(a)The Poisson distribution truncated at zero has probability function givenbyf (x; θ) = exp(−θ)θxA(θ)x! , x = 1, 2, ... (θ > 0),whereA(θ) = 1 − exp(−θ).Let X1, . . . , Xn be a random sample of size n from this truncated Poissondistribution.(i) Show that this distribution belongs to the regular exponential distri-bution and give its canonical parameter c(θ) and the correspondingsufficient statistic T (X1, ..., Xn).
Question
(a)The Poisson distribution truncated at zero has probability function givenbyf (x; θ) = exp(−θ)θxA(θ)x! , x = 1, 2, ... (θ > 0),whereA(θ) = 1 − exp(−θ).Let X1, . . . , Xn be a random sample of size n from this truncated Poissondistribution.(i) Show that this distribution belongs to the regular exponential distri-bution and give its canonical parameter c(θ) and the correspondingsufficient statistic T (X1, ..., Xn).
Solution
The Poisson distribution truncated at zero can be written in the form of a regular exponential family. The regular exponential family has the probability density function (pdf) given by:
f(x; θ) = h(x) * exp{η(θ)T(x) - A(θ)}
where h(x) is the base measure, η(θ) is the natural parameter, T(x) is the sufficient statistic, and A(θ) is the log partition function.
For the given truncated Poisson distribution, we can write the pdf as:
f(x; θ) = [exp(-θ) * θ^x / x!] / (1 - exp(-θ))
This can be rewritten as:
f(x; θ) = exp{x log(θ) - θ - log(x!) - log(1 - exp(-θ))}
Comparing this with the pdf of the regular exponential family, we can identify:
h(x) = 1/x!, η(θ) = log(θ), T(x) = x, A(θ) = θ + log(1 - exp(-θ)).
So, the canonical parameter c(θ) is log(θ) and the corresponding sufficient statistic T(X1, ..., Xn) is ∑Xi (the sum of all Xi from i=1 to n).
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