How to achieve the conditional probability distribution function in this case? I am looking this sample Bayesian problem.  I do not get how they get this formula in the second case.
$f_{X\mid\Theta}(x\mid\theta ') = 1/ \theta^n$
Shouldn't it be just $1/\theta_{max}$?  Please help me understand.  TIA!

 A: $f_{X|\Theta}(x|\theta)$ is the likelihood function of observing $X$ conditioned on the fact that distribution is uniform with parameter $\theta$. Now consider the likelihood of observing $n$ variables $X_1,X_2,\cdots,X_n$, when $\forall i: X_i$'s are i.i.d. (identically independently distributed). because of independence we have
$$f_{X_1,\cdots,X_n|\Theta}(x_1,\cdots,x_n|\theta) = f_{X_1|\Theta}(x_1|\theta)\times \cdots \times f_{X_n|\Theta}(x_n|\theta)$$
Here a subtle fact is that all $f_{X_i|\Theta}(x_i|\theta)$ must be non-zero and the parameter $\theta$ is shared. For $X_i$ we have 
$$f_{X_i|\Theta}(x_i|\theta) \begin{cases} \frac{1}{\theta}, \qquad \text{if} \quad 0\le x_i \le\theta \\ 
0, \qquad \text{otherwise}\end{cases}$$
When we consider all of $x_i$'s this way, the likelihood equation given emerges. In other words $\forall i: 0\le x_i \le \theta \le 1$ which is equivalent to $0 \le \max\{ x_1,\cdots,x_n \} \le \theta \le 1$.
Now down to integration, $c$ is some normalizing constant, therefore the integration in denominator is unnecessary to calculate, the only thing we need to calculate is $f_{\Theta}f_{X|\Theta}$, i.e. the nominator of fraction which with the given explanations is obvious.
A: From the uniformity of the conditional distribution of $\ X_i\ $ over $\ [0,\theta]\ $ given $\ \Theta=\theta\ $, we have
$$
P\left(X_i\le x_i\right)=\cases{\frac{x_i}{\theta}&if $ 0\le x_i\le\theta\ $\\
0 &otherwise}\ ,
$$
so
$$
f_{X_i|\Theta}\left(x_i |\theta\right)=\cases{\frac{1}{\theta}&if $ 0\le x_i\le\theta\le1\ $\\
0 &otherwise}\ ,
$$
and from the conditional independence of $\ X_1,X_2,\dots,X_n\ $ given $\ \Theta=\theta\ $, we get
\begin{align}
f_{X|\Theta}\left(x|\theta\right)&= f_{X_1|\Theta}\left(x_1|\theta\right) f_{X_2|\Theta}\left(x_2|\theta\right)\dots f_{X_n|\Theta}\left(x_n |\theta\right)\\
&=\cases{\frac{1}{\theta^n}&if $ 0\le x_i\le\theta\le1\ $ for all $i=1,2\dots n\ $\\
0 &otherwise}\ .
\end{align}
