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Phng php hp l cc i - Bi ton c lng khong 1. Phng php hp l cc i nh ngha 1.1. Gi s (X1, X2,, Xn) l mu ngu nhin c lp t phn phi f(x, Hm L(X/ ) = f(X1, )f(X2, ) f(Xn, ) c gi l hm hp l. nh ngha 1.2. Thng k L(X/ (X) *(X) = L(X/ ) vi mi . c gi l c lng hp l cc i ca ), U.

nu

c gi l c lng hp l cc i ca hm tham s t( ).

Trng hp mt tham s.

tm c lng hp l cc i, ta c th s dng phng php tm cc i hm L(X/ ) m chng ta tng quen bit. Ta bit rng cho hm L(X/ ) c cc tr a phng ti = iu kin cn l . Gii phng trnh ny, tm cc nghim ca n sau ta xt du ca o hm hng nht hay hng hai tm cc i hm L(X/ ). V d 1.3. Gi s (X1, X2,, Xn) l mu ngu nhin c lp t phn phi Poisson vi tham s > 0. Tm c lng hp l cc i ca . Gii. Phn phi ca Xi l

P[Xi = xi] = Hm hp l

; xi = 0, 1, 2,

=> lnL(X, =>

) = (ln

)

-n

- ln

Vy nu

=>

Ta li c

"

.

Vy ti ra

th

tc l hm L(X, .r)

) t cc i. T suy

l c lng hp l cc i ca Trng hp tham s l mt vect =(1,,

Lm tng t nh trng hp 1 tham s. Ta gii h phng trnh

(*) Gii h ny ta tm c

t

. Nu ma trn

l xc nh khng m th ti

=

0

hm hp l L(X,

) t cc i.2

V d 1.4. Gi s (X1, X2,, Xn) l mu ngu nhin t phn phi chun N(a; hp l cc i ca (a; 2).

). Tm c lng

Gii. Ta c

=> Lnf(Xi, a,

2

)=

=>

v

Thay vo h (*) ta c

=>

l c lng hp l cc i ca (a;

2

).

2. c lng khong nh ngha 2.1. Khong ( cy 1 nu P[1(X) 1(X), 2(X))

c gi l khong c lng ca tham s

vi tin

t2] =

. Khong c lng ca

2

vi

V d 2.4. xc nh chiu cao trung bnh ca cc cy bch n trong mt khu rng, tin hnh o ngu nhin 35 cy v thu c kt qu sau Chiu cao X (m) S cy (ni) 6,5 7,0 2 7,0 7,5 4 7,5 8,0 10 8,0 8,5 11 8,5 9,0 5 9,0 9,5 3

Gi thit chiu cao ca cc cy bch n l i lng ngu nhin c phn phi chun. Vi tin cy 95%, xc nh khong c lng cho phng sai DX. Gii. Ta c v . = 16,8 v t2 = = 47. T

Tra bng tm c t1 =