Given that $$F(x)= (\theta +1)x^\theta\qquad \text{for} \qquad 0<x<2$$ find maximum likelihood estimator.
My progress:
$L(\theta) = \prod_{i=1}^{n}(\theta +1)(x_i)^\theta$
$L(\theta) = (\theta +1)\prod_{i=1}^{n}(x_i)^\theta$
edit* i think this step is wrong and i multiply the left part by $n$ but the solution is still strange.
$l(\theta) = \log( (\theta +1)\prod_{i=1}^{n}(x_i)^\theta)$
$l(\theta) = \log( (\theta +1) + \sum_{i=1}^{n}\log((x_i)^\theta))$
$= \log( (\theta +1) + \theta\sum_{i=1}^{n}\log((x_i)))$
$l'(\theta)= \frac{1}{\theta + 1} + \theta $
now if i equate this to 0 and try to solve i get complex roots, i dont know if ive done something wrong or i am meant to continue?
$\theta$ is strictly positive
Have you verified that the proposed function is a probability distribution function? Here, we still need to verify that $\int_{\Bbb R}F(x)\,\text dx =1$, which isn't true. Thus, I'll assume that the function's support is $[0,1]$ (instead of $[0,2]$). Anyway, your computation of the likelihood is incorrect. Indeed, $$ L(\theta) = \prod_{i=1}^n F(x_i) \\ = (\theta+1)^n\left( \prod_{i=1}^n x_i\right)^\theta $$ with $0<x_i<1$ for all $i$. The log-likelihood $$ \ell(\theta) = n\log(\theta+1) + \theta\sum_{i=1}^n \log x_i $$ is differentiated as follows $$ \ell'(\theta) = \frac{n}{\theta+1} + \sum_{i=1}^n \log x_i \, . $$ This derivative vanishes at $$ \theta = -1 -n\bigg/\sum_{i=1}^n \log x_i . $$ If this local extremum corresponds to a maximum of the likelihood, then the previous equation defines the MLE. To avoid similar mistakes in future, expand the product $\prod_{i=1}^n$ for $n=1,2, \dots$