Page 348 - Probability and Statistical Inference

6. Sufficiency, Completeness, and Ancillarity 325

But, observe that P {W = w | U = u} must be free from θ since U is a
θ
sufficient statistic for θ. So, let us write g(u) = P {W = w | U = u}. Now,
θ
E [g(U)] = P (W = w) which we had denoted earlier by h(w), and this is free
θ
θ
from θ. Hence, we have verified that {g(U) - h(w)} is a genuine statistic.
Now, we note that E [g(U) – h(w)] ≡ 0 for all θ ∈ Θ and use the fact that
θ
the statistic U is complete too. Thus, by the Definition 6.6.3 we must have
g(u) – h(w) = 0 w.p.1, that is, g(u) ≡ h(w) for all w ∈ W, u ∈ U. In other
words, we have shown the validity of (6.6.10). !
Example 6.6.12 Let X , ..., X be iid N(µ, σ ) with n ≥ 2, (µ, σ) ∈ ℜ × ℜ +
2
n
1
where µ is unknown, but σ is known. Let U = which is complete sufficient
statistic for µ. Observe that W = S is an ancillary statistic for µ. The fact that
2
W is ancillary can be claimed from its explicit distribution or by appealing to
the location family of distributions. Now, by Basu’s Theorem, the two statis-
tics and S are then independently distributed. The Theorem 4.4.2 showed
2
this via Helmert transformation. !
Example 6.6.13 (Example 6.6.12 Continued) Let V = X - X , the sample
n:n
n:1
range. Then, and S/V are independently distributed. Also, and (X – )
n:n
are independent. In the same spirit, and (X – ) are independent. Is
2
n:1
independent of | X – | /S? The ancillarity of the corresponding statistics
n:n
can be verified by appealing to the location family of distributions. We leave
out the details as exercises. !
Example 6.6.14 Let X , ..., X be iid N(µ, σ ) with n ≥ 2, (µ, σ) ∈ ℜ × ℜ +
2
n
1
2
where σ is unknown, but µ is known. Let U = which is a
complete sufficient statistic for σ . Observe that W = (X – )/U is ancillary
2
1
for σ. Immediately we can claim that and W are independently
distributed. Also, and (X – )/S are independent where S is
2
n:1
the customary sample variance. Here, ancillarity of the corresponding statis-
tics can be verified by appealing to the scale family of distributions. We leave
out the details as exercises. !
2
In the N(µ, s ) case with µ, s both unknown, Basu’s Theorem can
be used to show that and S are independent. See the Example
2
6.6.15. Compare its elegance with the brute-force approach
given in the Example 4.4.9 via Helmert transformations.
2
Example 6.6.15 Suppose that X , ..., X are iid N(µ, σ ) with n ≥ 2, (µ,
1 n
σ) ∈ ℜ × ℜ where µ and σ are both assumed unknown. Let us see how
+
we can apply Basu’s Theorem to show that and S are independently
2
distributed. This clever application was originated by D. Basu. Let us have