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Unformatted text preview: 20 Some Special Distributions 3.3.1
(1 — my6 = (1 — zenW? => 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 — 202 = (1 — 204/9,
s X is X2(4).
3.3.6 P(Y s y) = 1 [P(X > 3)]3 — (ey)3 =1 — e39 = 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 eZaﬁ‘u — pair“, for all t < l/ﬁ. The minimum of the right side, say K (t), can be found by
W) = eQﬂﬁwamu — 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
vmr 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 ~ 115). the independence of Z and I1_E, and
1&6 = 116, we get E(W) = 0(1 — e) + ac0[1 — (1 — 6)] = o
Var(W) = E(W2)
= 4221,24 + 2a6221,_,(1 — 11_£)+ 0322(1 — 1132]
= (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.,
(ieps) *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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 Fall '08
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