pphap hop ly cuc dai btoan uoc luog khoang

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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,

    ), U. Hm L(X/ ) = f(X1, )f(X2, ) f(Xn, ) c gi l hm hp l.

    nh ngha 1.2. Thng k c gi l c lng hp l cc i ca

    nu

    L(X/ (X) L(X/ ) vi mi .

    *(X) = 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] = ; xi = 0, 1, 2,

    Hm hp l

    => lnL(X, ) = (ln ) - n - ln =>

    Vy nu =>

    Ta li c

    " .

    Vy ti th tc l hm L(X, ) t cc i. T suy

    ra l c lng hp l cc i ca .

  • Trng hp tham s l mt vect = ( 1,, r)

    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.

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

    Tm c lng hp l cc i ca (a; 2).

    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 ( 1(X), 2(X)) c gi l khong c lng ca tham

    s vi tin cy 1 - nu

    P[ 1(X) < < 2(X)] = 1- .

    Khong ( 1(X), 2(X)) c gi l khong tin cy. Gi tr 1- gi l tin cy.

    Hiu 1- 2 gi l chnh xc ca c lng.

    Ch : Thng thng ngi ta chn ( 1, 2) sao cho l nh nht.

  • a. Khong c lng ca xc sut p trong phn phi nh thc

    Gi s k l s ln xut hin bin c A trong dy n php th Bernoulli. Gi thit

    xc sut bin c A xut hin trong mi php th l p. Xt xc sut;

    .

    vi 1 - l tin cy cho trc. Bin i v tri ca ng thc trn ta c

    t x = v thay biu thc di cn bc hai p bng th

    (1)

    Mt khc, theo nh l Laplace ta c

    Trong

    Suy ra

  • T ta c .

    Vy nu cho ta tnh c v tra bng phn phi chun N(0;1) ta tm c

    x .

    T (1) ta c . Bin i suy ra khong c lng cho p l

    V d 2.2. Trong t vn ng bu c, phng vn 1600 c tri c bit 960 ngi

    trong s s b phiu cho ng c vin A. Vi tin cy 95% , ti thiu ng c

    vin A s chim c bao nhiu phn trm phiu.

    Gii. Ta c = v x =1,96. Thay vo cng thc trn ta c