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Given a non-normal distribution (a table) and a specific sample, how do I find the $MLE$ of the mean using the given sample?

To be more specific I was given the following distribution:

$$ \begin{array}{|c|c|c|c|} \hline \text{Number of Weeks} & 1 & 2 & 3 \\ \hline Prob. & 2\theta & 3\theta & 1-5\theta \\ \hline \end{array}$$

And the sample says that for 5 different people the results were: $1,2,3,2,3$ weeks. I know that $\bar{x} = \frac{11}{5}$ and that $E(x) = 1\cdot2\theta + 2\cdot3\theta + 3\cdot(1-5\theta) = 8\theta -15\theta +3 = -7\theta + 3$. I also know that $E(\bar{x}) = E(x)$.

What I don't know is how do I estimate the MLE using the above information.

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Hint: Given this data, the likelihood is $$\mathcal{L}(\theta |x) \propto Pr(X|\theta) = (2\theta)^1 (3\theta)^2 (1-5\theta)^2 $$

Now find the $\theta$ which maximises this, with the constraint that $2\theta$, $3\theta$ and $(1-5\theta)$ must be all be non-negative as they are probabilities.

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    $\begingroup$ @Quaker In your question, you calculate the mean in terms of $\theta$. And MLE's have a property called functional invariance. $\endgroup$
    – Gregor Thomas
    Commented Jan 28, 2014 at 20:21
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    $\begingroup$ @shujaa I was asked to find the MLE of $\Theta$ in one of the former tasks, in this one I was asked to find the MLE of the mean not the parameter. There is a possibility that the question is not written properly and I would love to know if this is the case. $\endgroup$
    – Quaker
    Commented Jan 28, 2014 at 20:30
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    $\begingroup$ @Quaker The mean is $\phi(\theta)$ for a function $\phi$ given by yourself in your question. That's why shujaa points out the functional invariance property. $\endgroup$
    – Stéphane Laurent
    Commented Jan 28, 2014 at 20:35
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    $\begingroup$ (+1: very clear). Quaker, another option to consider--to help with the insight--is to rewrite the likelihood directly in terms of the mean, which you note equals $3-7\theta$. What the others are saying is, it doesn't matter whether you obtain the MLE for $\theta$ or for the mean, because the MLE of the mean must equal $3$ minus $7$ times the MLE of $\theta$. (Such nice relationships among estimators do not hold for most other estimation procedures.) $\endgroup$
    – whuber ♦
    Commented Jan 28, 2014 at 20:45
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    $\begingroup$ Yes. Part of this is obvious and part is subtle. The obvious thing is that whatever value of $\theta$ maximizes the likelihood will correspond to the value of $3-7\theta$ that maximizes the likelihood, because the function $\theta\to 3-7\theta$ is one-to-one. You're not changing the likelihood; you're just changing how you express the parameter on which it depends--kind of like converting between pounds and kilos. The subtle parts in real applications are (1) you better check that the correspondence really is one-to-one and (2) confidence intervals may change in more complicated ways. $\endgroup$
    – whuber ♦
    Commented Jan 28, 2014 at 21:24

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