Question

Let X1, X2,...,Xn be a random sample from a Poisson distribution with parameter λ .
Find an estimator of λ using
i) the method of moments;
ii) the method of maximum likelihood.
Also, compare the estimators obtained in parts i) and ii).

Answer 1

For this distribution

$$E(X_i)=Var(X_i)=\lambda$$

Hence by comparing the first and second population and sample moments we get two different estimators of the same parameter,

$$\widehat{\lambda}_1=\bar{X}$$

$$\widehat{\lambda}_2={1 \over n}\displaystyle\sum_{i=1}^n{X_i}^2-{\bar{X_i}}^2$$

The likelihood is

$$L(\lambda|x)=\displaystyle\prod_{i=1}^n{\lambda^{x_i}e^{-\lambda} \over x_i!}={\lambda^{\textstyle\sum_{i=1}^nx_i}e^{-n\lambda} \over \textstyle\prod_{i=1}^nx_i!}$$

We need to find the value of λ which maximizes the likelihood. This value will also maximize

$$l(\lambda|x)=logL(\lambda|x),$$

which is easier to work with. Now, we have

$$l(\lambda|x)=\displaystyle\sum_{i=1}^nx_ilog\lambda-n\lambda-\displaystyle\sum_{i=1}^nlog(x_i!).$$

The value of  which λ maximizes l (λ | x) is the solution of dl /dλ =0. Thus, solving the equation

$${dl \over d\lambda}={\textstyle\sum_{i=1}^nx_i \over \lambda}-n=0$$

yields the estimator 

$$\widehat{\lambda}=T(X)=\textstyle\sum_{i=1}^nX_i/n=\bar{X},$$

which is the same as the Method of Moments estimator. The second derivative is negative for all λ hence,

$\widehat{\lambda}$ indeed maximizes the log-likelihood.