Exercise 3.8 - MLE for the uniform distribution

The Bayesian analysis for the uniform distribution seems to be of less significance since uniform distribution appears to appear less frequently than other continuous distributions. But the exercises remain good introductory examples.

The likelihood for the uniform distribution is a truncated function, whose domain is $[-a,a]$, so we must have $a\ge \underset{i}{\max <!--\; FUNCTION\; APPLICATION\; -->}\left\{|x_i|\in 𝒟\right\}$. Then the likelihood lookes like:

or generally:

For question (a), in order to maximize this value with $a\ge \underset{n}{\max <!--\; FUNCTION\; APPLICATION\; -->}\left\{|x_n|\in 𝒟\right\}$, the outcome is:

For question (b), if $|x_{n+1}|>\underset{i=1}{\overset{n}{\max <!--\; FUNCTION\; APPLICATION\; -->}}\left\{|x_i|\right\}$ then $p(x_{n+1})$ is zero. Otherwise the probability is $\frac{1}{2\cdot a_{\text{MLE}}}$.

For question (c), we believe that MLE for the uniform distribution is not fluent enough since when $x_{n+1}$ passes $\pm \underset{i=1}{\overset{n}{\max <!--\; FUNCTION\; APPLICATION\; -->}}\left\{|x_i|\right\}$, the predicted probability drops as a step function, which is undesired for a continuous distribution.