Find mle of theta from some mixed density

Question

I'm trying to find a mle from:

$$ P_\theta(x) = (1 + \theta) I(0 \leq x \leq 1/2) + (1 - \theta) I(1/2 < x \leq 1)$$

Then,

\begin{align*} L(\theta) &= \Pi_{i=1}^n P_\theta(x_i) = (1+\theta)^k (1-\theta)^{n-k} \\ l(\theta) &= klog(1+\theta) + (n-k)log(1-\theta) \\ \hat{\theta}^{mle} &= -1 + \frac{2k}{n} \end{align*} where $k= \# \{x_i : 0\leq x_i \leq 1/2, i=1,...n\}$.

Do I get mle right?

If I assume that my solution is correct, How could I derive the consistency of $\hat{\theta}^{mle}$?

Also, can I derive the limit law $\sqrt{n}(\hat{\theta}^{mle} - \theta)$ as $n \rightarrow \infty$?

Answer 1

1. Well, yes, your calculations are correct.

2. Consistency follows from the compactness of the parameter space: $\theta \in [-1, 1] $

3. Hint:

$$ P( |\theta_*-\theta| < \epsilon)=P\Big( \frac{n}{2}(1 + \theta - \epsilon) < k < \frac{n}{2}(1 + \theta + \epsilon)\Big) $$

where $\theta_*$ is an MLE estimation of $\theta$.

With respect to the variable $k$ your process is described via binomial distribution with probability of success $(1+\theta) / 2$ and of failure $(1-\theta) / 2$ since from the density function we get: $$ P(x_i \in [0, 1/2] ) = \ (1 + \theta) / 2 \\ P(x_i \in [1/2, 1] ) = (1 - \theta) / 2 $$