Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 1. 1, page 2, in Table of Contents: Section 33 is Specific Examples of Markov Processes.

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1 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 1 1, page 2, in Table of Contents: Section 33 is Specific Examples of Markov Processes. 1, page 24, solution ot the exercise: (60%)(100) + (40%)(200) = , p.61, sol. 2.62: = 1, p.64, sol. 2.69: θ 2 2 = θ 2 = 1.5. n-1 1, p.96: Comparing the two results for t = 1: Γ[n ; λ] = 1 - e - λ λ i / i! = e - λ λ i / i!. i=0 i=n 1, page 135, last two lines: S(t) dt = (e t + e t - e t - e t + e t ) dt = 0 0 1/ /0.03-1/0.08-1/ /0.10 = = , page 213, solution 6.5 is correct but needs much more explanation: Prob[1 claim good] = 0.05 e = Prob[1 claim bad] = 0.1 e -0.1 = By Bayes Theorem, the probability that an insured that had one claim in year one is good is: Prob[Good] Prob[1 claim Good] Prob[Good] Prob[1 claim Good] + Prob[Bad] Prob[1 claim Bad] = (2 / 3)( ) (2 / 3)( ) + (1/ 3)( ) = 51.25%. By Bayes Theorem, the probability that an insured that had one claim in year one is bad is: (1/ 3)( ) = 48.75% = %. (2 / 3)( ) + (1/ 3)( ) Prob[2 claims good] = e /2 = Prob[2 claims bad] = e -0.1 /2 = Therefore, this insuredʼs probability of having 2 claims in year 2 is: (51.25%)( ) + (48.75%)( ) = 0.281%. Comment: Bayes Theorem is in Mahlerʼs Guide to Bayes Analysis and Conjugate Priors. 1, p.311, very last line of sol : An Exponential with mean θ, has a second moment of 2θ 2. 1, p. 343, solution to the first exercise: (82-37)/100 = 45%. 1, p. 346, Q. 12.7: a Poisson Process

2 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 2 5 1, p. 365, solution to the exercise: m(5) = (4 + t) dt = /2 = , p. 385: Solutions 4.19 and 4.20 are reversed in how I numbered of them. 1, p. 395, 4th line from the bottom: = {-e e (0-2)} + e 0.3 (e e -0.8 ) = e -0.5 = The final solution to the exercise is OK. 1, p. 396, 2nd line: 2-2e e e -0.8 = e -0.5 = , p. 399, last line of solution ot the first exercise, there is a missing decimal place: ( ) 1, p. 401, 3rd line from the bottom, a missing closing bracket: Φ[( )/ ] 1, p. 414, solution to the exercise at the bottom: on (0, 5) rather than (0, T). 1, p. 443, next to last paragraph: (0, 1, 0)P = (0.1, 0.75, 0.15) = probabilities after one day, having started in state 2. (0, 1, 0)P 2 = (0.1, 0.75, 0.15)P = (0.155, , ) = probabilities after two days. 1, p.447, in the solution to the exercise: (0.75)(0.25) = , p.481, in solutions : (0.1265, , ) P = ( , , ). ( , , ) P = ( , , ). 1, p. 544, Sol : B. (0.2)(25) + (0.5)(50) + (0.3)(100) = 60 1, p. 544, Sol : the letter should be A. Also, (0.54, 0.35, 0.11)P = (0.4525, 0.443, ). 1, p. 610, in the exercise: the transition matrix should be Also the first balance equation should be: 0.8π π 3 = π 1.

