Find likelihood function and MLE for hypothesis I'm currently reading about statistics and I'm trying to understand likelihood function and MLE (maximum likelihood estimation).
I have come across an exercise which is basically like this:
I'm given four numbers including $x_1, x_2, x_3$ and $n$, where $n$ is the total number of trials. The numbers for $x_1, x_2$ and $x_3$ are all different.
Then I'm given a model $M_0$ which states that $(X_1, X_2, X_3)$ is distributed as the multinomial distribution with parameters $n$ and $(\pi_1, \pi_2, \pi_3)$ where $(\pi_1, \pi_2, \pi_3) \in \Pi^3$.
Finally, I have to consider the hypothesis $H_1: (\pi_1, \pi_2, \pi_3) = (1 - 2p, p, p)$, $p \in (0, 0.5)$
My task is then to show that the likelihood function for the hypothesis $H_1$ is given by:
$$
L(p) ={{n} \choose {x_1,x_2,x_3}}\cdot (1 - 2p)^{x_1}\cdot p^{x_2+x_3}
$$
and show that the maximum likelihood estimation for p is given by
$$
\hat{p} = \frac{x_2 +x_3}{2n}.
$$
I have been reading and watching a lot of videos about the likelihood function and MLE in order to try to understand it, but I don't even know where to begin in this case. Also, I'm a bit confused whether $p \in (0, 0.5)$ means that $p$ is either $0$ or $0.5$ since the brackets used aren't the normal set brackets { and }.
Can someone point me in the right direction?
 A: Recall that for a multinomial distribution the PDF is
$$
f(x_1, \ldots, x_k; n, \pi_1, \ldots, \pi_k) = \frac{n!}{x_1!\cdots x_k!}\pi^{x_1} \cdots \pi^{x_k} = {{n} \choose {x_1,\ldots,x_k}}\pi^{x_1} \cdots \pi^{x_k}
$$
with
$$
\sum_{i= 1}^k x_i = n
$$
In your case $k = 3$. The likelihood evaluated at the point ${\bf p} = (1 - 2p, p, p)$ is then
$$
L({\bf p}) = {{n} \choose {x_1,x_2,x_3}}(1- 2p)^{x_1} p^{x_2} p^{x_3} = {{n} \choose {x_1,x_2,x_3}}(1- 2p)^{x_1} p^{x_2+ x_3} = L(p)
$$
Now, to find the value of $p$ for which $L(p)$ we need to solve $dL/dp = 0$ 
\begin{eqnarray}
\frac{dL}{dp} &=& (1-2 p)^{x_1} (x_2 + x_3) p^{x_2 + x_3-1}-2
   x_1 (1-2 p)^{x_1-1} p^{x_2 + x_3} \\
&=& (1 - 2p)^{x_1}p^{x_2 + x_3} \left(\frac{x_2 + x_3}{p} - 2\frac{x_1}{1 - 2p} \right)
\end{eqnarray}
Therefore the value of $\hat{p}$ for which $dL(\hat{p})/dp = 0$ is
$$
\frac{x_2 + x_3}{\hat{p}} - 2\frac{x_1}{1 - 2\hat{p}} = 0 ~~~\Rightarrow ~~~ \hat{p} = \frac{x_2 + x_3}{2(x_1 + x_2 + x_3)}
$$
or equivalently
$$
\hat{p} = \frac{x_2 + x_3}{2n}
$$
A: The notation $p\in(a,b)$ means $a<p<b$, and $p\in[a,b]$ means $a\le p \le b$, and $p\in(a,b]$ means $a<p\le b$ and $p\in[a,b)$ means $a\le p<b.$ The notation $p\in\{a,b\},$ on the other hand, means either $p=a$ or $p=b.$
You have
$$
L(p) ={n \choose x_1,x_2,x_3} (1 - 2p)^{x_1}\cdot p^{x_2+x_3}.
$$
Thus
$$
\ell(p) = \log L(p) = \text{constant} + x_1 \log(1-2p) + (x_2+x_3)\log p.
$$
Therefore
\begin{align}
\ell\,'(p) & = \frac{-2x_1}{1-2p} + \frac{x_1+x_2} p \\[10pt] & = \frac{(x_2+x_3) -2p(x_1+x_2+x_3)}{p(1-2p)} \quad \begin{cases} > 0 & \text{if } 0<p<\dfrac{x_2+x_3}{2(x_1+x_2+x_3)}, \\[10pt] < 0 & \text{if } \dfrac{x_2+x_3}{2(x_1+x_2+x_3)} < p < 1. \end{cases}
\end{align}
The problem ought to have said $p\in[0,1/2]$ rather than $p\in(0,1/2).$ The probability of an endpoint maximum is clearly more than $0$ and the MLE would be at an endpoint in that case.
