I'm really stuck with this problem:
Suppose a single sample $X$ is taken from a discrete distribution with pmf given by
$p(x;\theta) = \left\{\begin{array}{lcc}\frac{1+\theta}{12}&\mbox{if}&x=0\\ \frac{2\theta}{12}&\mbox{if}&x=1\\ \frac{11-3\theta}{12}&\mbox{if}&x=2\\ \end{array}\right.$
where $\theta = \{1,2\}$. Give the MLE of $\theta$ based from $x$.
From what I understand, getting $L(\theta)$ means multiplying the pmf of each $x$. I ended up with
$L(\theta) = (\frac{1+\theta}{12})(\frac{2\theta}{12})(\frac{11-3\theta}{12})$
Then taking the $ln$...
$lnL(\theta) = ln(1+\theta) + ln(2\theta) + ln(11-3\theta) - 3ln(12)$
I was proceeding to differentiating but I feel that the equation must have an $x$ in it. I can't figure out though. I kinda sure I have misunderstood something.
If $x=0$ then $p(0,1)=1/6$ and $p(0,2)=1/4$. So given the value of $x$ the MLE would be $\theta=2$ as this is more likely. Just repeat for the other values of $x$.
The general case is as below. If there are $n_i$ cases of $x=i$, then $log_L(\theta)= n_0*Log(1+\theta)+ n_1*Log(2\theta) + n_2*Log(11-3 \theta)$. If $\theta$ was continuous you would differentiate this and solve. However all you need to do is feed in $\theta=1$ and $\theta=2$ and see which is largest.