Student Notes To Accompany MS4214: STATISTICAL INFERENCE

Example 2.13 (Poisson). Let X = (X1, . . . , Xn) be independent and Poisson dis-
tributed with mean λ so that
f(x; λ) =
n�
i=1
λxi xi! e−λ = λΣxi
�
i xi! e−nλ .
Take k { � xi; λ} = λ Σxi e −nλ and h(x) = ( � xi!) −1 , then t(x) = �
i xi is sufficient. �
Example 2.14 (Binomial). Let X = (X1, . . . , Xn) be independent and Bernoulli dis-
tributed with parameter θ so that
f(x; θ) =
n�
θ xi 1−xi Σxi n−Σxi
(1 − θ) = θ (1 − θ)
i=1
Take k { � xi; θ} = θ Σxi (1 − θ) n−Σxi and h(x) = 1, then t(x) = �
i xi is sufficient. �
Example 2.15 (Uniform). Factorization criterion works in general but can mess up if
the pdf depends on θ. Let X = (X1, . . . , Xn) be independent and Uniform distributed
with parameter θ so that X1, X2, . . . , Xn ∼ Unif(0, θ). Then
f(x; θ) = 1
θ n
0 ≤ xi ≤ θ ∀ i.
It is not at all obvious but t(x) = max(xi) is a sufficient statistic. We have to show
that f(x|t) is independent of θ. Well
Then
So
f(x|t) =
f(x, t)
fT (t) .
P (T ≤ t) = P (X1 ≤ t, . . . , Xn ≤ t)
n�
�
t
= P (Xi ≤ t) =
θ
i=1
FT (t) = tn
θn ⇒ fT (t) = ntn−1
θn Also
f(x, t) = 1
θn ≡ f(x; θ).
Hence
f(x|t) =
1
,
ntn−1 and is independent of θ. �
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� n