Master s Examination Solutions Option Statistics and Probability Fall 2011

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1 Master s Examination Solutions Option Statistics and Probability Fall (STAT 41) Suppose that X, Y and Z are i.i.d. Uniform(,1). Let t > be a fixed constant. (i) Compute P ( X Y t). (ii) Compute P (XY t). (iii) Compute P ( XY Z Solution: XY t). (Hint: P ( Z XY t) = E(P ( t Z)). ) Z (i) Since Y >, P (X/Y t) = P (X ty ) = P (X/t Y ). If t 1, P (X/Y t) = P (X ty ) = If t > 1, P (X/Y t) = P (X/t Y ) = (ii) When t > 1, P (XY t) = 1; when t 1, P (XY t) = t 1 dxdy + 1 ty t/y t x/t dxdy = t/2; dydx = 1 1/(2t). dxdy = t t log t. (iii) Since X, Z >, P (XY/Z t) = P (XY tz) = P (Y/Z t/x). If t 1, P (XY/Z t) = P (XY tz) = = 1 1 (tz tz log(tz))dz = 3t/4 t log t/2; If t > 1, P (XY/Z t) = P (Y/Z t/x) = = 1 1 P (XY tz)dz (Based on part (ii)) P (Y/Z t/x)dx (1 x/(2t))dx = 1 1/(4t). (Based on part (i)) 1

2 2. (STAT 411) Suppose X N (θ, 1), where θ is unknown and X 1,..., X n is a random sample from this distribution. Consider the parameter g (θ) = P (X 2) (i) Find an unbiased estimator for g (θ) that is a function of X 1 only. Justify your answer. (ii) Show that X is complete sufficient for θ. (iii) Given the fact that the conditional p.d.f. of X 1 given X = x is N ( ) x, n 1 n, find the minimum variance unbiased estimator (mvue) of g (θ). Justify your result, and in particular mention the result that you use to establish the mvue property. Solution: (i) For a set A let us denote the indicator function { 1, if x A I A (x) =, otherwise. Then an unbiased estimator of g (θ) is h (X 1 ) = I [2, ) (X 1 ). (ii) Follows from the fact that the normal distribution belongs to the exponential family. (iii) Use Rao-Blackwell Theorem to get the mvue. Since X is complete sufficient, the mvue is the conditional expectation E ( h (X 1 ) X ) = E ( I [2, ) (X 1 ) X ) ( ( = P Y 2 Y N X, n 1 )) n ( ) n (2 x) = 1 Φ, n 1 where Φ is the standard normal distribution function. 3. (STAT 411) Suppose that X 1,..., X n are i.i.d. beta(µ, 1) and Y 1,..., Y m are i.i.d. beta(θ, 1). assume that the X i s and the Y j s are independent. (Hint: the pdf of beta(α, β) is f(x) = Γ(α+β) Γ(α)Γ(β) xα 1 (1 x) β 1, when < x < 1.) Also 2

3 (i) Construct a likelihood ratio test for H : µ = θ vs H a : µ θ. (ii) Express the LRT in (i) in terms of n T = log X i n log X i + m j=1 log Y, j and specify the distribution of T under H. Solution: (i) The likelihood function is ( n ) µ 1 ( m ) θ 1 L(µ, θ) = µ n x i θ m y j, j=1 and then the MLE s for µ and θ are ˆµ = i n, and ˆθ = m log x i j. Under H, the log y j likelihood function is ( n m ) µ 1 L(µ) = µ n+m y j, and the MLE for µ is ˆµ = n+m i log x i+ j log y j x i j=1. The test statistic for the LRT is Λ = sup ( n )ˆµ ˆµ ( H L sup H Ha L = ˆµn+m m ˆµ nˆθ x i m j=1 y j )ˆµ ˆθ, and we reject H if Λ c. (ii) Substituting ˆµ, ˆθ and ˆµ yields that ( n x i) µ µ ( m j=1 y j) µ θ = 1 and Λ = ˆµn+m (m + n)m+n = (1 T ) ˆµ nˆθ m T n. m m m n n Therefore, Λ c is equivalent to T c 1 or T c 2. Note that simple transformation implies that log X i exp(µ) and log Y j exp(θ), so T = U/(U + V ) where U and V are independent, U Gamma(n, 1/µ) and V Gamma(m, 1/θ). Then under H, T beta(n, m). 3

