2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.