 Lecture 7
CR upper bound (cont) Then: U=I(q)(S- q) + b Take expected values, b=0 Hence U= I(q)(S- q) if bound is attained. Thus if bound is attained, we must have S=I(q)-1U + q. This can’t be an estimate unless it does not involve q. 730 Lecture 7
CR upper bound (cont) Conversely… Easy to see that bound is attained if U= I(q)(S- q) . Because…… I =Cov(S,U) =Cov(S,I(q)(S- q)) =Cov(S)I(q) 730 Lecture 7
Example • Use this to see that no unbiased estimate of the Normal variance can attain the bound. • From example 14 on p 27, Var (S) ³ n/(2s4). • If bound is attained, then we must have S=(1/n)S(Xi-m)2 • Not an estimate!!! Bound not much help. 730 Lecture 7
How else? • How else can we find umvues? • Suppose we have a sufficient statistic T, and we can find an unbiased estimate U. • Then E(U|T) is a function of T, is unbiased, and has smaller variance than U. (or at least no bigger). 730 Lecture 7
Memory jog…. • E(E(X|Y))=E(X) • Var(E(X|Y)) £ Var(X). 730 Lecture 7
Complete statistics • A statistic T is complete if the only function g for which E(g(T))=0 for all q is g=0. • If a sufficient statistic T is complete, there is only one function of T that is unbiased. • Because…. • If u(T) and v(T) both have expectation q, then g(T)=u(T)-v(T) has expectation 0 and so g=0 ie u=v. 730 Lecture 7
Rao-blackwellisation • The process of taking the conditional expectation of an unbiased statistic given a sufficient statistic is called…… Rao-Blackwellisation (invented by Rao and Blackwell) 730 Lecture 7
Lehmann-Scheffe theorem • If T is a complete sufficient statistic, and U is unbiased, then E(U|T) is the umvue. • For suppose S is unbiased and VarS£Var E(U|T). Then VarE(S|T) £ VarS £ Var E(U|T). • But E(S|T) = E(U|T) by completeness. • Hence Var S=VarE(U|T). No unbiased estimator can have a smaller variance than E(U|T) so it must be the umvue. 730 Lecture 7
Example • For normal, mean is sufficient and complete. Since it is unbiased, it must be the umvue. • Same argument for exponential. • Same argument for the Poisson. • CR lower bound argument works as well. 730 Lecture 7
Example • Find the umvue for Pr(X=0) =e-q where X is Poisson. • Put S=1 if X1=0, and 0 otherwise. Then S is unbiased. • Consider T=X1+…+ Xn . It is a complete sufficient statistic. 730 Lecture 7
Example (cont) • We need E(S|T)=P(X1=0|T=t). Consider 730 Lecture 7
Can write as Example (cont) Thus the umvue is ((n-1)/n)T Note: This is biased but has smaller MSE for some values of q. 730 Lecture 7
Example • Exponential. What is umvue of 1/q? • Might suspect that 1/mean estimates 1/q, since the mean estimates q. 730 Lecture 7
Recall…. • If X1,…,Xn are Exponential with mean q, then T=X1+…+Xn is Gamma(n,q) • Density is 730 Lecture 7
Example (cont) We have Thus (n-1)/T is unbiased, so is the umvue. 730 Lecture 7
Example • Suppose X1,…,Xn are iid N(m,s2) • We have seen that (SXi, SXi2) is sufficient. It is also complete (hard). • Find the umvue of m/s 730 Lecture 7
Solution • Try the sample version:ie Sample mean/sample sd • Note that for the Normal, mean and sd are independent (310 stuff) • Thus 730 Lecture 7
Solution (cont) 730 Lecture 7
Solution (cont) Thus is an unbiased estimate of m/s that is a function of a complete sufficient statistic and so is the umvue. 730 Lecture 7
A final example • Unbiased is not the whole story! • Sample variance, normal data – divide by n-1, n or something else? • Let’s try dividing by n+c. What should c be???? 730 Lecture 7
Sample variance(continued) • We have already seen ( see assignment 1) that Chisquare distn!! 730 Lecture 7
Sample variance(continued) Thus 730 Lecture 7
Sample variance(continued) Calculate MSE= Var + Bias2 For what c is MSE minimised? 730 Lecture 7
Sample variance(continued) Differentiate Answer is 730 Lecture 7
Sample variance(continued) Hence minimum when c=1. Does not depend on n!!! NB c¹-1!! Thus the estimate with minimum MSE is (for all n!!) 730 Lecture 7
