咕咕 asked in 教育與參考考試 · 1 decade ago

統計suppose a box has 3 balls...

suppose a box has 3 balls labeled 1, 2, and 3. Two balls are selected without replacement from the box. Let X be th number on the first ball and Y be the number on the second ball : Compute Cov(X,Y) and p(X,Y)

1 Answer

Rating
  • 1 decade ago
    Favorite Answer

    (X,Y)抽出不放回有這6個可能

    (1,2)(1,3)(2,1)(2,3)(3,1)(3,2)

    X 1 1 2 2 3 3

    Y 2 3 1 3 1 2

    ΣX=1+1+2+2+3+3=12

    X-BAR=12/6=2

    ΣY=2+3+1+3+1+2=12

    Y-BAR=12/6=2

    ΣXY=1*2+1*3+2*1+2*3+3*1+3*2=22

    ΣX^2=1^2+1^2+2^2+2^2+3^2+3^2=28=ΣY^2

    σx^2=[(ΣX^2)-(6*X-BAR^2)]/6=(28-6*2^2)/6=0.66=σy^2

    COV(X,Y)

    =1/N*(ΣXY-ΣX*ΣY/N)

    =1/6*(22-12*12/6)

    =-0.33 #

    ρ=COV(X,Y)/(σx^2*σy^2)^0.5

    =-0.33/0.66

    =-0.5 #

    這次應該沒問題了,感謝統計老兵翻譯再加上指正-.-"

    2008-01-07 20:08:19 補充:

    (X,Y)抽出不放回有這6個可能

    (1,2)(1,3)(2,1)(2,3)(3,1)(3,2)

    X 1 2 3 fy(Y)

    Y 1 0 1/6 1/6 1/3

    2 1/6 0 1/6 1/3

    3 1/6 1/6 0 1/3

    fx(X) 1/3 1/3 1/3 1

    2008-01-07 20:08:37 補充:

    E(X)=E(Y)=1*1/3+2*1/3+3*1/3=2

    E(X^2)=E(Y^2)=1^2*1/3+2^2*1/3+3^2*1/3

    V(X)=V(Y)=(1^2*1/3+2^2*1/3+3^2*1/3)-2^2

    E(XY)=1*2*1/6+1*3*1/6+2*1*1/6+2*3*1/6+3*1*1/6+3*2*1/6

    COV(X,Y)=E(XY)-E(X)*E(Y)

    =(1*2*1/6+1*3*1/6+2*1*1/6+2*3*1/6+3*1*1/6+3*2*1/6)-2*2

    2008-01-07 20:08:44 補充:

    ρ=COV(X,Y)/(V(X)*V(Y))^0.5

    =COV(X,Y)/V(X) <=本題V(X)=V(Y)

    =[(1*2*1/6+1*3*1/6+2*1*1/6+2*3*1/6+3*1*1/6+3*2*1/6)-2*2]/[(1^2*1/3+2^2*1/3+3^2*1/3)-2^2]

    =-0.5

    2008-01-07 20:10:31 補充:

    聯合機率表的(X=1,Y=1)(X=2,Y=2)(X=3,Y=3)機率0

    其他都是1/6

    X,Y的邊際機率都是1/3

Still have questions? Get your answers by asking now.