## STB – 2019

• 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 X1, . . . , Xn 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 x1,x2,…,xn 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.