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PSTAT 120B Probability and Statistics - Week 5 Fang-I Chu University of California, Santa Barbara May 2, 2013 Fang-I Chu PSTAT 120B Probability and Statistics Announcement Office hour: Tuesday 11:00AM-12:00PM Please make use of office hour or email if you have question about hw problem. Put a circle around your name on roster if you bring your two blue books and hand them to me (after section). If you haven’t received any group email from me, please put your email down in the back of roster. Fang-I Chu PSTAT 120B Probability and Statistics Topic for review Exercise #8.15 Exercise #8.13 Exercise #9.26 Hint for homework problem 1(#7.58) Hint for homework problem 4(#8.8) Hint for homework problem 5(#8.12) Hint for homework problem 8(#9.5) Fang-I Chu PSTAT 120B Probability and Statistics Exercise 8.15 8.15 Let Y1 , Y2 , . . . , Yn denote a random sample of size n from a population whose density is given by 3β 3 y −4 β≤y f (y ) = 0 elsewhere where β > 0 is unknown. (This is one of the Pareto distributions introduced in Exercise 6.18.) Consider the estimator β̂ = min(Y1 , Y2 , . . . , Yn ) (a)Derive the bias of the estimator β̂. (b)Derive MSE(β̂) Fang-I Chu PSTAT 120B Probability and Statistics #8.15 8.15(a) (a)Derive the bias of the estimator β̂. Solution: Known: Y has pdf as f (y ) = 3β 3 y −4 , β ≤ y Definition: bias of estimator β̂ is written as E (β̂) − β Denote β̂ = Y(1) Goal: Derive the bias of the estimator β̂. . Fang-I Chu PSTAT 120B Probability and Statistics #8.15 8.15(a) (a)Derive the bias of the estimator β̂. Solution: Way to approach: Find cdf of y using given pdf Use Theorem 6.5 on page 336, obtain fy(1) (y ) = 3nβ 3n y −(3n+1) , y ≥ β Using definition not expected value, obtain E (Y(1) ) = 3n 1 Bias(β̂) = E (Y(1) ) − β = 3n−1 β − β = ( 3n−1 )β Fang-I Chu PSTAT 120B Probability and Statistics 3n 3n−1 β #8.15 8.15(b) (b) Derive MSE(β̂) Known: Y has pdf as f (y ) = 3β 3 y −4 , β ≤ y Definition: MSE of estimator β̂ is written as MSE (β̂) = E (β̂ − β)2 Denote β̂ = Y(1) Goal: Derive the MSE of the estimator β̂. . Fang-I Chu PSTAT 120B Probability and Statistics #8.15 8.15(b) (b) Derive MSE(β̂) Way to approach: 3n )β and By definition, use pdf, we find E (Y(1) ) = ( 3n−1 3n 2 2 E (Y(1) ) = 3n−2 β MSE(β̂) = E (β̂ − β)2 2 = E (Y(1) ) − 2βE (Y(1) ) + β 2 3n 3n β 2 − 2β · ( )β + β 2 3n − 2 3n − 1 2 = β2 (3n − 1)(3n − 2) = Fang-I Chu PSTAT 120B Probability and Statistics Exercise 8.13 8.13 We have seen that if Y has a binomial distribution with parameters n and p, then Yn is an unbiased estimator of p. To estimate the variance of Y, we generally use n( Yn )(1 − Yn ). (a) Show that the suggested estimator is a biased estimator of V (Y ) (b) Modify n( Yn )(1 − Yn ) slightly to form an unbiased estimator of V (Y ). Fang-I Chu PSTAT 120B Probability and Statistics #8.13 #8.13(a) (a) Show that the suggested estimator is a biased estimator of V (Y ) Solution: 1. Information: Y has a binomial distribution with parameters n and p Y n is an unbiased estimator of p. 2. Goal: Show that n( Yn )(1 − Y n ) is a biased estimator of V (Y ) Fang-I Chu PSTAT 120B Probability and Statistics #8.13 Solution: Way to approach: E (Y ) = np and V (Y ) = npq so E (Y 2 ) = npq + (np)2 E {n( Y 1 Y ) 1− } = E (Y ) − E (Y 2 ) n n n = np − pq − np 2 = (n − 1)pq #8.13(b) (b) Modify n( Yn )(1 − V (Y ). the estimator V (Y ) Y n ) slightly to form an unbiased estimator of n2 Y n−1 ( n )(1 − Fang-I Chu Y n ) is an unbiased estimator of PSTAT 120B Probability and Statistics Exercise 9.26 #9.26 It is sometimes relatively easy to establish consistency or lack of consistency by appealing directly to Definition 9.2, evaluating P(|θ̂n − θ| ≤ ) directly, and then showing that limn→∞ P(|θ̂n − θ| ≤ ) = 1. Let Y1 , Y2 . . . , Yn denote a random sample of size n from a uniform distribution on the interval (0, θ). If Y(n) = max(Y1 , Y2 , . . . , Yn ), we showed in Exercise 6.74 that the probability distribution function of Y(n) is given by 0 ( y )n F(n) (y ) = θ 1 Fang-I Chu y <0 0≤y ≤θ y >θ PSTAT 120B Probability and Statistics Exercise 9.26 #9.26 (a)For each n ≥ 1 and every > 0, it follows that P(|Y(n) − θ| ≤ ) = P(θ − ≤ Y(n) ≤ θ + ). If > θ verify that P(θ − ≤ Y(n) ≤ θ + ) = 1 and that, for every positive , we n . obtain P(θ − ≤ Y(n) ≤ θ + ) = 1 − (θ−) θ 1. Information: F(n) (y ) = ( yθ )n for 0 ≤ y ≤ θ P(|Y(n) − θ| ≤ ) = P(θ − ≤ Y(n) ≤ θ + ) 2. Goal: show when > θ, we have P(θ − ≤ Y(n) ≤ θ + ) = 1 and, n for > 0, we obtain P(θ − ≤ Y(n) ≤ θ + ) = 1 − (θ−) θ Fang-I Chu PSTAT 120B Probability and Statistics Exercise 9.26 #9.26 (a)For each n ≥ 1 and every > 0, it follows that P(|Y(n) − θ| ≤ ) = P(θ − ≤ Y(n) ≤ θ + ). If > θ verify that P(θ − ≤ Y(n) ≤ θ + ) = 1 and that, for every positive , we n . obtain P(θ − ≤ Y(n) ≤ θ + ) = 1 − (θ−) θ 3. Bridge: when > θ ,F(n) (θ + ) = 1 and F(n) (θ − ) = 0 (why?) n when < θ ,F(n) (θ + ) = 1 and F(n) (θ − ) = ( θ− θ ) (why?) 4. Fine tune: Therefore, for > θ, P(θ − ≤ Y(n) ≤ θ + ) = 1 n For < θ, P(θ − ≤ Y(n) ≤ θ + ) = 1 − ( θ− θ ) Fang-I Chu PSTAT 120B Probability and Statistics Exercise 9.26 #9.26 (b)Using the result from part (a), show that Y(n) is a consistent estimator for θ by showing that, for every > 0, limn→∞ P(|Y(n) − θ| ≤ ) = 1. 1. Information: From (a), > 0, n P(θ − ≤ Y(n) ≤ θ + ) = 1 − ( θ− θ ) 2. Goal: Y(n) is a consistent estimator for θ i.e. or every > 0, limn→∞ P(|Y(n) − θ| ≤ ) = 1 3. Bridge: limn→∞ P(θ − ≤ Y(n) ≤ θ + ) = limn→∞ 1 − θ− n θ 4. Fine tune: we have got our proof! Fang-I Chu PSTAT 120B Probability and Statistics =1 Exercise 7.58 7.58 Suppose that X1 , X2 , . . . , Xn and Y1 , Y2 , . . . , Yn are independent random samples from populations with means µ1 and µ2 and variances σ12 and σ22 , respectively. Show that the random variable Un = (X̄ − Ȳ ) − (µ1 − µ2 ) q (σ12 +σ22 ) n satisfies the conditions of Theorem 7.4 and thus that the distribution function of Un converges to a standard normal distribution function as n → ∞. (Hint: Consider Wi = Xi − Yi , for i = 1, 2 . . . , n.) Fang-I Chu PSTAT 120B Probability and Statistics #7.58 Hint: Our goal is to show Un ∼ Z = W̄ −µW̄ Var(W̄ ) when n → ∞. Find E (W̄ ) and Var(W̄ ) Use the given facts that Xi ’s and Yi ’s are independent when deriving Var(W̄ ) Fang-I Chu PSTAT 120B Probability and Statistics Exercise 8.8 #8.8 Suppose that Y1 , Y2 , Y3 denote a random sample from an exponential distribution with density function 1 −y (θe θ y >0 f (y ) = 0 elsewhere Consider the following five estimators of θ: θ̂1 = Y1 , θ̂2 = Y1 + Y2 , θ̂3 = Y1 + 2Y2 , θ̂4 = min(Y1 , Y2 , Y3 ), θ̂5 = Ȳ (a) Which of these estimators are unbiased? (b) Among the unbiased estimators, which has the smallest variance? Fang-I Chu PSTAT 120B Probability and Statistics # 8.8 Hint: Use linear combination property in expected value to find E (θ̂1 ), E (θ̂2 ), and E (θ̂3 ). Use the result from exercise 6.81 to find E (θ̂4 ) and Var(θ̂4 ) Use the given fact that Y1 , Y2 and Y3 are random sample. i.e. they are independent to each other. Fang-I Chu PSTAT 120B Probability and Statistics Exercise 8.12 #8.12 The reading on a voltage meter connected to a test circuit is uniformly distributed over the interval (θ, θ + 1), where θ is the true but unknown voltage of the circuit. Suppose that Y1 , Y2 , . . . , Yn denote a random sample of such readings. (a) Show that Y is a biased estimator of θ and compute the bias. (b) Find a function of Y that is an unbiased estimator of θ. (c) Find MSE(Y ) when Y is used as an estimator of θ. Fang-I Chu PSTAT 120B Probability and Statistics #8.12 Hint: Y ∼ Uniform(θ, θ + 1) i.e. fy (y ) = E (Y ) = θ + 1 2 1 2 for θ < y < θ + 1 (why?) (why?) Use the pdf of Y look up the table, what is the variance of Y ? Alternative formula for MSE: MSE(Y ) = Var(Y ) + (Bias(Y ))2 Fang-I Chu PSTAT 120B Probability and Statistics Exercise 9.5 #9.5 Suppose that Y1 , Y2 , . . . , Yn is a random sample from a normal distribution with mean µ and variance σ 2 .Two unbiased estimators of σ 2 are n σˆ12 = S 2 = 1 1 X (Yi − Ȳ )2 and σ̂22 = (Y1 − Y2 ) n−1 2 i=1 Find the efficiency of σ̂12 relative to σ̂22 Fang-I Chu PSTAT 120B Probability and Statistics #9.5 Hint: Using Theorem 7.3, we have (n−1)S 2 σ2 ∼ χ2n−1 2 E ( (n−1)S ) = n − 1 (why?) σ2 2 ) = 2(n − 1) (why?) V ( (n−1)S σ2 Use definition of relative efficiency in section 9.2. Fang-I Chu PSTAT 120B Probability and Statistics