3 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 3 Good 1, p. 615, Q : the transition matrix should be: Typical Poor , p. 620, Q , the 3rd and 4th lines are wrong: ʻIf it did not rain two days ago but rained yesterday, the probability that it will rain today is 60%. If it rained two days ago but not yesterday, the probability that it will rain today is 50%. Also, choice E should be B C M B 1/ 2 1/ 2 0 1, p. 623, Q , the transition matrix is mislabeled: C 1/ 4 1/ 2 1/ 4 M 0 1/ 3 2 / 3 1, p. 634, Sol : π 3 = 12/43. Final answer is OK. 1, p. 655, next to last line: Alternately, (0, 0, 0, 1, 0) P = 1, p. 671, 4th paragraph: For j < N, let I j = 1 1, pp , solutions : S = (I - P T ) -1 = 4 2. The final solutions are OK , p. 682, solution 23.13: 4.10/6.02 = The final solutions are OK. 6 1, p. 687, last line, the first summation should go to infinity: k, π2 k2 k=1 1, p. 690, solution to the second exercise: P = , p. 712: In the 3rd row and last column of the matrix, there is a missing decimal point: , p. 712, in the line below the big matrix: P n goes to a matrix 1, p. 713, line 2: (P 32 ) 2,

4 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 4 1, p. 713, line 5: (P 32 ) 1,5 1, p. 726, solutions 26.4: 1 - (0.52 / 0.48)23. Final solution is OK. 1 - (0.52 / 0.48) 25 1, p. 749, second to last line, i and j need to be subscripts: P ij 1, p. 750, second to last line, i and j need to be subscripts: P ij = Q ij 1, p. 761, next to last line: 5. The new 2nd element is: = , p. 767, 4th paragraph: Min[1, b( )/b( )] 1, p. 788, solution to exercise, missing decimal place: 0.45P 2 = 5 1, p. 791, next to last line, t missing in the exponent of the integrand: e t dt 0 1, p. 795, solution to last exercise: The chance of exactly one visit to the SNF = (chance of transition to SNF permanent from ILU) + (chance of transition ILU SNF temp.) (chance of transition SNF temp. SNF perm.) + (chance of transition ILU SNF temp.) (chance of transition SNF temp. Death) + (chance of revisit to ILU)(chance of death in ILU) = (5/25) + (12/25)(7/24) + (12/25)(12/24) + (1/10)(8/25) = The is OK. 1, p. 798, second bullet: A new resident starts off in an ILU. 1, p. 825, near the bottom: d P 21(t) dt = q 21 P 11 (t) - q 21 P 21 (t). 1, p. 826, near the bottom, missing minus signs: Thus, d P 12(t) dt = -14be -14t. -14be -14t = 1, p. 831, third paragraph: q 12 = , p. 838, third paragraph:

5 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 5 1, p. 838, last page: 0.5v 1 = (λ - 0.2)(λ - 0.7)v 1 /0.3. 1, p. 846 & p. 855, given matrices in Q to do not have their rows add to 1, change them to: P ij (1) =. For example, P (1) = P ij (2) = For example, P (2) = Then the solution is: P ij (3) = , p. 853, solution 32.16: (1, 0, 0) P = (0.965, 0.025, 0.01) 1, p. 861, solution to exercise: d P ij (t) dt = (0.08)(j-1)P i,j-1 (t) + (0.05)(j+1)P i,j+1 (t) j P i,j (t) 2 1, p. 871, sol. 33.6: P 1,2 (20) = exp[-λ k 20] = exp[-λ 1 20] k=1 λ 2 λ 2 - λ 1 + exp[-λ 2 20] λ m λ - exp[-λ λ m - λ k 20] m=1, m k k m λ m - λ k k=1 m=1, m k λ 1 λ 1 - λ 2 - exp[-λ 1 20] = e -(20)(0.02) (7/5) + e -(20)(0.07) (-2/5) - e -(20)(0.02) = 16.95%. Alternately, the chance that there is a birth at time 0 t < 20 and no birth thereafter by time 20 is: e t e (20 - t) 20 dt = 0.02 e -1.4 e 0.05 t dt = 0.02 e -1.4 (e 1-1)/0.05 = 16.95%. 0 0

