Page 322 - Probability and Statistical Inference

6. Sufficiency, Completeness, and Ancillarity 299

0 < σ < ∞. Here, we write θ = µ, χ = ℜ and Θ = ℜ. It is easy to verify that the
statistic is minimal sufficient for θ. Now consider the statistic U
= X X + X and suppose that the question is whether U is sufficient for θ.
2
1
3
Assume that U is sufficient for θ. But, then must be a function of U,
by the Definition 6.3.1 of minimal sufficiency. That is, knowing an observed
value of U, we must be able to come up with a unique value of T. One can
proceed in the spirit of the earlier Example 6.3.2 and easily arrive at a contra-
diction. So, U cannot be sufficient for θ. !
Example 6.3.6 (Example 6.2.13 Continued) Suppose that X , ..., X are iid
n
1
+
Uniform(0, θ), where θ (> 0) is unknown. Here, χ = (0, θ) and Θ = ℜ . We
wish to find a minimal sufficient statistic for θ. For two arbitrary data points
x = (x , ..., x ) and y = (y , ..., y ), both from χ, we have:
1 n 1 n

Let us denote a(θ) = I(0 < x < θ)/I(0 < y < θ). Now, the question is
n:n
n:n
this: Does the term a(θ) become free from θ if and only if x = y ?
n:n n:n
If we assume that x = y , then certainly a(θ) does not involve θ. It
n:n
n:n
remains to show that when a(θ) does not involve θ, then we must have x =
n:n
y . Let us assume that x ≠ y , and then show that a(θ) must depend on the
n:n
n:n
n:n
value of θ. Now, suppose that x = 2, y = .5. Then, a(θ) = 0, 1 or 0/0 when
n:n
n:n
θ = 1, 3 or .1 respectively. Clearly, a(θ) will depend upon the value of θ
whenever x ≠ y . We assert that the term a(θ) becomes free from θ if and
n:n
n:n
only if x = y . Hence, by the theorem of Lehmann-Scheffé, we claim that
n:n n:n
T = X , the largest order statistic, is minimal sufficient for θ. !
n:n
Remark 6.3.1 Let X , ..., X be iid with the pmf or pdf f(x; θθ θθ θ), x ∈ χ ⊆ ℜ,
n
1
k
θ θ θ θ θ = (θ , ..., θ ) ∈ Θ ⊆ ℜ . Suppose that a statistic T = T(X) = (T (X), ...,
k
1
1
T (X)) is minimal sufficient for θθ θθ θ. In general can we claim that r = k? The
r
answer is no, we can not necessarily say that r = k. Suppose that f(x; θ)
corresponds to the pdf of the N(θ, θ) random variable with the unknown
parameter θ > 0 so that k = 1. The reader should verify that
is a minimal sufficient for θ so that r = 2. Here,
we have r > k. On the other hand, suppose that X is N(µ, σ ) where µ and σ 2
2
1
2
are both unknown parameters. In this case one has θθ θθ θ = (µ, σ ) so that k = 2.
But, T = X is minimal sufficient so that r = 1. Here, we have r < k. In many
1
situations, of course, one would find that r = k. But, the point is that there is
no guarantee that r would necessarily be same as k.
The following theorem provides a useful tool for finding minimal sufficient