MLE for Uniform(-A,A) - exam today

[nolita_day]

So my prof. has not replied to my e-mail, so I was wondering if someone here can help me understand why the MLE for a random variable X~Unif(-θ,θ) is max|X_i|. Attached is the problem as well as my attempt for the solution.

Here is my thought process:

Upon sketching the graph, I thought the answer would be min( |min(X_i)|, |max(X_i)| ) because there are two different tails and the one closer to zero should have the highest value for L(θ). So if |min(X_i)| < |max(X_i)|, then L(θ = min(X_i)) > L(θ = max(X_i)), which would make min(X_i) the MLE for θ.

Whether or not this gets answered in time before my exam, I'd still be curious to know the reasoning for the correct answer :)

Thanks a bunch!

 

[Stephen Tashi]

In my opinion, your last graph is slightly confused. You are attempting to show the interval $[-\theta, \theta]$ on the graph. You should only show L as a function of $\theta$.

For L to be greater than 0, the conditions on $\theta$ are:
$\theta \gt 0$ (or else the interval $[-\theta,\theta]$ would have zero length)
$max(X_1,X_2,...X_n) \le \theta$
$-\theta \le min(X_1,X_2,...X_n)$

The last condition is equivalent to $\theta \ge - min(X_1,X_2,...X_n)$

So the graph should show L > 0 when $\theta$ is larger than the largest of the two values $-min(X_1,X_2,...X_n)$ and $max(X_1,X_2,...X_n)$.

It doesn't look pretty, but you can denote that condition by
$\theta \ge max( -min(X_1,X_2,...X_n), max(X_1,X_2,...X_n) )$