ST3239: Survey Methodology - The Department of Statistics and ...

Question: What is the distribution of X? (Hint: Classify the people into “Yes” and “Not
Yes”)
Theorem 2.5.1
E(X) = np 1 , E(Y ) = np 2 , E(Z) = np 3 ,
V ar(X) = np 1 q 1 , V ar(Y ) = np 2 q 2 ,
Cov(X, Y ) = −np 1 p 2 .
Proof. X = number of people saying “YES” ∼ Bin(n, p 1 ). So EX = np 1 , V ar(X) = np 1 q 1 .
Now Cov(X, Y ) = E(XY ) − (EX)(EY ) = E(XY ) − n 2 p 1 p 2 . But
E(XY ) =
=
=
=
∑
x,y≥0,x+y≤n
∑
x,y≥1,x+y≤n
∑
x,y≥1,x+y≤n
∑
x,y≥1,x+y≤n
xyP (X = x, Y = y)
xyP (X = x, Y = y, Z = n − x − y)
n!
xy
x! y! (n − x − y)! px 1p y 2p n−x−y
3
n!
(x − 1)! (y − 1)! (n − x − y)! px 1p y 2p n−x−y
3
= n(n − 1)p 1 p 2
∑
x−1,y−1≥0,(x−1)+(y−1)≤(n−2)
(n − 2)!
(x − 1)! (y − 1)! ((n − 2) − (x − 1) − (y − 1))! px−1 1 p y−1
2 p (n−2)−(x−1)−(y−1)
3
∑
= n(n − 1)p 1 p 2
x 1 ,y 1 ≥0,x 1 +y 1 ≤(n−2)
(n − 2)!
(x 1 )! (y 1 )! ((n − 2) − x 1 − y 1 )! px 1
1 p y 1
2 p (n−2)−x 1−y 1
3
= n(n − 1)p 1 p 2 = n 2 p 1 p 2 − np 1 p 2 .
Therefore, Cov(X, Y ) = E(XY ) − n 2 p 1 p 2 = −np 1 p 2 .
Theorem 2.5.2
E(ˆp 1 ) = p 1 , E(ˆp 2 ) = p 2 ,
V ar(ˆp 1 ) = p 1 q 1 /n, V ar(ˆp 2 ) = p 2 q 2 /n,
Cov(ˆp 1 , ˆp 2 ) = −p 1 p 2 /n.
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