hw8-stat516 - 20 Some Special Distributions 3.3.1(1 — my6...

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Unformatted text preview: 20 Some Special Distributions 3.3.1 (1 — my6 = (1 — zen-W? => X is x202). Fiom tables, P(X < 5.23) = 0.05. 3.3.4 2:2: 392732 4!23t3 M(t) — 1 + + 3! + 3! = 1+ 2(2t) + 3(2):)2 + 4(2t)3 + - -- = (1 — 20-2 = (1 — 20-4/9, s X is X2(4). 3.3.6 P(Y s y) = 1- [P(X > 3)]3 — (e-y)3 =1 — e-39 = G(y), 9(11) = 0(9) = 3e‘3", 0 < y < 00. 3.3.7 f’(:z:) = Elie—2m + Elite—I/W—l/fl) = 0; hence, a: = ,6 which is given as 2. Thus X is x2(4). 3.3.9 P(X 2 204;) g e-Zafi‘u — pair“, for all t < l/fi. The minimum of the right side, say K (t), can be found by W) = e-Qflfiwamu — fit)“’“ + rem—2am - m)” = 0 which implies that (1 —fit)‘1 — 2 = 0 and t =1/2fi. That minimum is K(1/2fl) = e"’(1 - (1/2))” = (2/6)"- 3.3.10 If r = 0, M(t) = 1 = em”, which is the mgf of a degenerate distribution at a: = 0. . 3.3.14 The differential equation requires log g(0,w) = —kw" + c. The boundary condition g(0,0) = 1 imlies that c = 0. Thus g(0,w) = exp{—kw"} and C(11)) = 1 — exp{—kw’} and G’(w) = krwr‘l exp{—kw"}, 0 < w < oo. 3.3.15 he joint pdf of X and the parameter is e‘mm“ flflmMWQ = fl mfm,$=mLan0<m<a) 2 oo 1+1 —2m 2 3+2 P<x=o.1.2) = 2/ fl+dm=sz+2>gL2>_ 1:0 0 11?- I=o 2:. 2 1 2 3 11 _ 2+2 _ _ _ _ = _ — §(m+1)(1/2) — 4+8+ 16 16. 3.3.16 C(21) = P(Y S y) = P(-2log X s y) = P(X 2 exp{—y/2}) 1 / (1)dm=1-,— exp{—y/2}, 0<y <00 exp(-y/2) ‘ G’(y) = (1/2) exp{*y/2}. 0 < y < 00; g(y) so Y is x2(2). 3.3.17 f(a:) = 1/(b —- a), a. < x < b, has respective mean and variance of flzi=sand$b+gfi=13 Solve for a, and b. Em P(a+mr(a+1>r(m _ a I‘(cx)l‘(fl)l“(oz + t! +1) _ a + fl“ E(X2) _ F(a + fi)1‘(av+ 2)I‘(fi) _ (a + 1):): ' ‘ l“(6!)1‘(fi)1‘(01+:6+2)_(0t+fi+1)(cv+fi)' afl “’2 “2)— WW = 3.3.20 3 1 = 6/ a:(3—a:)4d:1:.Letm=3y,—g§=3. 1 . c A (3y)(3 — 3y)43dy = 3%“?555); so c = 6- 5/36 = 10/35. 1 3.3.21 If a = [3, show that f + z) = f — 2:). 3.3.22 Note that k—l 77' 1» n—w 7"! — "— D‘[‘ = (11920”) “1‘” k' 100 22 Some Special Distributions 3.3.26 (a). Using the mean value theorem, we have P(a:sX<a:+A) = m 00A AP(X 2 27:) 1(3): lim A_.o m, Ado where 5 —» a: as A —) 0. Letting A —» 0, we then get the desired result. ((1) The pdf of X is mm) = exp {:0 — ebo} from which the desired result follows. I the integral for cl§(—z), let w = —v and it follows that ‘I>(—z) = 1 — @(z). 0.90 P(X'#<89?#) = a a~ P<X—p<94—p) ___ 0' 0 Thus 39—0—5 = 1.282 and 94—;5 = 1.645. Solve for p and a. 0.95. 3.4.5 2 Thus if c = 1/[x/2_1r‘/1/(2 log 2) , we would have 3 N(0, 1/(2 log 2)) distribu- lOIl. 62—12 = Ce—zglog 2 = cexP {_M}. E(IX -ul) 00 2/ (a: — ,u.) 1 «3'(:“‘“)2/""’2 dz # 27m 2 exp {—(rv — #)2/202}]: = mg- 3.4.8 /3 exp{—(x — 02/20/11)} dm 2 3 1 v-mr m %[¢(31—7§)[email protected](21+§)l= %[ Of course, X is N(3, 16). P [00004 < “€551,— < 3.84] and w is 98(1), so, the answer is 0.95 — 0.05 = 0.90. 1 2 exp{—(90 — 3)2/2(1/4)} dz <1>(—2)] . 23 3.4.13 P(1<X2<9) = p(—3<X< —1)+P(1<X<3) ~1 -1 —3 — 1 3 — 1 - H 2 >-¢( 2 )l+[¢<—2 imi- 3.4.15 I l I 2 I M(t) = 1+0+ Mg”? +0+ 4'“; )2't4 +~- t2 2 t2 2 2 = 1+1—/'+( 5') +--- =exp{t2/2}; so X is N(0,1). 3.4.20 '-' . 02132 (71,1330 [exp {M + — expW}, which is the mgf of a degenerate distribution at a: = p. ‘ 3.4.22 (1 / wen/F(b) dy = —f(b)/F(b)- —OO Multiply both sides by F(b) then differentiate both sides with respect to b. This yields, bf(b) = f’(b) and -—(b2/2) + c = log f(b). Thus 2 f(b) = ale—5 /2, which is the pdf of a N (0, 1) distribution. 3.4.25 Using W = Z11.E + @270 ~ 11-5). the independence of Z and I1_E, and 1&6 = 11-6, we get E(W) = 0(1 — e) + ac0[1 — (1 — 6)] = o Var(W) = E(W2) = 4221,24 + 2a6221,_,(1 — 11_£)+ 0322(1 — 11-32] = (l—e)+a§[l—(1—e)], . which is the desired. 3.4.26 If R or SPLUS is available, the code on page 168, i.e., (i-eps) *pnorm (w) +eps*pnorm (w/sigc) evaluates the contaminated normal cdf with parameters eps and sigc. Using R, the probability asked for in Part (d) is ...
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