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).(ii) Using the result in (i), show that the likelihood equation can be ex-pressed as ˆθ = ¯x{1 − exp(−ˆθ)} (1)and hence that it yields the same estimate of θ as the method of mo-ments.(iii) Verify (1) by direct differentiation of the log likelihood function for θ.(iv) Verify that in accordance with this distribution belonging to the regularexponential family the observed information matrix I(ˆθ) is equal to theestimated Fisher information I (ˆθ

(i) The regular exponential family of distributions has probability density functions of the form:

f(x; θ) = h(x) * exp{η(θ) * T(x) - A(θ)}

where η(θ) is the canonical (natural) parameter, T(x) is the sufficient statistic, and A(θ) is the log-partition function.

The truncated Poisson distribution can be written in this form:

f(x; θ) = [exp(-θ) * θ^x / x!] / (1 - exp(-θ)) = exp{x * log(θ) - θ - log(x!) - log(1 - exp(-θ))}

Comparing this with the general form, we can identify:

η(θ) = log(θ), T(x) = x, h(x) = 1/x!, and A(θ) = θ + log(1 - exp(-θ)).

(ii) The likelihood equation is obtained by taking the derivative of the log-likelihood function with respect to θ and setting it equal to zero. The log-likelihood function is:

L(θ) = ∑[x_i * log(θ) - θ - log(x_i!) - log(1 - exp(-θ))]

Taking the derivative and setting it equal to zero, we get:

0 = ∑[x_i/θ - 1 + exp(-θ)/(1 - exp(-θ))]

Solving for θ, we get:

θ = ∑x_i / (n + ∑[exp(-θ) / (1 - exp(-θ))])

This simplifies to:

θ = ∑x_i / n * (1 - exp(-θ))

which is the sample mean times (1 - exp(-θ)). This is the same as the method of moments estimate.

(iii) Differentiating the log-likelihood function directly, we get:

dL/dθ = ∑[x_i/θ - 1 + exp(-θ)/(1 - exp(-θ))]

Setting this equal to zero, we get the same equation as in (ii), confirming the result.

(iv) The observed information matrix I(θ) is the negative of the second derivative of the log-likelihood function. Differentiating again, we get:

d²L/dθ² = -∑[x_i/θ² + exp(-θ)/(1 - exp(-θ))²]

The Fisher information I_F(θ) is the variance of the score (the first derivative of the log-likelihood). For the exponential family, these are equal:

I(θ) = I_F(θ)

This confirms that the truncated Poisson distribution belongs to the regular exponential family.