Let $(X_1, X_2, X_3, X_4)$ be a random vector from a multinomial
$$\left[n,\frac{1-2\theta+\theta^2}{5},\frac{\theta(2-\theta)}{5},\frac{\theta(2-\theta)}{5},\frac{(1-\theta)^2}{5}\right]$$
find the ML estimate of $\theta$
This is a homework problem, and I don't want the answer, but I'm hoping for guidance on how to obtain the ML. The trouble I am running into is after taking the partial derivative wrt $\theta$ and attempting to solve for the parameter. I get a monster of an equation and I'm unable to algebraically isolate $\theta$. If anyone has run into a problem like this before, could you please provide insight on a different way to obtain the parameter estimate?
Thanks!
At this time the question says $$ \left[n,\frac{1-2\theta+\theta^2}{5},\frac{\theta(2-\theta)}{5},\frac{\theta(2-\theta)}{5},\frac{(1-\theta)^2}{5}\right]. $$ Not that $1-2\theta+\theta^2$ is the same thing as $(1-\theta)^2,$ so the first and fourth probabilities are the same.
Since the sum of the numerators is $2,$ I'm guessing this ought to have said $$ \left[n,\frac{(1-\theta)^2} 2,\frac{\theta(2-\theta)} 2, \frac{\theta(2-\theta)} 2, \frac{(1-\theta)^2} 2\right]. $$ Then the likelihood function will be $$ L(\theta) = \text{constant}\times (1-\theta)^{2x_1} \cdot \big(\theta(2-\theta)\big)^{x_2} \cdot \big(\theta(2-\theta)\big)^{x_3} \cdot (1-\theta)^{2x_4}, $$ where we should remember that in this context, "constant" means not depending on $\theta,$ so in particular the multinomial coefficient is a constant.
So we have \begin{align} & \ell(\theta) = \log L(\theta) \\[6pt] = {} & 2(x_1+x_4)\log(1-\theta) + (x_2+x_3)(\log\theta + \log(1-\theta)). \end{align} The value of $\theta$ that maximizes $L(\theta)$ is the same one that maximizes $\ell(\theta)$ because $\log$ is an increasing function. Next: $$ \ell\,'(\theta) = \frac{-2(x_1+x_4)-(x_2+x_3)}{1-\theta} + \frac{x_2+x_3} \theta. $$ Write that as one fraction with a common denominator and use the result to figure out for which values of $\theta$ this derivative is positive and for which it is negative.