Page 309 - Probability and Statistical Inference

286 6. Sufficiency, Completeness, and Ancillarity

which is free from λ. In other words, is a sufficient statistic for the
unknown parameter λ. !
Example 6.2.3 Suppose that X has the Laplace or double exponential pdf
given by f(x; θ) = 1/2θe –|x|/θ I(–∞ < x < ∞) where θ (> 0) is an unknown
parameter. Let us consider the statistic T = | X |. The difference between X
and T is that T provides the magnitude of X, but not its sign. Conditionally
given T = t (> 0), X can take one of the two possible values, namely t or –t,
each with probability 1/2. In other words, the conditional distribution of X
given T = t does not depend on the unknown parameter θ. Hence, | X | is a
sufficient statistic for θ. !
Example 6.2.4 Suppose that X , X are iid N(θ, 1) where θ is unknown, –
1
2
∞ < θ < ∞. Here, χ = ℜ and Θ = ℜ. Let us consider the specific statistic T =
X + X . Its values are denoted by t ∈ = ℜ. Observe that the conditional pdf
1
2
of (X , X ), at (x , x ) would be zero if x + x ≠ t when T = t has been
2
1
2
1
2
1
observed. So we may work with data points (x , x ) such that x + x = t once
2
2
1
1
T = t is observed. Given T = t, only one of the two X’s is a free-standing
variable and so we will have a valid pdf of one of the X’s. Let us verify that T
is sufficient for θ by showing that the conditional distribution of X given T =
1
t does not involve θ. Now following the Definition 4.6.1 of multivariate nor-
mality and the Example 4.6.1, we can claim that the joint distribution of (X ,
1
T) is N (θ, 2θ, 1, 2, ), and hence the conditional distribution of X given
2 1
T = t is normal with its mean = θ + = 1/2t and conditional
2
variance = 1 – ( ) = 1/2, for all t ∈ ℜ. Refer to the Theorem 3.6.1 for the
expressions of the conditional mean and variance. This conditional distribu-
tion is clearly free from θ. In other words, T is a sufficient statistic for θ. !
How can we show that a statistic is not sufficient for θ?
Discussions follow.
If T is not sufficient for θ, then it follows from the Definition 6.2.2 that the
conditional pmf or pdf of X , ..., X given T = t must depend on the unknown
1
n
parameter θ, for some possible x , ..., x and t.
1 n
In a discrete case, suppose that for some chosen data x , ..., x ,
n
1
the conditional probability P{X = x , ..., X = x | T = t},
n
1
1
n
involves the parameter θ. Then, T can not be sufficient for θ.
Look at Examples 6.2.5 and 6.2.7.
Example 6.2.5 (Example 6.2.1 Continued) Suppose that X , X , X are
3
1
2
iid Bernoulli(p) where p is unknown, 0 < p < 1. Here, χ = {0, 1}, θ = p,