統計suppose a box has 3 balls...

Question

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)

Answer 1

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(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