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376 7. Point Estimation
statistic g(U) is the UMVUE for T(θθ θθ θ). Then, we have:
Proof Since the statistic U is assumed both complete and sufficient, by
virtue of the Lehmann-Scheffé Theorems it follows that E [T | U] must be the
θ
best unbiased estimator of T(θθ θθ θ). But g(U) is the unique UMVUE of T(?) and
hence the result follows immediately. !
Example 7.5.15 Suppose that X , ..., X are iid N(µ, σ ) where µ is un-
2
1 n
known but σ is known with n = 4, −∞ < µ < ∞, 0 < σ < ∞ and χ = ℜ. Let
2
which is unbiased for T(µ) = 2µ and U = . But, U is
complete sufficient for µ and hence g(U) = 2U is the unique UMVUE for T(µ).
Thus, in view of the Theorem 7.5.5 we can immediately write E [T | .2
µ
2
2
2
Instead if we had T = (X + X ) , then E [T] = 4µ + 2σ = T(µ). But, the
1
µ
2
unique UMVUE of µ was earlier (Example 7.5.7) found to be
2
Hence in view of the Theorem 7.5.5, we can immediately write
!
Suppose that X , ..., X are random samples from Poisson(λ),
n
1
0 < λ < ∞. Among other things, the following example shows
easily that V (S ) > V for all λ.
2
λ
λ
Example 7.5.16 Suppose that X , ..., X are iid Poisson(λ) with 0 < λ < ∞
1 n
unknown and n ≥ 2. Let us denote T = S , the sample variance, and U =
2
Obviously, E [T] = λ = T(λ). But, U is complete sufficient and so U is the
λ
unique UMVUE for T(λ). Thus, in view of the Theorem 7.5.5, we can write
E [T | ] = .
λ
2
Now, we use the Theorem 3.3.1, part (ii) to rewrite V (S ) as
θ
which exceeds V [ ] , for all λ, because V(S | ) is a positive random
2
λ
variable whose expectation is positive. For the direct calculation of V (S ),
2
θ
however, refer to the expression given by (5.2.23). !
Example 7.5.17 Let X , ..., X be iid Uniform (0, θ) where θ(> 0) is the
n
1
unknown parameter with n ≥ 2. Then, U = X is complete sufficient for θ.
n:n
Consider Obviously, E [T] = ½θ+1/3θ = T (θ). Now, since U
2
θ
-1
n-1 -n
has its pdf given by nu θ I (0 < u < θ), we have E [X ] = n(n+1) θ and
θ n:n
Hence, the unique UMVUE of T(θ) is given by
Thus, in view of the Theorem 7.5.5,