4 4. (STAT 416) A study gives body weights of diabetic and normal mice. Diabetic 42, 44, 38, 52 Normal 34, 43, 35, 33, 26 Given the significant level.5, the research interested if there is evidence to show that diabetic mice are heavier than normal mice. (a). If use Median Test procedure, specify your test hypotheses and test statistic formula. (b). Is there any significant evidence between the median body weights of the two groups based on the procedure in (a)? (c). If use Control Median Test, which one should be selected as a control sample? What is the observed test statistic for control median test? Solution: Denote medians of X and Y by M X, M Y respectively. (a). Hypotheses: H : M Y = M X vs. H 1 : M Y < M X (left-tailed test) Sign-test: U = m I {Xi <M z } where M Z is the pooled median, M Z = 38. (b). Sample size m = 4, n = 5, N = = 9, [9/2] = 4 Observed statistic u o = 4 I {Xi <38} =, then p-value is ( 4 ) ( 5 ) P (U ) = 4 ( ) = P-value.4 < α =.5, reject H, i.e. conclude that the diabetic mice is heavier than regular mice. (c). If controlled test is used, sample Y should be the control group, M Y = Median{34, 43, 35, 33, 26} = 34. Controlled median test statistic: V = m I {Xi <M Y } = 4 I {Xi <34} =. 4

5 5. (STAT 431) Let the size of a finite population for the sample survey be N = 4. A- What will be the HT estimator and its unbiased variance estimator of the population mean if we use SRS (4, 2) sampling plan. B- Suggest a homogeneous linear unbiased estimator of the population mean which is NOT the HT estimator. Provide an unbiased variance estimator of your proposed estimator of the pollution mean. C- Suppose upon implementation of SRS (4, 2) the sample 3, 4 is chosen for the survey. What will be the forms of your 4 estimators in A and B above for the survey data Y3 and Y4? Answer: A- HT estimator of the population mean will be the sample mean and its variance will be (1 2/4)S 2 /2 = (1/2 1/4)S 2, and an unbiased estimator of this variance will be (1/2 1/4)s 2, where s 2 will be the sample variance. B- Use the idea in Theorem 2.4 in the Book by Hedayat and Sinha and set up the following system of linear equations. Based on sample {1,2}...a Y1 + b Y2 with prob. 1/6 Based on sample {1,3}...b Y1 + c Y3 with prob. 1/6 Based on sample {1,4}...d Y1 + e Y4 with prob. 1/6 Based on sample {2,3}...f Y2 + g Y3 with prob. 1/6 Based on sample {2,4}...h Y2 + i Y4 with prob. 1/6 Based on sample {3,4}...j Y3 + k Y4 with prob. 1/6 Now find solutions for the unknown coefficients a,b,...,k such that Y1 (a+b+d)/6 = Y1/4,... and Y4 (e+i+k)/6 = Y4/4. There are lots of solutions beyond the ones by provided by HT estimator. C- Just use the data and pug them into the solutions provided. Of course in exam you were supposed to give the details. 6. (STAT 461) There is a grocery store. During each period, the number of customers entering in the store follows the distribution of P (ξ = ) = 1/4 and P (ξ = 1) = 3/4. The total number of periods that each customer stays in the store follows the distribution of P (Z = k) = α(1 α) k 1, k = 1, 2,.... Specify the transition probabilities for the Markov chain whose state is the number of customers in the store at the start of each period. 5

