Using method of maximum likelihood find the estimator for $\mathcal N(m,1)$-normal distribution and $\mathcal U(\theta, 1), \theta<0$

From what I understand, if the parameter is negative it is done a little differently than, if it were positive... I don't see how or why, because normally the estimator is $$m=\frac{1}{n}\sum_{k=1}^{n}X_k$$ for $\mathcal N(m,1).$

For $\mathcal U(\theta, 1), \theta<0$ I was thinking that $$L\left(\theta|{\bf x}\right)=\frac{1}{(1-\theta)^n}$$ and since $\theta<x_1,...,x_n$ that the estimator should be $$\min\{0,Y_1\}$$ where $$Y_1=\min_{1 \leq k \leq n}\{x_k \}$$ I'm supposed to check also whether this is centered($\min\{0,Y_1\}$).(I was also told that this estimator is correct, just not sure about $L\left(\theta|{\bf x}\right).$) Thought this was the best place to ask. :D

  • $\begingroup$ Please add the [self-study] tag (you will need to remove one of the existing tags) & read its wiki. $\endgroup$ – gung Jan 11 '16 at 0:02
  • $\begingroup$ check! ${}{}{}{}$ $\endgroup$ – Jerry West Jan 11 '16 at 0:03
  • $\begingroup$ The expression for $L(\theta|x)$ cannot be correct--it's not even defined for $\theta=-1$! $\endgroup$ – whuber Jan 11 '16 at 0:03
  • $\begingroup$ How ${}$ about now? $\endgroup$ – Jerry West Jan 11 '16 at 0:04
  • 3
    $\begingroup$ 1. "From what I understand" -- can you explain from where this understanding arises? You're asking for comment on something you don't explain at all. $\:$ 2. To my recollection, each of these problems (or at least very similar ones, similar enough to be helpful) are discussed a number of times on site. Some site-searches may help you solve your problems. $\endgroup$ – Glen_b Jan 11 '16 at 0:06

Writing the likelihood function:

$$ L(\theta|X) = \begin{cases} \frac{1}{(1-\theta)^n} & \theta <0; \theta\leq X_{(1)} \leq X_{(2)} \leq \dots X_{(n)} \leq 1\\ 0 & \text{otherwise} \end{cases} $$

which is a decreasing function of $\theta$ and hence the MLE(given the constraint that $\theta < 0$) is given by: $\min(0, X_{(1)})$ where $X_{(i)}$ represents $i^{th}$ order statistic.

  • $\begingroup$ I have a slight confusion: If the parameter space is $\theta<0$, how can the MLE ever be $0$ (when $X_{(1)}>0$) ? If however the parameter space was $\theta\le 0$, then I can see that the MLE can be $0$. $\endgroup$ – StubbornAtom May 4 '18 at 10:17

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