Question A.3: Find the maximum likelihood estimator for the parameter μ in......

Assume n data points,\ x_1 , x_2, . . . , x_n . Since the Poisson is a discrete distribution and there is only one unknown parameter μ, m = 1, and the likelihood function is defined by Equation (A.8) as

\ L(\theta_1, … , \theta_m)=Pr[X_1=x_1,X_2=x_2, … , X_n=x_n]\\ = Pr[X=x_1] Pr[X=x_2] … Pr[X=x_n]\\=p(x_1;\theta_1, … , \theta_m) p(x_2;\theta_1, … , \theta_m) … p(x_n;\theta_1, … , \theta_m)\\=\prod\limits_{i=1}^{n}{p(x_i;\theta_1, … , \theta_m)}.               (A.8)

\ L(\mu)=(e^{-\mu}\frac{\mu^{x_1}}{x_1!}) (e^{-\mu}\frac{\mu^{x_2}}{x_2!}) … (e^{-\mu}\frac{\mu^{x_n}}{x_n!})\\[0.3cm]=e^{-\mu n}\frac{\mu^{x_1+x_2+ … + x_n}}{(x_1!) (x_2!) … (x_n!)}\\[0.3cm]=e^{-\mu n}\frac{\mu^ {\sum\limits_{i=1}^{n}{x_i}}}{\prod\limits_{i=1}^{n}{x_i!}},

where it should be noted that L is a function only of μ. It is best to take the logarithm of both sides before differentiating:

\ln  L(\mu)=-\mu n+(\sum\limits_{i=1}^{n}{x_i})  \ln  \mu  –  \ln(\prod\limits_{i=1}^{n}{x_i!}),\\ \frac{1}{L}\frac{d}{d\mu}L(\mu)=-n+\frac{1}{\mu}\sum\limits_{i=1}^{n}{x_i}.

The maximum occurs at dL/dμ = 0. It follows that

\ -n+\frac{1}{\mu}\sum\limits_{i=1}^{n}{x_i}=0

and

\ \hat{\mu}=\frac{1}{n}\sum\limits_{i=1}^{n}{x_i}.                   (A.10)

The second derivative confirms this to be a maximum. Therefore, by Equation (A.10), the best estimate for μ turns out to be the average of the data points. This discovery justifies our parameter choice in Example A.2 of using the computed mean estimate as opposed to using the variance.