𝑢(∑ 𝑥𝑖
𝑛
𝑖=1 ) = ∑ 𝑥𝑖
𝑛
𝑖=1
𝑔(𝑢(∑ 𝑥𝑖
𝑛
𝑖=1 ); 𝜃) = ( 𝜃2
1−𝜃)𝑛
(1−𝜃
𝜃 )∑ 𝑥𝑖
𝑛
𝑖=1
ℎ(𝑥1, 𝑥2, ... , 𝑥𝑛) = 1
Therefore, ∑ 𝑥𝑖
𝑛
𝑖=1 is suffient statistics for 𝜃.
(c) Method of mments
𝑋̅ = 𝐸(𝑋)
𝑋̅ = ∑ 𝑥 𝑝(𝑥)2
𝑥=1
𝑋̅ = ∑ 𝑥{𝜃𝑀𝑀
2−𝑥(1 − 𝜃𝑀𝑀)𝑥−1}2
𝑥=1
𝑋̅ = 𝜃𝑀𝑀 + 2(1 − 𝜃𝑀𝑀) Make 𝜃𝑀𝑀 subject
𝜃𝑀𝑀 = 2 − 𝑋̅
(d) If the 𝜃𝑀𝑀 is unbiased then 𝐸(𝜃𝑀𝑀) should be equal to 𝜃
𝐸(𝜃𝑀𝑀) = 𝐸(2 − 𝑋̅ )
𝐸(𝜃𝑀𝑀) = 2 − 𝐸(𝑋̅ )
𝐸(𝜃𝑀𝑀) = 2 − 𝜃 ≠ 𝜃
Hence, 𝜃𝑀𝑀 is biased estimator of 𝜃.
(e) From (a) we have
𝐿(𝑥1, 𝑥2, ... , 𝑥𝑛; 𝜃) = 𝜃2𝑛−∑ 𝑥𝑖
𝑛
𝑖=1 (1 − 𝜃)∑ 𝑥𝑖
𝑛
𝑖=1 −𝑛 Obtain log of the equation
𝑙(𝜃) = (2𝑛 − ∑ 𝑥𝑖
𝑛
𝑖=1 ) ln(𝜃) + (∑ 𝑥𝑖
𝑛
𝑖=1 − 𝑛) ln(1 − 𝜃) obtain score function
𝑆(𝜃) = (2𝑛−∑ 𝑥𝑖
𝑛
𝑖=1 )
𝜃 − (∑ 𝑥𝑖
𝑛
𝑖=1 −𝑛)
1−𝜃 equate to zero and solve
(2𝑛−∑ 𝑥𝑖
𝑛
𝑖=1 )
𝜃 − (∑ 𝑥𝑖
𝑛
𝑖=1 −𝑛)
1−𝜃 = 0 multiply by 𝜃(1 − 𝜃)
(2𝑛 − ∑ 𝑥𝑖
𝑛
𝑖=1 )(1 − 𝜃) − (∑ 𝑥𝑖
𝑛
𝑖=1 − 𝑛)𝜃 = 0
(2𝑛 − ∑ 𝑥𝑖
𝑛
𝑖=1 ) − ((2𝑛 − ∑ 𝑥𝑖
𝑛
𝑖=1 ) + (∑ 𝑥𝑖
𝑛
𝑖=1 − 𝑛))𝜃 = 0

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