How to find maximum likelihood estimator of this pdf? A random sample $X_1,X_2\ldots X_n$ is drawn from a population with pdf
$$f(x|\theta)=\frac{1}{2}(1+\theta x),\quad -1<x<1,-1<\theta<1$$
I know how to find MLE, but I feel confused in this pdf. It is obvious that it is an increasing function in terms of $\theta,$ if hold $x$ fixed. But $\theta$ is strictly less than $1.$ So I have to stop.
 A: 
It is obvious that it is an increasing function in terms of $\theta,$ if hold $x$ fixed.

Wrong, because $x$ can be negative. Let's write the loglikelihood:
$$L(\theta) = \log  \prod_{i=1}^{n} \frac12 (1 + \theta x_i)  \\
= - n  \log 2 + \sum_{i=1}^{n} \log(1 + \theta x_i) \tag1$$
with the domain $[-1 \le \theta \le 1]$.
(Notice that the values $x_i=0$ do not contribute, so we can exclude them).
Then $L(\theta)$ is differentiable inside $(-1,1)$, its critical points are given by $g_1(\theta)=0$ where
$$ g_1(\theta) \triangleq \frac{\partial L(\theta) }{\partial \theta}= \sum_{i=1}^{n} \frac{x_i}{1 + \theta x_i}=0 \tag2$$
Further
$$ g_2(\theta) = \frac{\partial^2 L(\theta) }{\partial \theta^2}= - \sum_{i=1}^{n} \frac{x_i^2}{(1 + \theta x_i)^2}<0 \tag 3$$
Then we can conclude that $g_1(\theta)=0$ has a single solution in the reals. Let's call it $\theta^*$.
Then the MLE is
$$\theta_{ML} = \begin{cases}
\theta^* & \text{if } -1\le \theta^* \le 1\\
-1 & \text{if } \theta^* < -1\\
1 & \text{if } \theta^* > 1\\
\end{cases} \tag 4
$$
To find $\theta^*$ we must resort to numerical methods. A simple procedure that seems to work is Newton–Raphson:
$$ \theta_{k+1} = \theta_{k} + \frac{\sum_{i=1}^{n} \frac{x_i}{1 + \theta_k x_i}}{\sum_{i=1}^{n} \frac{x_i^2}{(1 + \theta_k x_i)^2}} \tag{5}$$
Starting with $\theta_0=0$, this seems to converge quickly.
