Maximum likelihood estimator on uniform distribution I try to be rational and keep my questions as impersonal as I can in order to comply to the community guidelines. But this one is making me mad. Here it goes. 
Consider the uniform distribution on $[0, \theta]$. The likelihood function, using a random sample of size $n$ is 
$\frac{1}{\theta^{n}}$.
Now $1/\theta^n$ is decreasing in $\theta$ over the range of positive values. Hence it will be maximized by choosing $\theta$ as small as possible while still satisfying  $0 \leq x_i \leq \theta$.The textbook says 'That is, we choose $\theta$ equal to $X_{(n)}$, or $Y_n$, the largest order statistic'.But if we want to minimize theta to maximize the likelihood, why we choose the biggest x? Suppose we had real  numbers for x like $X_{1} = 2, X_{2} = 4, X_{3} = 8$.If we choose 8, that yields $\frac{1}{8^{3}}=0.001953125$. If we choose $\frac{1}{2^{3}}=0.125$. Therefore why we want the maximum in this case $X_{n}$ and not $X_{1}$, since we`ve just seen with real numbers that the smaller the x the bigger the likelihood? Thanks!

$\theta$ is the parameter to estimate, which corresponds to the upper bound of the $U(0,\theta)$. The observed samples are $x_1=2,\,x_2=4$, and $x_3=8$. The likelihood function to maximize is $\mathcal{L}(\theta|X) = \frac{1}{\theta^n}$ with $X$ corresponding to the observed values (a vector, really), $\theta$ the upper margin of the interval for which this uniform distribution is defined, and $n$ the number of samples.
In order to "see" this intuitively, it's important to realize that $\theta$ has to be at least as big as your largest sampled value ($8$) to avoid leaving samples out of the interval for which your probability density function is defined.
Picking out $8$ would render the value of the $pdf$ at any point equal to $\frac{1}{8}$, and the joint probability distribution (you are sampling three observations - a vector), $\frac{1}{8^3}$. That would certainly the maximum $\mathcal{L}(\theta|X)$, because it is the largest denominator possible (compare to $\frac{1}{2^3}$ or $\frac{1}{4^3}$).
