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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*} $\textbf{Please have I done the right thing ?}$

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How do you justify your answers?

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 $e^{-\hat\theta}$.

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  • $\begingroup$ What about the maximum likelihood estimate of the probability of no goals in a game? $\endgroup$ – steve mike Mar 8 at 15:52
  • $\begingroup$ What about it ? $\endgroup$ – StubbornAtom Mar 8 at 15:54
  • $\begingroup$ Would it be correct for me to say the maximum likelihood estimate of the probability of no goals in a game is 0 ? $\endgroup$ – steve mike Mar 8 at 15:58
  • $\begingroup$ 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. $\endgroup$ – StubbornAtom Mar 8 at 16:06

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