6 Solution: Since P (Z = k + 1 Z > k) = α for any k, a customer leaves the store with probability α at the next period no matter how long he/she has been in the store. If X n = i, then there will be m of them leave with probability of ( ) i α m (1 α) i m := l im. m At the same time, there will be one customer coming in with probability of 3/4 and none with probability 1/4. Denote P ij = P (X n+1 = j X n = i). (i) For i = and j, we have P j = P (ξ = j). (ii) For i 1 and j = i + 1, we have P ij = 3/4(1 α) i. (iii) For i 1 and 1 j i, we have P ij = 1/4l i,i j + 3/4l i,i j+1. (iv) For i 1 and j =, we have P ij = α i /4. 7. (STAT 471) A carpenter sends orders to Furniture Wholesale stores to supply him 35 planks of length 7 feet, 48 planks of length 5 feet and 14 planks of length 4 feet. The Furniture Wholesale stores sends just N planks of length 17 feet and says you should be able to meet your order by cutting them appropriately into planks of smaller sizes to meet your demand and this will be the cheapest way to meet your orders. Before cutting the planks, the carpenter thought of the following. With each 17 feet plank cut two 7 feet plank and throw the rest, cut three 5 feet plank and throw the rest, cut four 4 feet plank and throw the rest. He found this needed in all ( ) = 38 planks of the 17 feet size. Not by trial and error, but only by using the revised simplex method and knapsack problem, find a new cutting pattern for the carpenter which replaces one of the above cutting patterns by the new cutting pattern which reduces the number of planks needed. Is it equal to the number N of planks they supplied? Solution: See pages 7,8,9. 8. (STAT 471) Write down the dual to the following LP problem: max 3x 1 5x 2 + 4x 3 + x 4 x 5 such that x 1 x 2 + x 3 = 7 4x 2 x 3 + 5x 4 5 5x 3 2x x 1 + 4x 3 8x 4 = 23 x 1 unrestricted, x 2, x 3, x 4 unrestricted Solution: See page 9. 6

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12 9. (STAT 473) Joe and Bob are eight year old trying to exchange their gifts accumulated during Halloween trick or treat walk in their neighborhood. Joe got 1 M&M chocolate packet, 1 peanut toffee. Bob got an apple toffee, and 1 orange. They value the items as follows (their utilities measured in each one s scales). M & M Peanut toffee Apple toffee Orange Joe Bob Plot the Nash bargaining set and find the Nash bargaining solution. Solution: See pages 1, (STAT 481) (1). In order to fit the data with a simple linear regression model Y i = β + β 1 x i + ε i, i = 1,..., n, where ε i s are i.i.d. N (, σ 2 ), write down the least square criterion, derive the normal equation and solve it to find the ordinary least square estimators, ˆβ, ˆβ 1. (2). A multiple regression model is Y n 1 = X n (p+1) β (p+1) 1 +ε n 1, where independent errors follow normal distributions, i.e. ε N n (, Σ = σ 2 I n ) and β Y 1 1 x 1,1... x 1,p β 1 Y =., X =......, β =.. Y n 1 x n,1... x n,p β p A least square criterion is defined as S (β) = (Y Xβ) (Y Xβ). Given) that p n, ) find the least square estimator ˆβ which minimizes S (β), and calculate E (ˆβ and V ar (ˆβ. (3). Suppose that the variance matrix of the errors in (2) is a diagonal matrix Σ = diag(σ 2 1,..., σ 2 n). To obtain unbiased estimators, weighted least square criterion is defined as S w (β) = (Y Xβ) W (Y Xβ), 12

