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

1. Nov 5, 2012

### 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|Xi|. 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(Xi)|, |max(Xi)| ) because there are two different tails and the one closer to zero should have the highest value for L(θ). So if |min(Xi)| < |max(Xi)|, then L(θ = min(Xi)) > L(θ = max(Xi)), which would make min(Xi) 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!

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2. Nov 5, 2012

### 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
$\theta \ge max( -min(X_1,X_2,...X_n), max(X_1,X_2,...X_n) )$