Page 351 - Probability and Statistical Inference

6. Sufficiency, Completeness, and Ancillarity 328

6.2.3 Suppose that X is distributed as N(0, σ ) where 0 < σ < ∞ is the
2
unknown parameter. Along the lines of the Example 6.2.3, by means of the
conditional distribution approach, show that the statistic | X | is sufficient for
σ .
2
6.2.4 Suppose that X is N(θ, 1) where –∞ < θ < ∞ is the unknown param-
eter. By means of the conditional distribution approach, show that | X | can
not be sufficient for θ.

6.2.5 (Exercise 6.2.2 Continued) Suppose that m = 2, n = 3. By means of
the conditional distribution approach, show that X + Y Y can not be suffi-
1 2
1
cient for p.
6.2.6 Suppose that X , ..., X are distributed as iid Poisson(λ), Y , ..., Y n
1
m
1
are iid Poisson(2λ), and that the X’s are independent of the Y’s where 0 < λ <
∞ is the unknown parameter. By means of the conditional distribution ap-
proach, show that is sufficient for ?.
6.2.7 (Exercise 6.2.6 Continued) Suppose that m = 4, n = 5. Show that X
+ Y can not be sufficient for λ. 1
1
6.2.9 (Example 6.2.5 Continued) Let X , ..., X be iid Bernoulli(p) where 0
1
4
< p < 1 is the unknown parameter. Consider the statistic U = X (X + X ) + X .
1
4
2
3
By means of the conditional distribution approach, show that the statistic U is
not sufficient for p.
6.2.10 Let X , ..., X be iid N(µ, σ ) where –∞ < µ < ∞, 0 < σ < ∞. Use the
2
1
n
Neyman Factorization Theorem to show that
(i) is sufficient for µ if σ is known;
(ii) is sufficient for σ if µ is known.
6.2.11 Let X , ..., X be iid having the common pdf σ exp{–(x – µ)/σ}I(x
–1
1
n
> µ) where –∞ < µ < ∞, 0 < σ < ∞. Use the Neyman Factorization Theorem
to show that
(i) X , the smallest order statistic, is sufficient for µ if σ is known;
n:1
(ii) is sufficient for σ if µ is known;
(iii) is jointly sufficient for (µ, σ) if both
parameters are unknown.
6.2.12 Let X , ..., X be iid having the Beta(α, β) distribution with the
1
n
parameters α and β, so that the common pdf is given by