? asked in 教育與參考考試 · 1 decade ago

統計問題 求助高手(15點)

有兩個問題解不出來,請求高手幫忙,越詳細越好,謝謝!

Q1:Let X have a poisson distribution with mean λ

Find a transformation u(x) so that var[u(x)] is about free of λ,for large values of λ

Q2:Let X1,X2 be independent randow variables,each with binominal p.m.f

X1~b(2.1/3),X2~b(2.1/3)

1.Find the joint p.m.f of Y=X1 and W=X1+X2

2.Find the joint p.m.f of w

3.compute the correlation cofficient of Y and W

2 Answers

Rating
  • Fwos
    Lv 4
    1 decade ago
    Favorite Answer

    1.X~Poisson(λ),

    suppose that Y1、Y2、…、Yλare the random sample from Poisson(1),

    then X= Y1+Y2+ … +Yλ=Σ(i=1...λ)Yi,

    according to the Central Limit Theorem(C.L.T.),

    we have (Σ(i=1...λ)Yi-λ) / √λ= (X-λ) / √λ → N( 0 , 1),

    ( "→" 代表分佈收斂,converge in distribution)

    (X-λ) / √λ → N(0 , 1)

    => (X-λ) → N( 0 ,λ)

    => X → N(λ,λ)

    => u(X) → N( u(λ),λ*[u'(λ)] ^ 2) (by delta-method)

    Var(u(X))= λ*[u'(λ)] ^ 2 is free ofλ

    => u'(λ)= c / √λ,where c is an arbitrary number

    => u(λ)= 2*c*√λ,

    for convenient,we take c= 1 / 2,then u(x)= √x

    2.<解法一>

    (1)Y=X1 and W=X1+X2 => X1=Y and X2= W-Y,

    the joint p.m.f. of X1 and X2 is

    f(x1, x2)= ( 2 ; x1)*( 2 ; x2)*(1/ 3)^(x1+x2)*(2/ 3)^(4-x1-x2)( (a; b)代表組合函數)

    the joint p.m.f. of Y and W is

    g(y, w)= f(y, w-y)= ( 2 ; y)*( 2 ; w-y )*(1/ 3)^(w)*(2/ 3)^(4-w), y= 0...2, w= y...y+2

    (2) the marginal p.m.f. of W is

    g(w)= Σ(y= 0...w) g(y, w)

    = (1/ 3)^(w)*(2/ 3)^(4-w)*Σ(y= w-2...w)( 2 ; y)*( 2 ; w-y )

    = (1/ 3)^(w)*(2/ 3)^(4-w)*(4 ; w)(高中好像有公式,但我忘了,所以硬做的)

    (3)E[YW]

    = Σ(y= 0...2)Σ(w= y...y+ 2) y*w*( 2 ; y)*( 2 ; w-y )*(1/ 3)^(w)*(2/ 3)^(4-w)

    = Σ(y= 0...2) y*( 2 ; y)*Σ(w= y...y+ 2) w*( 2 ; w-y )*(1/ 3)^(w)*(2/ 3)^(4-w)

    = Σ(y= 0...2) y*( 2 ; y)*(9*y+ 6) / 9 *(1/ 3)^ y*(2/ 3)^(2-y)

    = (1/ 9)*Σ(y= 0...2) (9*y^2+ 6*y)*( 2 ; y)*(1/ 3)^ y*(2/ 3)^(2-y)

    = (1/ 9)*[9*(4/ 9 + 4/ 9)+ 6*(2/ 3)]= 12 / 9

    Y= X1~ b(2 , 1/ 3) and W= X1+X2 ~ b(4 , 1/ 3),

    E[Y]= 2/ 3 ,E[W]= 4/ 3, Var(Y)= 4/ 9, Var(W)= 8/ 9,

    Cov(Y,W)= E[YW]-E[Y]*E[W]= 12 / 9-(2/ 3)(4/ 3)= 4/ 9

    ρ= Cov(Y,W) / √(Var(Y)*Var(W))= (4/ 9) / √[(4/ 9)(8/ 9)]= 1/ √2

    <解法二>

    (1)同<解法一>

    (2)X1~ b(2 , 1/ 3),X2~ b(2 , 1/ 3),p相同,所以根據二項分佈的加成性可知

    W= X1+ X2 ~ b(4 , 1/ 3)

    (3)Cov(Y , W)= Cov( X1 , X1+X2)= Cov(X1 , X1)+ Cov(X1 , X2)= Var(X1)= 4 / 9

     後面作法同<解法一>

    (建議字型->適中,排版比較好看)

    2006-02-27 10:22:16 補充:

    1.一般而言,當W和Y有函數關係時,最好不要乎略了他們的關係,此題雖然結果會一樣,但是,在連續函數的marginal p.d.f.求法中,不會一樣

    2.解法二是我們老師教的,逐一討論各變數的covariance,

    Cov(X1,X1+X2)

    =E[X1*(X1+X2)]-E[X1]*E[X1+X2]

    =E[X1^2]+E[X1*X2]

    -(E[X1])^2-E[X1]*E[X2]

    =E[X1^2]-(E[X1])^2

    +E[X1*X2]-E[X1]*E[X2]

    =Cov(X1)+Cov(X1,X2)

    =Var(X1)+Cov(X1,X2)=Var(X1)

  • 1 decade ago

    請問(2)的marginal p.m.f of W 中的g(w)= Σ(y= 0...w) g(y, w)

    為何不是g(w)= Σ(y= 0..."2") g(y, w)

    還有解法二

    Cov(Y , W)= Cov( X1 , X1+X2)= Cov(X1 , X1)+ Cov(X1 , X2)= Var(X1)= 4 / 9 不太懂,請問是參考哪本書的?

    2006-03-01 13:20:02 補充:

    謝謝你啦,我後來有寫出來!

Still have questions? Get your answers by asking now.