How to calculate MLE when given a sample set that includes negative values?

Question

Let $X_1, X_2, ..., X_n$ ~ $U(-\theta, \theta)$, it is known that $\hat{\theta}_{MLE} = X_{max}$ of the sample set.

The given sample set is $X_1 = -0.9$, $X_2 = -0.3$, $X_3 = -0.1$, $X_4 = 0.2$, $X_5 = 0.4$, $X_6 = 0.6$, and $X_7 = 0.8$

I'm not sure which of the following is correct...

$$\hat{\theta}_{MLE} = |X_k| \hspace{1cm} or \hspace{1cm} \hat{\theta}_{MLE} = X_k$$

and equivalently in this case...

$$\hat{\theta}_{MLE} = |-0.9| = 0.9 \hspace{1cm} or \hspace{1cm} \hat{\theta}_{MLE} = 0.8$$

This is different from similar problems since it looks specifically at calculating these estimators when given a sample of numeric values, which also contains negatives.

Your question boils down to asking "what is $\max{-0.9,-0.3,-0.1,0.2,0.4,0.6,0.8}$?" The max is just the biggest element. — Clement C., Apr 03 '19 at 01:11
$0.9$ is not an element of the set. So under your premise, no. But your premise is wrong (the MLE is $\max_i |X_i|$), so 0.9 is the right answer. — Clement C., Apr 03 '19 at 01:19
Yes, I know it is not an element of the set, but $\hat{\theta}{MLE} = X{max} = max_i |X_i| = |-0.9| = 0.9$ right? This is what I was trying to say, I apologise that my math writing wasn't correct, I didn't quite know how to write it best until your comment. — LegendOfKass, Apr 03 '19 at 01:26
If you define Xmax as the absolute value of the largest element, then yes. That's... Not a standard notation by any means. — Clement C., Apr 03 '19 at 01:28
As for the MME, do I take just take the sum of the sample values, or do I take the sum of the absolute values? The $\tilda{\theta} = \frac{n+1}{n^2}\sum{X}$ — LegendOfKass, Apr 03 '19 at 01:28
Possible duplicate of Maximum Likelihood Estimator for $X_1,\dots, X_n ; \sim U(-\theta,\theta)$ — Masoud, Apr 03 '19 at 08:19

score 2 · Answer 1 · 2019-04-03T02:16:53.483

2

Given a sample $x\equiv \{x_i\}_{i=1}^n$, the likelihood is $$ L(\theta\mid x)=(2\theta)^{-n} \times1\{\theta\ge M(x)\}, $$ where $M(x):=\max_{1\le i\le n}|x_i|$. Then the MLE of $\theta$ is $\hat{\theta}_n=M(x)$.

The first moment of $X_1$ is zero. However, one may consider the second moment: $\mathsf{E}X_1^2=\theta^2/3$. Then the MME of $\theta$ is $\hat{\theta}_n=\sqrt{3\sum_{i=1}^n x_i^2/n}$.

edited Apr 03 '19 at 02:16

answered Apr 03 '19 at 01:46

How would I calculate $\theta_{MME} = \frac{n+1}{n^2}\sum{X}$? Would I take the sum of the values as given, or the sum of the absolute values? – LegendOfKass Apr 03 '19 at 01:53
I'm not following how you got those moments. I found that $E(X_{max}^k) = \frac{n\theta^k}{n+k}$, thus the first moment would be $E(X_{max}) = \frac{7\theta}{8}$ – LegendOfKass Apr 03 '19 at 23:22
How is $E(X_{max}^k)$ related to the method of moments estimator (MME)? – Apr 03 '19 at 23:35
Wouldn't you use $E(/hat{/theta})$? – LegendOfKass Apr 03 '19 at 23:40
See https://en.wikipedia.org/wiki/Method_of_moments_(statistics) – Apr 03 '19 at 23:42
Oh ok. Maybe I got that mixed up with bias and mse. I would only care about $f(x;/theta)= /frac{1}{2/theta}$, not $X_{max}$ for the MME? – LegendOfKass Apr 03 '19 at 23:48
For the MME one would care about the moments of $X_i$. – Apr 03 '19 at 23:54

How to calculate MLE when given a sample set that includes negative values?

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