6 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 6 1, p. 884: As t approaches infinity, S(t) approaches a Normal Distribution. 2, page 7, middle of page: Var[X 1 ] + Var[X 2 ] + Var[X 3 ] = 3 VaR[X]. 2, page 19, middle of page: (α+1) = 2.393α. 2, page 50, line 8: with α = 4 2, p.66, 2nd line from the bottom, there is a missing summation sign: ln[f(x i )] θ 2, p.68, solution to the last exercise: µ^ = lnx i N. σ2 = (lnx i - µ^ ) 2 N. n 2, p.69, above the exercise: 0 = n/a + ln[x i ] - n ln[θ]. n/a = - ln[x i / θ]. a^ = i=1 n i=1 -n. ln[x i / θ] 2, p.70, fifth line: n ln[a] + n ln[a+1] + (a-1) ln[x i ] + ln[1 - x i / θ] - n a ln[θ]. 2, page 73, next to last paragraph, there are should not be an n before ln(3): This differs by: 2ln(x) + ln(3), from the log density of the Weibull (for τ = 3) of: -θ -3 x 3 + 2ln(x) - 3ln(θ) + ln(3). 2, page 78, reword Q. 4.10: What is the maximum likelihood estimate of θ? 2, page 79, reword Q. 4.14: What is the fitted value of α? 2, page 123, solution 4.36: ln f(x) = -(x - θ) 2 /2 + ln(2/π)/2. loglikelihood is: -Σ(x i - θ) 2 /2 + n ln(2/π)/2. 2, page 151, last line of the fourth paragraph: 84,500/θ 2-2/θ = 0. 2, page 182, sol. 5.14: ln[f(x)/s(1000)] = -x/θ - ln(θ) /θ. Loglikelihood = -1500/θ /θ -5000/θ - 15,000/θ - 4ln(θ) + (4)(1000/θ) = -20,500/θ - 4ln(θ).

7 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 7 2, p. 210, sol. to first exercise: likelihood = f(x 1 )f(x 2 )... f(x n ) = x 1 2 x x n 2 exp[-σx i /θ] / {2 n θ 3n } 2, p. 212, 7th line from bottom: = -0.5Σx i 2 /σ 2 + µσx i /σ 2-0.5n µ 2 /σ 2 - nln(σ) - (n/2)ln(2π). 2, p. 216, sol. 6.5&.6.6: loglikelihood = {µσx i /σ 2-0.5nµ 2 /σ 2 } - 0.5Σx i 2 /σ 2 - nln(σ) - (n/2)ln(2π). 2, p. 229, line 5: If each X i has mean µ and variance σ 2, then X n has mean µ and variance σ 2 /n. 2, p. 257, the values on the x-axis of the graph should be 5.8, 7, and 8.2. Density % 2.28% x 2, p.278, 3rd paragraph: F(6.63) = , p.280, second paragraph: 1 exp[-y 2 / 2] dy = 1-2t - 2π 1 1-2t. Thus the m.g.f. of a Chi-Square Distribution with one degree of freedom 2, p.281, footnote: M(t) = 1/(1-2t) ν /2

8 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 8 2, p.282, sol. to first exercise: = e -x/2 x , p.283: More generally, given a sample of size n from a Normal Distribution with known mean µ, one confidence interval for σ 2 covering probability 1 - α is: Σ(X i - µ) 2 / χ 1 2 α/2 (n) to Σ(X i - µ) 2 / χ 2 α/2 (n), where χ α/2 2 (n) is the value at which the Chi-Square Distribution with n d.f. is α/2. 0 to Σ(X i - µ) 2 / χ 2 α (n), and Σ(X i - µ) 2 / χ 1 2 α (n) to are also each confidence intervals for σ 2 covering probability 1 - α. 2, p.284, middle of page: Coefficient of Variation = 2ν + 4λ / (ν + λ). 2, p.304, third paragraph: X = -(Y 1 + Y 2 )σ. 2, p.304, fourth paragraph: = (4E[X 1 2 ]/9 + E[X 2 2 ]/9 + E[X 3 2 ]/9 2, p.304, last paragraph: (-2E[X 1 2 ]/9-2E[X 2 2 ]/9 + E[X 3 2 ]/9 2, p.306, 6th line from the bottom, no square on y 1 : exp[-(y 1-2µ) 2 2, p.308 and p.1225: Given a sample of size n from a Normal Distribution, a confidence interval for σ 2 covering probability 1 - α is: (n-1)s 2 / χ 1 2 α/2 (n 1) to (n-1)s 2 / χ 2 α/2 (n 1), where χ α/2 2 (n 1) is the value at which the Chi-Square Distribution with n - 1 d.f. is α/2. 0 to (n-1)s 2 / χ 2 α (n 1), and (n-1)s2 / χ 1 2 α (n 1) to are also each confidence intervals for σ 2 covering probability 1 - α. 2, p.309, final line: (100/ )S 2 = 0.804S 2 to (100/ )S 2 =1.283S 2. 2, p.321, sol , final line, there should be no 3 in front of S 2 : Prob[S ] = , p. 374, final line of solution 12.9: χ 2 = {(O 1 - np 1 )/ np 1 (1- p 1 ) } 2 approaches 2, p. 398, 6th line from the bottom, the last probability is wrong: 2800/8600

