Suppose that $X$ is a discrete random variable with \begin{pmatrix} 0 & 1 &2&3 \\ 2θ/3& θ/3 & 2(1-θ)/3 &(1-θ)/3 \end{pmatrix}
where $0\le θ\le 1$ is a parameter.
10 independent observations were taken: $( 3,0,2,1,3,2,1,0,2,2)$
1)What is the maximum likehood estimate of $θ$?
2) What is the approximate standard error of the maximumlikehood estimate.
(Question similar to this.)
1): $$L(\theta) = P(X=0| \theta)^2 P(X=1| \theta)^2 P(X=2| \theta)^4 P(X=3| \theta)^2= \bigg( \frac{2}{3} \theta \bigg)^2 \bigg( \frac{1}{3} \theta \bigg)^2 \bigg( \frac{2}{3} (1- \theta) \bigg)^4 \bigg( \frac{1}{3} (1-\theta) \bigg)^2\\$$
\begin{align*} l(\theta) &=2 ln( \frac{2}{3} \theta) + 2 ln( \frac{1}{3} \theta)+ 4 ln( \frac{2}{3} (1- \theta) )+ 2 ln ( \frac{1}{3} (1-\theta) )\\ \end{align*}
$$l'(\theta)= \frac{2}{\theta}+ \frac{2}{\theta}- \frac{4}{1-\theta}- \frac{2}{1-\theta} = \frac{4}{\theta}- \frac{6}{1-\theta}$$
We say first derivate is zero and we get the estimate: \begin{align*} \hat{\theta} &= \frac{4}{10} = \frac{2}{5} \end{align*}
So far, everything is correct.
2): I am stuck on this part as I get the wrong result, and I don't know why. Here is my path to solution:
$$l''(\theta)= \frac{-4}{\theta^2}- \frac{6}{(1-\theta)^2}.$$
\begin{align*} E(l''(\theta)) &= \frac{-4}{\theta^2}- \frac{6}{(1-\theta)^2} \end{align*}
We calculate standard error (we insert $\hat{\theta}=0.4$): $$se_{\hat{\theta}}=\frac{1}{\sqrt{-E(l''(\hat{\theta}))}} = \sqrt{\frac{\hat{\theta}^2(1-\hat{\theta})^2}{4 - 8\hat{\theta}+10\hat{\theta}^2}} = 0.024.$$
Where did I do the mistake?