13 please choose an appropriate weight matrix ) W. Is the weighted least square ) estimator ˆβ w unbiased? Calculate its variance V ar (ˆβw and compare it with V ar (ˆβ. Solution: (1). Least squre criterion: S (β + β 1 ) = n {Y i β β 1 x i } 2. Normal Equation: Least square estimator: { S(β,β 1 ) β = S(β,β 1 ) β 1 = n nβ + β 1 x i = n y i n n x i + β 1 x 2 i = n x i y i β ˆβ 1 = n (x i x) y i n (x i x) 2, ˆβ = ȳ ˆβ 1 x. (2) Take derivative w.r.t. β : S (β) β = { (Y Xβ) (Y Xβ) } = 2X Y+2 (X Xβ) β Normal equation S (β) β thus OLS solution: ˆβ = (X X) 1 X Y. = X Xβ = X Y E (Y) = E (Xβ + ε) = Xβ,V ar (Y) = V ar (ε) = Σ =σ 2 I n ) E (ˆβ = (X X) 1 X E [Y] = (X X) 1 X E (Xβ + ε) = β ) V ar (ˆβ = (X X) 1 X V ar (Y) [ (X X) 1 X ] = σ 2 (X X) 1. (3). W = Σ 1 = diag(1/σ 2 1,..., 1/σ 2 n). S w (β) β = 2X WY+2 (X WXβ) = ˆβ w = (X WX) 1 X WY. and ) E (ˆβw = (X WX) 1 X WE (Y) = (X WX) 1 X W Xβ = β, 13

14 ) V ar (ˆβw [ ] = (X WX) 1 X WV ar (Y) (X WX) 1 X W = (X WX) 1 X W Σ WX (X WX) 1 = (X WX) 1 X WX (X WX) 1 because W = 1 = (X WX) 1 ) In (2), W = Σ 1 = σ 2 I n, and V ar (ˆβ = (X σ 2 I n X) 1 = σ 2 (X X) (STAT 481) In a study of the compressive strength of concrete, five batches of concrete are mixed. Each batch is just large enough to produce three molded concrete cylinders. Three treatments corresponding to three different drying methods are randomly assigned to the three concrete cylinders from each batch. The measured compressive strengthes (1 pounds per square inch) of the 15 concrete cylinders are listed in the table below. Batch Treatment Treatment Mean A B C Batch Mean Although great care is taken to achieve uniformity among the batches, one can expect variability from batch to batch. A fixed-effect model is assumed as follows for the observations: Y ij = µ + τ i + β j + ϵ ij, i = 1, 2, 3; j = 1, 2,..., 5, where µ is an overall mean, τ 1, τ 2, τ 3 are the treatment effects satisfying 3 τ i =, β 1, β 2,..., β 5 are the batch effects satisfying 5 j=1 β j =, and ϵ ij are independent N(, σ 2 ) random variables. (a) Estimate the parameters µ, τ 1, τ 2, τ 3, β 1, β 2,..., β 5. (b) Complete the ANOVA table below. Source SS df MS F Treatment Batch Error 46.8 Total

15 Determine the value of the statistic used to test if all the treatment effects are the same (null hypothesis) and specify the distribution of the statistic under the null hypothesis. Solution: (a) With usual notations, the estimates ˆµ = Ȳ = 1 ( ) = ˆτ 1 = Ȳ1 Ȳ = = 1.2 ˆτ 2 = Ȳ2 Ȳ = = 3.4 ˆτ 3 = Ȳ3 Ȳ = = 2.2 ˆβ 1 = Ȳ 1 Ȳ = = 7.6 ˆβ 2 = Ȳ 2 Ȳ = = 1.6 ˆβ 3 = Ȳ 3 Ȳ = = 2.4 ˆβ 4 = Ȳ 4 Ȳ = =.6 ˆβ 5 = Ȳ 5 Ȳ = = 7.4 (b) First, Y 1 = = 236, Y 2 = = 259, Y 3 = = 231, i = 726. So The complete ANOVA table is SS Treatment = 1 5 ( ) = 89.2 SS Block = = Source SS df MS F Treatment Batch Error Total j Y ij = Here 7.62 is the value of the F -statistic used for testing if all treatment effects are the same. Under null hypothesis, the distribution of the F -statistic follows F (2, 8) distribution. 15