4
The likelihood of your sample writes as
$$\begin{align*}
\prod_{i=1}^3 \{p\mathbb{I}_1(x_i)+(1-p) \mathbb{I}_{-1}(x_i)\}&=\prod_{i=1}^3 p^\frac{1+x_i}{2}(1-p)^\frac{1-x_i}{2}\\&=[p(1-p)]^{3/2}
p^\frac{\sum_{i=1}^3 x_i}{2}(1-p)^\frac{-\sum_{i=1}^3x_i}{2}\\
&=[p(1-p)]^{3/2} \sqrt{p/(1-p)}^{\sum_{i=...
