hw9-stat516 - 24 Some Special Distributions> eps =.25>...

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Unformatted text preview: 24 Some Special Distributions > eps = .25 > sigc = 20 > w = -2 > w2 = 2 > (l—eps)*pnorm(w)+eps*pnorm(w/sigc) + (1-(1—eps)*pnorm(w2)-eps*pnorm(w2/sigc)) [1] 0.2642113 3.4.28 Note X1 — X2 is N(—1,2). Thus P(X1 — X2 > 0) = 1 — <1>(1N§) = 1 — <I>(0.707). 3.4.30 The distribution of the sum Y is N (43, 9), so P(Y < 40) = 11> (4°; 43) = @(—1). 3.5.1 For Part (b), 13(le = 3.2) = 110 + (0.6)61£4(3.2 — 2.8) = 116. Var(Y|a: = 3.2) = 100(1 — 0.36) = 64 Answer = ‘1) (Lil—Ki) - <1) (5)?) = @(1) — <IJ(—-1.25). 3.5.3 31/! _ aMO91J2) 8—t2 — T/M(t1»t2) 621p 82M(t1,t2) 6M(t1,t2) 6M(t1,t2) 2 at1at2 _ [M(t1,t2) at16t2 - atq atl J /M(t"t2) 821p a2M(o,o) _ aM(0, 0) aM(o, 0) 61316132 ¢1=t2=0 6t16t2 61:1 6132 , because M (0, 0) = 1. This is the covariance. @Because E(Y|:1: = 5) = 10 + p(5/1)(5 — 5) = 10, this probability requires that 16-10 —2 i=1—p2, andp=%., 3.5.8 (1:) = f; f(a:,y)dy = (l/x/21r)exp{—a:2/2}, because the first term of he integral is obviously equal to the latter expression and the second term integrates to zero as it is an odd function of y. Likewise 1 f2(y) = \/—2=1r Of course, each of these marginal standard normal densities integrates to one. e><p{—y2/2}- 3.5.9 Similar to 3.5.8 as the second term of / my. 2) da: —00 equals zero because it is an integral of an odd function of :v. 3.5.10Write 'en apply Theorem 3.5.1. . X1 Y1 = 3 1 “2 = X2 = Bx. §;) ya 1 -—5 1 ){3 ,,- Evaluate By. and BVB’ . I l ,7 :w/ 35.16 vvrue ' 1 1 Jr (X1+X2,X1—X2)= [ 1 —1.HX;] Then apply Theorem 3.5.1. 25 3.5.21 This problem requires statistical software which at least returns the spectral decomposition of a matrix. The following is from an R output where the variable amat contains the matrix 2. > sum(diag(amat)) [1] 1026 Total Variation > eigen(amat) $values [1] 925.36363 60.51933 25.00226 15.11478 The first eigen is the variance first principal $vectors First column is first principal [.1] [.2] [.3] [.4] [1,] -0.5357934 0.1912818 0.7050231 -0.4234138 [2,] -0.4320336 0.7687151 -0.3416228 0.3251431 [3,] -0.5834990 -0.4125759 -0.5727115 -0.4016360 [4,] -0.4310468 -0.4497438 0.2413252 0.7441044 > 925.36363/1026 [1] 0.9019139 Over 90% value of the component. the component. VOW 3 0:2“;(‘ I; 0<l z \ \ <9. YMMO- ‘0 W yl>1e-3é): - zéhfléfl 10.05— 7744? . _’\</( X‘s 2 \ i x; i Yam Y2; .g»‘( ““9"” %(%“3\1 a V3 2, 6, § 2 Suns/Hmég _’ QM) WM MW a W” “ Lforcwm 26 Some Special Distributions e F = U then ,i, = %—, which has an F-distribution with 7‘2 and n ' grees of freedom. 3.6.10 Note T2 = WQ/(V/r) = (W2/1)/(V/T)- Since W is N(0,1), then W2 is x2(1), Thus T2 is F with one and 1‘ degrees of freedom. 3.6.12 The change-of-variable technique can be used. An alternative method is to observe that 1 V 1+(U/V) =V+U’ where V and U are independent gamma. variables with respective parameters (1‘2/2, 2) and (Tl/2, 2). Hence, Y is beta with a = 13/2 and fl = r1/2. Y: 3.6.14 For Part (a), the inverse transformation is :1:1 = (y1y2)/ (1 + 111) and 9:2 = yg/(l + yl). The space is y.- > 0, i=1,2. The Jacobian is J = y2/(1 +1102. It is easy to show that the joint density factors into two positive functions, one of which is a function of y1 alone while the other is a function 312 alone. Hence, Y1 and Y2 are independent. 3.7.3 Recall from Section 3.4, that we can write the random variable of interest as X = IZ+3(1 — I)Z, where Z has a N (0, 1) distribution, I is 0 or 1 with probabilities 0.1 and 0.9, respectively, and I and Z are independent. Note that E(X) = O and the variance of X is given by expression (3.4.13); hence, for the kurtosis we only need the fourth moment. Because I is 0 or 1, I k = I for all positive integers Is. Also I (I — 1) = 0. Using these facts, we see that E(X“) = .9E(Z“) + 34(.1)E(z4) = E(Z“)(.9 + (1)34). Use expression (1.9.1) to get E(Z“). 3.7.4 The joint pdf is m = _ m—l 1101+!” 11—] __ p-l Integrating out 0, we have s 1 “OPP/3) a+1—1 _ p+z—1—1 fx(a:) 0 —I‘(a)I‘(fi)0 ‘ (1 9) d0 F(a + fi)l"(a +1)I‘(fl + :1: — 1) I‘(a)I‘(,B)I‘(a + fl + 11:) ' 3.7.7 Both the pdf and cdf of the Pareto distribution are given on page 193 of the text. Their ratio (h(m)/ (1 — H (2:))), quickly gives the result. ...
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