# Determining sufficient statistic for single random variable

just need some hint to tackle this problem:

So, we are given a random variable X with pdf: $f_X(x;\theta) = \frac{1}{2\theta}$ for $-\theta < x < \theta$, and zero otherwise.

Then, we are asked whether $|X|$ is a sufficient statistics for $\theta$.

My thoughts: First, I'm confused, since for MLE for example, we usually deal with $n$ samples of $X$, however, I'm convincing myself that we may have $n=1$ and so, no much a big deal. We can still estimate $\theta$. Then, what follows is finding the MLE of this particular likelihood function, and since this seems like a uniform distribution $\cal{U}$$\sim [0,2\theta]$, I know the maximizing value comes from taking a look to the support of the pdf and not from the typical method of derivative, etc.

Therefore, considering that we want to maximize $\frac{1}{2\theta}$, we can chose the estimation of $\theta$ to be some value of X. To maximize such ratio, we may chose the smallest possible value of $X$, which is $-\theta$, but that will give us a negative number. Then, the next non-negative smallest value would be zero, but that will blow up $\frac{1}{2\theta}$.

So, kinda stuck and confused from here... any help?

Thanks!

• You have that $f(x|\theta) = \frac{1}{2\theta}I_{x\in(-\theta,\theta)}$. The MLE is whatever value of $\theta$ maximizes this. However, if $\theta > |X|$ then the indicator function becomes $0$, which does not maximize the likelihood. The competing goal here is that, when the indicator is $1$, larger values of $\theta$ decrease the multiplier of the indicator $\frac{1}{2\theta}$. The smallest value of $\theta$ such that the indicator is $1$ is $|X|$. Thus $|X|$ is the maximum likelihood estimator. Commented Mar 21, 2018 at 8:44
• Also, there's nothing special about having $n>1$, a lot of statistics and probability problems are provided with $n=1$. Your estimate might not be as good as with a larger sample, but strictly speaking you can still come up with an MLE in most situations. Sometimes the MLE might not exist without a larger sample size (for example the MLE for the covariance in a multivariate normal model), but otherwise, in situations like this, it's still something you can obtain. Commented Mar 21, 2018 at 8:45
• Great, thanks for your replies. So that means that having $|X|$ is enough, and actually what we must have, for the MLE. So I can conclude that indeed, $|X|$ is a sufficient statistics? I think so. Commented Mar 21, 2018 at 22:40

$$f_\theta(x) = \frac{\mathbb{I}(-\theta < x < \theta)}{2 \pi} = \frac{\mathbb{I}(|x| < \theta)}{2 \pi} = g_\theta(|x|).$$
This establishes the requirements of the theorem, which proves sufficiency. The implication for the MLE is that it will be a function of $$x$$ only through $$|x|$$.