| Homework | Question | Due Day | Solution | Grader |
| 1 |
Ch 1. #2, #59, #65, #68, #78(a)(b) Ch 2. #33, #40 Ch 3. #7(also examine whether X, Y are independent random variables and explain why), #8, #21, #25. |
Mar 19 | sol | 沈子翔, 曹詠勛 |
| 2 |
Ch 3. #44, #64 [Hint: the pdf of exponential distribution with parameter λ is λexp(-λx), for x≥0 and 0, otherwise.], #70, #77, Ch 4. #13, #36, #49, #50, #54, #60. |
Mar 26 | sol | 顏珮雅, 劉必翔 |
| 3 |
Ch 2. #21 (Note. This is called memoryless property), #30 [Hint: First of all, try to find k0 such that p(k)/p(k+1)≥1 for any k≥k0 and p(k)/p(k+1)<1 for any k<k0, where p(·) is the pmf of Poisson.] Ch 3. #22 [Hint: You can first show that: If X1~Poisson(λ1), X2~Poisson(λ2), and X1, X2 are independent, then the conditional distribution of X1 given that X1+X2=n is binomial distribution B(n, λ1/(λ1+λ2)).], #26 [Hint: use beta function in LNp.75], Ch 4. #61, #75 (For practice purpose, you must use the law of total expectation and variance decomposition to find the mean and variance, respectively. Other answers are NOT acceptable.), #77, #80, #92 [Hint. Use the law of total expectation (check LNp.67 and LNp.73 for the mgfs of Poisson and gamma, respectively) to find the mgf of α+X, and compare it with the mgf of negative binomial distribution given in LNp.63], #100 [Hint. Apply the δ-method to find the approximate mean and variance.] |
Apr 07 | sol | 周右林, 陳信諺 |
| 4 |
Ch 2. #61, Ch 4. #20 [Hint: try sum-to-one method], #89 (For practice purpose, you must use moment generating function to show it and to find its mean and variance. Other solutions are not acceptable.) Ch 5. #1 [Hint: use Chebyshev's inequality and imitate the proof in LNp.96], #5 [Hint: use the theorem: if an→a, then (1+an/n)n→ea. The mgfs of binomial and Poisson distributions are given in LNp.60 and LNp.67, respectively.], #16, #17 [Hint: use CLT], #28 Ch 6. #8 [Hint: You can use mgf to relate the distributions of 2X and 2Y to the chi-square distribution], #11 |
Apr 14 |
sol |
沈子翔, 曹詠勛 |
| 5 |
Ch 8. #7(a)(b) [Hint: the 1st moment of geometric distribution is given in LN, ch1-6, p.62], #9 [Hint: check the diagram in LNp.12], #16(a)(b) [Hint: The 1st moment is zero, and you can use gamma pdf and gamma function given in LN, ch1-6, p.73&74, to find the 2nd moment], #19(a)(b), #21(a)(b) [Hint: For (a), let Y=X-θ, then Y~exponential(1), i.e., 1=E(Y)=E(X-θ). For (b), note that xi ≥ θ for i=1,...,n, i.e., x(1) ≥ θ.], #26 [Hint: Try to use the hypergeometric distribution given in LN, ch1-6, p.68, to model the data.], #50(a)(b) [Hint: Try to use gamma pdf and gamma function given in LN, ch1-6, p.73&74, to find the 1st moment.], #51 [Hint: In this case, median is the (m+1)th smallest observation, i.e., X(m+1). You can first show that if X(m+2)≥θ≥X(m) , then Σ|Xi-θ|=|X(m+1)-θ|+(X(m+2)-X(m))+(X(m+3)-X(m-1))+...+(X(2m+1)-X(1)). Then, try to generalize the result to obtain a general statement.], #60(a)(b)(c)(d) [Hint: For (b) and (d), use the results given in LN, ch1-6, p.73&74] |
Apr 21 | sol | 顏珮雅, 劉必翔 |
| 6 |
Ch 8. #7(c), also, obtain the Fisher information of X and identify the asymptotic sampling distribution of the MLE [Hint: the mean of geometric distribution is given in LN, ch1-6, p.62], #16(c), also, obtain the Fisher information of the i.i.d. sample and identify the asymptotic sampling distribution of the MLE [Hint: you can use gamma pdf and gamma function given in LN, ch1-6, p.73&74, to find E|Xi| or E(Xi2).], #18(d), also show that the pdfs form an exponential family and find a sufficient and complete statistic, #21(c), also show that X(1) is complete by definition and answer whether the pdfs form an exponential family, #47(a)(b)(c), also, obtain the Fisher information of the i.i.d. sample and identify the asymptotic sampling distribution of the MLE, #49, #50(c), also, obtain the Fisher information of the i.i.d. sample and identify the asymptotic sampling distribution of the MLE [Hint: Try to use gamma function and gamma pdf given in LN, ch1-6, p.73&74, to find E(Xi2).], #58(a)(b), also, identify the asymptotic sampling distribution of the MLE [Hint: the log-likelihood function is given in LN, CH8, p.24-25 and notice that the marginal distributions are X1~B(n, (1-θ)2), X2~B(n, 2θ(1-θ)), X3~B(n, θ2).], #69, #71, also show the pdfs form an exponential family and find a sufficient and complete statistic, #72, also show that the gamma distribution form a 2-parameter exponential family and show that ΠXi and ΣXi are sufficient and complete. |
Apr 28 | sol | 陳信諺, 廖芳翊 |
| 7 | May 05 | sol | 沈子翔, 曹詠勛 | |
| 8 | May 14 |
sol |
顏珮雅, 劉必翔 | |
| 9 |
Ch 9. #1, #2, #3, #4(a)(b) [Hint: apply Neyman-Pearson lemma.], #5, also if false, please explain why, #7 [Hint: apply Neyman-Pearson lemma.], #19(b)(c)(d) [Hint: apply Neyman-Pearson lemma.], #29, #30. |
May 26 | 陳信諺, 廖芳翊 | |
| 10 | Jun 02 | 沈子翔, 曹詠勛 | ||
| 11 |
Ch 9. #12, #13(a)(b)(c), #24
#26, also if false, please explain why. #37 [Hint: you may model the numbers of deaths as independent random variables, each distributed as Poisson P(λi), i=1,..., 12, and examine whether λi's are equal], #40 [Hint: p1+p2=1 and X1+X2=n], #43 [Hint: you can use GLR test for goodness-of-fit or Pearson's Chi-square test, and compare your answer with the solution given in textbook, page A40], #44 [Hint: Let O1, O2, and O3 be the numbers of AA, Aa, and aa, respectively. Then, (O1, O2, O3)~Multinomial(n, (1-θ)2, 2θ(1-θ), θ2). The MLE of θ, where 0<θ<1, is (2X3+X2)/(2n) (see LN, CH8, p.25). Use O1, O2, O3 to derive the likelihood ratio test statistic. By the way, you may compare this question and the Example 7.19 given in LN, CH9, p.43. Try to find the differences on their H0 and HA (or the differences on their Ω0 and Ω).]. |
Jun 09 | ||
| 12 |
textbook (2nd ed.) Ch 15. (Problem statements) (turn-in questions) Ch 15. #1, #11, #13, #15, #23, #24, #28, #29.
(no-need-to-turn-in question) Ch 15. #8, #9. |
Jun 16 |
sol |
陳信諺, 廖芳翊 |
,