Then we know that ln(X). takes the valuesa 1 = (b 1 −m) 2 ,... , ar= (br−m) 2. HenceFY(y) = and 1. P((X−3)/2)≤(1−3)/2 = P(Z≤−1) = P(Z≥1) = 0.1587. HenceFW(w) = 476 Full solutions from MIPS: DO NOT DISTRIBUTE, 8.4 aLetFbe the distribution function ofX, andGthat ofY. V. Then we know thatF(x) = 0 forx < 1 , F(x) =xfor 0≤x≤1, andF(x) = 1 distribution. Book solutions "A Modern Introduction to Probability and Statistics", Copyright © 2020 StudeerSnel B.V., Keizersgracht 424, 1016 GC Amsterdam, KVK: 56829787, BTW: NL852321363B01, Upgrade naar Premium om het volledige document te bekijken, Elaboration Book Modern Intro to Probability and Statistics Understanding Why and How - Coverage, Kraaikamp and more, Tentamen 8 november 2013, Vragen en antwoorden, Robust and Multivariable Control Design - Assignments - AL1 2015, Werkstuk/essay, Operations Maintenance, Offshore Wind Energie, Cijfer 7,5, Tentamen 14 maart 2014, vragen en antwoorden, A Modern Introduction to Probability and Statistics. Then fory >0 : FY(y) = P(Y≤y) = P(ln(X)≤y) = P(X≤ey) =FX(ey). Then applying. 10 ,| 100 − 100 |= 0,| 110 − 100 |= 10 and| 120 − 100 |= 20. This is of course what should happen:Z= 1/Y= 1/(1/X) =X, soZandXhave We see that the values are ThenZ= (X−3)/2 is anN(0,1) distributed random variable, so that P(X≤1) = Since E[U] = Var(V) = 0, part 8.4 bLetFbe the distribution function ofX, andGthat ofY. 1 −e−(λw). distribution function of anExp(α) distribution. atells us that 0 =a 1 = (b 1 −m) 2 ,... ,0 =ar = (br−m) 2. Meld je aan of registreer om reacties te kunnen plaatsen. P(V= E[V]) = 1. we use thatr >0), if and only ifs≤y≤s+r. entiating we obtainfY(y) =fX(−y) for ally. But this is only 1 −FX(−y) for allY (where you use thatXhas a continuous distribution). 8.8 LetXbe any random variable that only takes positive values, and letW= Si− 1 if 0. P(X≤y/λ) = 1−e−y.We recognizeGas the distribution function of anExp(1) 8.7 LetXbe any random variable that only takes positive values, and letY = 8.5 cWe simply differentiateFY:fY(y) =dydFY(y) = 3y 3 − 3 y 5 /2 for 0≤y≤, 8.6 aCompute the distribution function ofY :FY(y) = P(Y≤y) = P, sinceXhas a continuous distribution. Furthermore:σ 2 = 4, soσ= 2. as the distribution function of aU(7,9) random variable. For instance, Similarly, we obtain for the two other values, 8.2 bThe values taken byZare− 1 , 0 ,and 1. FurthermoreFX(b) = 0 forb <0, andFX(b) = 1 forb >2. dyFY(y) =, 8.6 bApplying partawithZ= 1/Y we obtainfZ(z) =z 12 fY( 1 z). 0 ,10 and 20, and the latter two occur in two ways: P(Y= 0) = P(X= 100) = 0.2, thatF(x) = 1−e−λxforx≥0, and we find thatG(y) = P(Y≤y) = P(λX≤y) = "A Modern Introduction to Probability and Statistics has numerous quick exercises to give direct feedback to the students. 8.2 cSince for anyαone has sin 2 (α) + cos 2 (α) = 1,Wcan only take the value 1, +prbr. 8.1 The random variableY can take the values| 80 − 100 |= 20,| 90 − 100 |= d abilities of thex-values to obtain the probability of they-value. 8.2 aFirst we determine the possible values thatY can take. We recognize this as the Differ- 8.9 If Y = −X, thenFY(y) = P(Y≤y) = P(−X≤y) = P(X≥−y) = IfXhas aPar(α) distribution, thenFX(x) = 1−x−αforx≥1. 7 ≤y≤9. but P(Y= 10) = P(X= 110) + P(X= 90) = 0.4; P(Y= 20) = P(X= 120) + Then forw >0 : IfXhas anExp(1) distribution, thenFX(x) = 1−e−xforx≥0. Then we know distribution. the same probability density function. provided 0≤(y−7)/ 2 ≤1 which happens if and only if 0≤y− 7 ≤2 if and only 29 modern introduction to probability and statistics full solutions february 24, 2006 dekking, kraaikamp, meester 458 full solutions from mips: do not Furthermore,G(y) = 0 ify <7 andG(y) = 1 ify > 9 .We recognizeG 8.3 aLetFbe the distribution function ofU, andGthe distribution function of We see thatGhas aU(s, s+r) In addition the book contains over 350 exercises, half of which have answers, of which half have full solutions. distribution. Then we investigate whichx-values lead to thesey-values and sum the prob- F(x) = 1−e−x/ 2 forx≥0, and we find thatG(y) = P(Y≤y) = P. P(X≤ 2 y) = 1−e−y.We recognizeGas the distribution function of anExp(1) Probability Introduction to Probability and Statistics Introduction to Probability and Statistics, 14th Edition Introduction to Probability and Statistics, 14th Edition 14th Edition | ISBN: 9781133103752 / 1133103758. introduction to probability and statistics - PDF Free Download Differentiating we obtain:fY(y) = A Modern Introduction to Probability and Statistics Full Solutions February 24, 2006 ©F.M.Dekking,C.Kraaikamp,H.P.Lopuha¨a,L.E.Meester.

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