9 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 9 2, p.405, Q.13.6, the values shown in the totals column are wrong; they should instead be: 111, ,015 2, p.442: The definition of a t-distribution with ν degrees of freedom 2, p.442: should be ν inside the square root in the denominator rather than n, Z χ2 / ν 2, p.443: Prob[t T] 2, p.444, line 4: f Y (y) = 2, p.444: Prob[t T] 2, p.444: f T (t) = y / ν f X (t y / ν) f Y (y) dy = y=0 1 ν y=0 y exp[-t2 y / (2ν)] 2π e - y / 2 y ν / ν / 2 Γ(ν / 2) dy 2, p.444, footnote 115: β[a, b] = Γ[a]Γ[b]} / Γ[a + b] 2, p.452, first line of solution 14.2: Prob[t < -3.5] > 2%/2 2, p. 471, line 7: there is 5% area above , p. 534, sol. 18.8: variance: σ 2 = σ X 2 + σ Y 2-2ρσ x σ y. Final solution is OK. 2, p. 571, footnote 155: sample size of 120 2, p. 572, 4 lines from the bottom: with mean: (10-7) / (σ/4) = 12/σ.

10 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 10 2, p. 573: Thus, the ratio (X - 7)/ (σ / 4) S / σ follows a noncentral t-distribution with ν = 15 and δ = 12/σ. (X - 7)/ (σ / 4) Prob[ S / σ > µ = 10] = Survival Function at of a noncentral t-distribution with ν = 15 and δ = 12/σ. Let us assume that the sample standard deviation is 5, so that we estimate δ = 12/5 = 2.4. Then using a computer, this probability and thus the power is 74.1%. Here is a graph of the power function for this test of H 0 : µ 7 versus H 1 : µ > 7: power mu As µ, the power approaches 1. As µ 7, the power approaches the significance of 5%. In general, the power is the Survival Function at the critical value of the t-test, of a noncentral t-distribution with ν degrees of freedom and δ = n (µ - µ 0 )/σ, where we estimate σ via S the sample standard deviation.

11 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 11 2, p. 691, line 5: since < 4 < 4.325, we reject at 20% and do not reject at 10%.. 2, p. 720: Analysis of Covariance is discussed in Section 20 of Mahlerʼs Guide to Regression. 2, p. 750,: IOA 01,9/01 Q. 11 should be numbered rather than Unfortunately, therefore all of the following problems need to have their numbers increased by one. So for example CAS S, 11/16, Q.39 should be numbered rather than , p. 781, 3rd line from the bottom: µ 1 > µ 0 or µ 0 > µ 1. 2, p. 782, 8th line: Reject when this difference is large: -5 + {ln(0.10) - ln(0.05)}σx i b. 2, p. 795, sol. 25.4: Loglikelihood for H 1 minus loglikelihood for H 0 is: +n + {ln(2) - ln(3)}σx i. Reject when this difference is large: +n + {ln(2) - ln(3)}σx i b. ( )Σx i (b - n). Final solution is OK. 2, p. 816, sol. ot the exercise :f(x) = τ(x/θ) τ exp(-(x/θ) τ ) /x. ln f(x) = ln(τ) + (τ-1)ln(x) - (x/θ) τ - τln(θ). The loglikelihood is: n ln(τ) + (τ-1)σln(x i ) - Σ(x i /θ) τ - nτln(θ). 2, p. 840, sol. 26.1, 2nd line of comment, no divided by sign: Prob[ X / n H 0 ] 2, p. 862, the last two paragraphs: 0.4 = F(Q 0.4 ) = 1 - exp[-q 0.4 /1000]. α = F(Q α ) = 1 - exp[-q α /θ]. 2, p. 865, first line: F -1 [ i n+ 1 )]. 2, p. 877, sol. 27.2: 0.37 = F(Q 0.37 ) = 1 - exp[-q 0.37 /400]. Q 0.37 = -400 ln[0.63] = , p. 887, solution to the next to last exercise: G 1 (x) = F(x) j S(x) 7 - j j j=0 2, p. 892, 2nd exercise: a uniform distribution on [d, u]. 2, p. 896, middle of the page: {Γ(s + 2)Γ(N s) / Γ(N + 3)} {Γ(r + 1)Γ(s - r) / Γ(s + 1)} {Γ(r)Γ(s - r)/ Γ(s)} {Γ(s)Γ(N s)/ Γ(N+ 1)}

