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5.2 Parameters and Identifiability 135
It follows that, for any x = 0, 1,...,
x ∞
η n n n
Pr(X = x) = exp(−λ) (1 − η) λ /n!
1 − η x
n=0
x ∞
η exp(−λ) n n
= (1 − η) λ /(n − x)!
1 − η x!
n=x
x
∞
η x x j j
= exp(−λ)(1 − η) λ (1 − η) λ /j!
1 − η
j=0
exp(−λ)
x
= (ηλ) exp{(1 − η)λ}
x!
x
= (ηλ) exp(−ηλ)/x!
so that X has a Poisson distribution with mean ηλ.
Hence, the model function is given by
x
p(x; θ) = (ηλ) exp(−ηλ)/x!, x = 0, 1,....
Since the distribution of X depends on θ = (η, λ) only through ηλ, the parameterization
given by θ is not identifiable; that is, we may have (η 1 ,λ 1 ) = (η 2 ,λ 2 ) yet η 1 λ 1 = η 2 λ 2 .
Suppose that instead we parameterize the model in terms of ψ = ηλ with parameter
space R . The model function in terms of this parameterization is given by
+
x
ψ exp(−ψ)/x!, x = 0, 1,...
and it is straightforward to show that this parameterization is identifiable.
Statistical models are often based on independence. For instance, we may have indepen-
dent identically distributed random variables X 1 , X 2 ,..., X n such that X 1 has an absolutely
continuous distribution with density p 1 (·; θ) where θ ∈ . Then the model function for the
model for (X 1 ,..., X n )isgiven by
n
p(x 1 ,..., x n ; θ) = p 1 (x j ; θ);
j=1
a similar result holds for discrete distributions. More generally, the random variables
X 1 , X 2 ,..., X n may be independent, but not identically distributed.
Example 5.7 (Normal distributions). Let X 1 ,..., X n denote independent identically dis-
tributed random variables, each with a normal distribution with mean µ, −∞ <µ< ∞
and standard deviation σ> 0. The model function is therefore given by
n
1 1 2 n
p(x; θ) = n exp − 2 (x j − µ) , x = (x 1 ,..., x n ) ∈ R ;
n
σ (2π) 2 2σ
j=1
+
here θ = (µ, σ) and = R × R .
Now suppose that X 1 ,..., X n are independent, but not identically distributed; specifi-
cally, for each j = 1, 2,..., n, let X j have a normal distribution with mean βt j and standard
deviation σ> 0, where t 1 ,..., t n are fixed constants and β and σ are parameters. The model