Maximum likelihood estimate of the parameter in Poisson distribution

Given that the number of goals scored per match by a football team can be assumed to be a Poisson random variable with mean $$\theta$$. In eight games, the team scores 3, 6, 2, 5, 4, 1, 4, 5 goals.

(a) Let $$X$$ that follows a Poisson distribution with mean $$\theta$$ be a random variable that represent the number of goals scored by a football team in a game.

The maximum likelihood estimate of $$\theta$$ is $$\hat{\theta}=\dfrac{1}{n}\sum_{i=1}^{5}X_{i}$$

(b)Considering the probability of $$k$$ number of goals scored in a game, we have $$p(X=k)=\dfrac{e^{-\theta}\theta^{k}}{k!}.$$

The probability of no goals in a match is $$p(X=0)=e^{-\theta}.$$ $$\textbf{Please have I done the right thing ?}$$

(c)The maximum likelihood estimate of the probability of no goals in a game is \begin{align*} \hat{\theta}&=\dfrac{1}{n}\sum_{i=1}^{5}X_{i}\\&=\dfrac{1}{n}\sum_{i=1}^{5}0\qquad \textrm{(for each X_{i}=0)}\\ &=0 \end{align*} $$\textbf{Please have I done the right thing ?}$$

Your sample size is $$n=8$$, the observed sample being $$(x_1,x_2,\ldots,x_8)=(3,6,2,5,4,1,4,5)$$.
• The maximum likelihood estimate of $$\theta$$ is $$\hat\theta(x_1,\ldots,x_8)=\frac{1}{8}\sum_{i=1}^8 x_i$$
• By invariance property of MLE, maximum likelihood estimate of the probability of no goals in a game is simply $$e^{-\hat\theta}$$.
• I answered that part. If $\hat\theta$ is the MLE of $\theta$, then $g(\hat\theta)$ is the MLE of $g(\theta)$ for any function $g$. Why zero? You don't justify your answers. – StubbornAtom Mar 8 at 16:06