Page 317 - Probability and Statistical Inference

294 6. Sufficiency, Completeness, and Ancillarity

In many examples and exercises the X’s are often assumed iid. But, if the
X’s are not iid, then all we have to do is to carefully write down the likelihood
function L(θ) as the corresponding joint pmf or pdf of the random variables
X , ..., X . The Neyman factorization would then hold.
1 n
Example 6.2.14 Suppose that X , X are independent random variables,
2
1
with their respective pdf’s f(x ; θ) = θe –θx 1 I(0 < x < ∞) and g(x ; θ) = 2θe –2θx 2
2
1
1
I(0 < x < ∞), where θ > 0 is an unknown parameter. For 0 < x , x < ∞, the
2 1 2
likelihood function is given by the joint pdf, namely
From (6.2.14) it is clear that the Neyman factorization holds and hence the
statistic T = X + 2X is sufficient for θ. Here, X , X are not identically
1 2 1 2
distributed, and yet the factorization theorem has been fruitful.!
The following result shows a simple way to find sufficient statistics when
the pmf or the pdf belongs to the exponential family. Refer back to the Sec-
tion 3.8 in this context. The proof follows easily from the factorization (6.2.9)
and so we leave it out as the Exercise 6.2.15.
Theorem 6.2.2 (Sufficiency in the Exponential Family) Suppose that
X , ..., X are iid with the common pmf or the pdf belonging to the k-param-
1
n
eter exponential family defined by (3.8.4), namely

with appropriate forms for g(x) ≥ 0, a(θθ θθ θ) ≥ 0, b (θθ θθ θ) and R (x), i = 1, ..., k.
i i
Suppose that the regulatory conditions stated in (3.8.5) hold. Denote the sta-
tistic j = 1, ..., k. Then, the statistic T = (T , ..., T ) is
k
1
jointly sufficient for θθ θθ θ.
The sufficient statistics derived earlier in the Examples 6.2.8-6.2.11 can
also be found directly by using the Theorem 6.2.2. We leave these as the
Exercise 6.2.14.

6.3 Minimal Sufficiency

We noted earlier that the whole data X must always be sufficient for the
unknown parameter θθ θθ θ. But, we aim at reducing the data by means of sum-
mary statistics in lieu of considering X itself. From the series of examples
6.2.8-6.2.14, we found that the Neyman factorization provided sufficient