統計問題 求助高手(15點)

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

Rating
• Fwos
Lv 4

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)

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!