Student Notes To Accompany MS4214: STATISTICAL INFERENCE

hence
E( √ � ∞ √ 1
Z) = z
2Γ(1) e−z/2dz = 23/2Γ(3/2) 2Γ(1) = � π/2 ,
0
hence ˜σ = 2ˆσ/ √ π is unbiased for σ. If we let d = (n − 1) we similarly find for
general n that
E( √ � ∞ √
Z) = z
so
0
zd/2−1 2d/2Γ(d/2) e−z/2dz = 2(d+1)/2Γ((d + 1)/2)
2d/2Γ(d/2) c =
makes cS an unbiased estimator of σ.
√ dΓ(d/2)
√ 2Γ([d + 1]/2)
7. First of all c(θ) −1 = � ∞
0 x2 e −θx dx = Γ(3)/θ 3 = 2/θ 3 . Next, we get E( ˜ θ) =
E(2/X) = θ 3 � ∞
0 xe−θx dx = θ 3 θ −2 Γ(2) = θ and so ˜ θ is an unbiased estimator of θ.
To get the variance we use E( ˜ θ 2 ) = 2θ 3 � ∞
0 e−θx dx = 2θ 2 so Var( ˜ θ) = 2θ 2 −θ 2 = θ 2 .
To find the Fisher information for θ, we first calculate ∂ ln f(x|θ) = −x + 3/θ,
∂θ
and then calculate − ∂2
∂θ2 ln f(x|θ) = 3/θ2 . Thus eff( ˜ θ) = θ2 /(3θ2 ) = 1
3 .
E(ˆµ) = θ3
6
� ∞
0 x3e−θxdx = θ3Γ(4) 6θ4 = 1/θ = µ so ˆµ is an unbiased estimator of µ.
Similarly E(ˆµ 2 ) = θ3
18
� ∞
0 x4 e −θx dx = θ3 Γ(5)
θ 5 = 4
3θ 2 so Var(ˆµ) = 1
3θ 2 . To calculate
the CRLB we find for g(θ) = 1/θ that g ′ (θ) = −θ −2 so the lower bound is
I(θ) −1 {g ′ (θ) 2 } = θ2
3 θ−4 = 1
= Var(ˆµ).
3θ2 8. The formal definition of a sufficient statistic (S) is that the conditional distribu-
tion of the sample X = (X1, X2, . . . , Xn) given the value of S, that is knowing
that S = s, does not depend on θ.
If f(x|θ) is the joint distribution of the sample X, the statistic S is sufficient for
θ if and only if there exists function g(s|θ) and h(x) such that, for all sample
points {xi} and all θ, the density factorises f(x|θ) = g(s|θ)h(x).
f(x|θ) =
n�
x
i=1
θ−1
i e−xi {Γ(θ)} n
= exp[(θ − 1) � ln(xi)]
{Γ(θ)} n
× e − � xi
↑ ↑
�� �
g ln(xi)|θ h(x)
Thus, by the factorisation theorem, S = n�
ln(Xi) is sufficient for θ.
47
i=1
,