What is MLE of $U[-k,7k]$?

Question

I need to find maximum likelihood estimator of uniform distribution $X : U[-k,7k]$. I think that the answer is $max\{x_1, ..., x_n \}$ . However, I think I have a mistake because if $X : U[0,k]$ the answer would be the same. So, it seems strange to me why MLE does not depend on the length of the interval.

Answer 1

Suppose $k = 1$ and your sample is $(-0.9, 0.2, 2.5, 3.7, 4.2)$. Based on the sample maximum $4.2$, your estimate of $k$ would be $(4.2)/7 = 0.6$. But based on the sample minimum, $-0.9$, your estimate would be $0.9$. Which one is better? Can you think of a way to write a mathematical expression to reflect this?

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Here is the sufficient statistic calculation.

Let the sample be $\boldsymbol x = (x_1, \ldots, x_n)$, and as usual, define the minimum and maximum order statistics $x_{(1)} = \min \boldsymbol x$, $x_{(n)} = \max \boldsymbol x$, respectively. Then we have $$f(\boldsymbol x \mid k) = \prod_{i=1}^n \frac{1}{8k} \mathbb 1 (-k \le x_i \le 7k) = (8k)^{-n} \mathbb 1 (-k \le x_{(1)} \le x_{(n)} \le 7k).$$ So by the Factorization theorem, a sufficient statistic is $T(\boldsymbol x) = (x_{(1)}, x_{(n)})$. This happens to be minimal, but it is not necessary to prove this. The point of finding $T$ is to show that any MLE of $k$ will be expressible as a function of $T$.

As for the MLE, we have the likelihood $$\mathcal L(k \mid x_{(1)}, x_{(n)}) \propto k^{-n} \mathbb 1(-k \le x_{(1)} \le x_{(n)} \le 7k).$$ Because $k^{-n}$ is a monotone decreasing function of $k$, $\mathcal L$ is maximized for the smallest possible positive choice of $k$ that keeps the indicator function equal to $1$; that is to say, we require that $k \ge x_{(n)}/7$ and $k \ge -x_{(1)}$. Equivalently, $$k \ge \max\left(\frac{x_{(n)}}{7}, -x_{(1)}\right).$$ Therefore, the desired MLE is $$\hat k = \max\left(\frac{x_{(n)}}{7}, -x_{(1)}\right).$$

Answer 2

Consider that for any $X_i$ you have that

$$-k\leq X_i\leq 7k$$

thus

$$\begin{cases} -k\leq X_{(1)} \\ 7k\geq X_{(n)} \end{cases}$$

Thus

$$k\geq \max\left(-X_{(1)};\frac{X_{(n)}}{7}\right)$$

Given that the likelihood ($1/(8k)^n$) is strictly decreasing in $k$

$$T=\max\left(-X_{(1)};\frac{X_{(n)}}{7}\right)$$

is MLE