- Consider a count variable X following a Poisson distribution with parameter θ > 0, where zero count (i.e., X = 0) is not observable. We have n observations X
_{1}, . . . , X_{n} from this distribution. Let denote the sample mean.

a) Derive the quantity for which is an unbiased estimator.

b) Suppose that the observed value of is strictly greater than 1. Show that the likelihood function of θ has a unique maximiser.

The pmf of a Poisson distribution with parameter is given by:

We know that X follows a Poisson distribution with parameter \theta and X=0 is not observable. Under such condition the probability mass function(pmf) of X is given by:

g(x | \theta ) =Pr[ X=x | X≠0] \\
\hskip{1.8cm}=Pr[X=x]/Pr[X≠0] \\
\hskip{0.8cm}=f(x)/(1-f(0) ) \\
\hskip{2.8cm}=\frac{θ^x e^{-θ}} {x!(1-e^{-θ} )} \quad ,x=1,2,3,…

**(a)** Now,

E (X)= ∑_{x=1}^∞ E\left[x g(x)\right]= ∑_{x=1}^∞E\left[x \frac{θ^x e^{-θ}} {x!(1-e^{-θ} )} \right]= ∑_{x=1}^∞\frac{θ^x e^{-θ}} {(x-1)!(1-e^{-θ} )} \\
= \frac{\theta e^{-θ}} {(1-e^{-θ} )}∑_{x=1}^∞\frac{θ^{x-1}} {(x-1)!} = \frac{\theta e^{-θ}} {(1-e^{-θ} )}∑_{x=0}^∞\frac{θ^{x}} {x!} = \frac{\theta} {(1-e^{-θ} )}

Also, as are random samples drawn from the distribution so,

E(\bar{X})= \frac{1}{n} \sum_{i=1}^n E(X_i) = \frac{\theta} {(1-e^{-θ} )} \quad
\left[ i.e. \bar{X} \text{is an u.e. of } \frac{\theta} {(1-e^{-θ} )} \right]

**(b)** The Likelihood function of x_{1},x_{2},…,x_{n} is given by:

L(θ)= ∏_{x=1}^ng(x_i|\theta ) \\
\hskip{2.4cm} =∏_{x=1}^n
((1-e^{-θ} )^{-1} \frac{ θ^{x_i} e^{-θ}}{(x_i !)} \\
\hskip{1.8cm}=\frac{1}{(e^θ-1)^n} \frac{θ^{\sum x_i }}{∏_{i=1}^{n}x_i !}

Taking ‘ln’ on both sides we get,

l(θ)=ln(L(θ))=-n log(e^θ-1)+n \bar{x} ln(θ)-ln(∏x_i !)

Differentiation wrt \theta would give,

\frac{d}{d\theta}l(\theta) = \frac{n}{1-e^{-\theta}} + \frac{n\bar{x}}{\theta} \\
\hskip{.5cm} \frac{d^2}{d\theta^2}l(\theta) = - \frac{n}{(1-e^{-\theta})^2}-\frac{n\bar{x}}{\theta^2}

Now,

\left[\frac{d}{d\theta}l(\theta) \right]_{\theta=\hat{\theta}} =0 \Rightarrow \frac{\hat{\theta}}{1-e^{-\theta}} = \bar{x} \\
\left[\frac{d^2}{d\theta^2}l(\theta) \right]_{\theta=\hat{\theta}}= -\frac{n\bar{x}^2}{\hat{\theta}^2}-\frac{n\bar{x}}{\hat{\theta}^2}<0

Thus, the likelihood function of θ has a unique maximiser.