12 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 12 2, p. 898, comment ot the first exercise: E[X (N) ] = N/(N+1). 2, p. 899, 3rd paragraph, missing superscript j: N N (x / T) j (1 - x / T) j N - j j=r 2, p. 899, solution to first exercise, missing superscript j: N N (x / T) j (1 - x / T) j N - j j=r 2, p. 901, second line: 2 F(x) f(x) = 2 (e -x/θ - e -2x/θ )/θ. 2, p. 902, next to last paragraph, there is an extra divided by 2: X (2) = X (1) + W 2 = W 1 /2 + W 2. 2, p. 902, last paragraph, there is an extra divided by 3: X (2) = X (1) + W 2 /2 = W 1 /3 + W 2 /2 2, p. 905, last line: 1/ N2 π2 / 6 = 6 N π 2, p. 936, Q : The lowest two values are excluded , p. 1017, sol , third line: 1 - Φ[ ] Final solution is OK , p. 1024, 4th line from bottom: = Var[X 1 + X 2 + X 3 + X 4 + X 5 ]/5 2 c ε 2, page 1032, 4th line from bottom, no minus sign: ε 2 f(ψ n ) dψ n + ε 2 f(ψ n ) dψ n - c+ε 2, p. 1045, 3rd paragraph: (α - x i) 2 α = 2 Σ(α x i ), 2, p. 1050, Q.30.14: estimator of ω. 2, p. 1115, line 5: -1 n E [ 2 ln f(x) / θ 2 ]

13 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 13 2, p. 1117, 6 lines from the bottom: 1 E [( lnl / θ) 2 ] 2, p. 1117, 3 lines from the bottom: -h'(θ) 2 n E [ 2 ln f(x) / θ 2 ] 2, p. 1138, line 6: -3σ2 σ4 + 1/σ 2 2, p. 1139, 3rd line from the bottom: 0.03 / (9.5)(0.0002) 2, p. 1141, Q. 32.6: D. at least but less than E. at least , p. 1185, line 8: C(x) = -ln(θ + x) 2, p. 1188, line 13: k r = d k r -1 dψ 2, p. 1193, sol. 34.5: (x+r) ln(1 + β) D(β) = -r ln(1 + β). 2, p. 1198, line 10: -1 n E [ 2 ln f(x) / θ 2 ] 2, p. 1239: -1 n E [ 2 ln f(x) / θ 2 ] 2, p. 1239: Variance of Estimated Single Parameters, Maximum Likelihood (Section 32) 3, p.130, first line: = {( ) - ( )} 3, p.171, sol , first line: λ > 0 3, p.184, last line: -e - t - t e - t t = x ] t = 0

