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Unformatted text preview: Chapter 3 Some Special Distributions 3.1.2 Sincen=9andp=1/3, ,u=3 and 02 :2. Hence, p—20’=3—2\/§ and
p+2a=3+2\/§and P(,u—2U<X<,u+2a)=P(X=1,2,...,5). 3.1.3
£13m = imp) = t1:
/_\
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ll $3M — mm] = ——“”( = L11
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i_"’.
ll 3.1.4p= P(X>1/2)= f11/23m2dzr=§andn=3. Thus (2) (Q2 (a =%. P(Y 2 1)— — 1 — P(Y— — 0)=1—(3/4)" 2 0.70. That' is, 0 30 2 (3/4)” which
can be solved by taking logarithms. Assume X and Y are independent with binomial distributions b(2, 1/2) and b(3,1/2), respectively. Thus we want
P(X>Y) = P(X=1,2andY=0)+P(X=ZandY=1) <3 (31(2) (1)2] i<%>“i+.<é>i+[3<%>i P(X21) = l—(l—p)2=5/9=>(1—p)2=4/9
P(Y21) = 1—(1—p)4=1—(4/9)2=65/81. 3.1.11 3.1.12 Let f(:1:) denote the pmfwhich is b(n,p). Show, for a: Z 1, that f(m)/f(:z:—1) =
1+ [(11 + 1)p — z]/a:(1— p). Then f(a:) > f(a; — 1) if (n + 1)p > a: and
f(a;) < f(a: — 1) if (n + 1)p < 3:. Thus the mode is the greatest integer
less than (n + 1)p. If (n + 1)p is an integer, there is no unique mode but
f[(n + 1)p]: f[( n + 1)p — 1] is the maximum of f(:1:). 17 18 Some Special Distributions 3.1.14
PM 2 3) = (1/3)(2/3>3 + <1/3)(2/3)4 +   = W = (2/3)?
_ (1/3)<2/3>== m:
p(:z:X23) _ (2/3)3 (1/3)(2/3) , 3,4,5,...
3.15 $2197?) @2 3.1.16 M(t) = 2:0(y::1)p'[(1  p)e‘]y = p"[1  (1 — p)e‘]", because the sum— mation equals p’(1 — 10)", where w = (1 — p)e‘. 5125115121 5 2 4 2 5 2 _ 6’
which is much different than 1/2 that some might have arrived at by letting
4 coins he heads and tossing the ﬁfth coin. 3.1.18 3.1.19 7! 1 7 7! 1 2 5 5 51 1 5
[2!1!11<E) i/[2T5’1(E) (E) J =1!1! <3) '
3.1.21
(a) E(X2) = 21:13:30 [$2 (2) (3)] g = ..:. %% = %
(b) f21($2$1) is b(a:1, 1/2) => E(X2:r1) = 2:1/2.
(c) E(:1:1/2) = 11/6.
3.1.22 p1 = 6(1/6)3 = 1/36, 172 = 6  5  3  (1/6)3 =_15/36. Thus X and Y are trinomial (n = 10,121 = 1/36,p2 = 15/36). cov(X, Y) =
—np1p2. Thus E(XY) = —np1p2 + nplnpg = 25/24. 3.1.25 se the mgf technique and independence to get E[82(X1—X2+ng)] = E[e£X]E[e—tX2]etnn (% + get)“ (1 + w
1 1 1 "1+"?
(5 + 58) . 19 3.2.1
e‘"p _ e‘”p2 1! 2! 3.2.4 iven p(z) = 4p(:I:—1)/a:, x = 1, 2, 3, . . .. Thus p(l) = 4p(0), p(2) = 42p(0)/2l,
p(3) = 43p(0)/3!. Use induction to show that p(m) = 4“’p(0)/:v!. Then =>p=2andP(X=4)=iz2r4. 1 = E1302) =p(0) film/3:! = p(O)e4 and p(az) = 4’e"4/a:!,:1: = 0,1,2,.. .. 1:0 1:0 3.2.6 For a; = 1, Dw[g(1,w)]+/\g(1,w) = Ae'A‘”. The general solution to Dw[g(1,w)]+
Ag(1,w) = 0 is g(1,w) = ce"\“’. A particular solution to the full diﬁerential
equation is Awe‘A‘”. Thus the most general solution is g(l,w) : Awewm' + ce‘A‘". However, the boundary condition g(1,0)requires that c = 0. Thus g(1,w) =
_ e‘A‘”. Now assume that the answer is correct for z = —1, and show that
s correct for a: by exactly the same type of argument used for a: = 1. P(X22)=1—P(X=00rX=1)=1——[e'”+e‘”p] 20.99.
Thus 0.01 2 (1 +,u)e"‘. Solve by trying several values of it using a calculator. 3.2.10
1: Z 6—33“; 2 0.99. 9;! 3:0 Horn tables, k = 8. — — f
3.2.11 ET? = ‘—%£ requires p2 = 6 and p = J5. Since e—ﬂzﬂ = Bea“5 > #, :1: = 2 is the mode. E(X!) = 2:10 m!% = 2:0 e"1 does not exist. 3.2.12 . Part (a), M(t1, t2) II no y e__2
Z Sammy——
x!(y ~ :3)! y=0 :I:=0 00 y y! n a: e—2etgy
Z[2W(e ll y! y=0 1:0 :0: e‘2[(1 + e“)et’]y y! y=0
6—2 exp[(1 + e“)et2]. ...
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 Fall '08
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