14 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 14 n 3, p.186, line 5: n! θ n+1 (x / θ ) {1 - i e - x / θ } i! i =0 3, p.186, line 5: x > 0 3, p.186, solution to the first exercise: = Γ(α+1) Γ(α) θ 3, p.194, 5 lines from the bottom: = λ n e - λ λα 1 e - λ /θ n! θ 0 α Γ(α) dλ 3, p.200, top of the page: Below the prior mixed Negative Binomial distribution (M) and the posterior mixed Negative Binomial distribution (P) are compared: 3, p.241, sol. 6.25: 1/θ = 3, p.251, sol. 6.63: (6)(0.04) = , p.327, first line: f(m) = exp[-(m- 7)2 / 8] 2 2π 3, p.329, last line: / 29 = 297/29 4, page 8: Corr[X - c, Y] = Cov[X - c, Y] / (StdDev[X - c] StdDev[Y]) = Cov[X, Y] / (StdDev[X] StdDev[Y]) = Corr[X, Y]. 4, p. 19, sol. 3.1: Using in the t-table the entries for 100 d.f < < Letter solution is OK. 4, page 23, middle of the page: the Pearson correlation coefficient 4, page 44, solution to the exercise: n c = 2 4, page 68, line two: greater than 7000 provides 4, page 84, sol. 6.8, line three: (1/2)(1/2)(70) = 17.5.

15 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 15 5, page 15, 3 lines from the bottom: (0.34)(4161) + (0.66)(6932) = , page 16, next to last paragraph: Thus a whisker extends from 2176 the third quartile to , page 43, 5 lines from the bottom: s X = , page 436, middle of the page: For the first observation, x = 1, µ = β 0 + β 1, and y = 2. For the third observation, x = 4, µ = β 0 + 4β 1, and y = 11. 5, page 89, 5 lines from the bottom: R 2 = ^β 2 x i 2 y i 2 5, page 91, lines 6 and 7: R 2 = 1 - RSS/TSS = 1 - (N - k - 1)s 2 /TSS. R 2 = k F p,n - k - 1 k F p,n - k (N- k -1). 5, page 124, 5th paragraph: E[R 2 ] = 25%. 5, page 157, sol. 6.2: 25-2 = 23 degrees of freedom 5, p. 217: XʼX = , , page 289, sol. 10.5: 4 and 18 degrees of freedom 5, page 386, top of page, subscript should be i rather than 2: s VIFi s X i N- 1. 5, p. 425: Q : Determine DFFITS 17. 5, p. 458, Q.16.3: Corr[ε t-1, ε t ] = 1, determine the expected value of the Error Sum of Squares. 5, p. 458, Q.16.4: If Corr[ε t-1, ε t ] = -1,

16 Errata, Mahler Study Aids for Exam S, 2017 HCM, 10/24/17 Page 16 5, p. 529, line 10: F = ( ) / / (12-6) 6, p.5, solution to the exercise: Then if a claim is reported at time t, -1 < t < 0, the probability that it is still not settled is: 1+t. The time since it was reported is -t. Thus the average time since reporting is: 0 0 (1+ t) (-t) dt = -t - t 2 dt = 1/2-1/3 = 1/ , pages 113 to 120: This is what Dobson and Barnett call Iterative Weighted Least Squares. 6, p.269: in last block of the table the labels should be tumor types rather than territories. 7, p. 35, solution 3.6: p p p p p p 105 = (l l l l l ) / l 105 = ( ) / 1668 = years. 7, p. 54, Q. 5.4: ȧ. x = 9. 8, p. 33, solution 2.16: - Prob[1, 2, 3, and 4 fail by time 15] 8, p. 91. line 10: For minimal cut set C j, its indicator function is 8, p. 92, line 7: Let C i be the minimal cut sets for a system, 9, p.15, sol. 2.2b: ( )/4 = , p.117, sol. 9.6: E. Prob(x t > 8) = Prob(w t + 1.1w t w t-2-0.3w t-3-0.4w t-4 > 8-5 = 3). E[w t + 1.1w t w t-2-0.3w t-3-0.4w t-4 ] = 0. Var[w t + 1.1w t w t-2-0.3w t-3-0.4w t-4 ] = (3) ( ) = 9.3. Prob(w t + 1.1w t w t-2-0.3w t-3-0.4w t-4 > 3) = 1 - Φ[(3-0) / 9.3 ] = 1 - Φ[0.984] = 16.3% 9, p.118, sol. 9.10: = β 2 / (1 + β β 2 2 ). Final answer is OK.