What is the posterior PDF in this problem?

Question

I am looking this sample Bayesian problem, and wish to calculate the posterior PDF in the second case.

The Bayes rule for the $\Theta$ continuous, $X$ continuous case is:

$f_{\Theta\mid X}(\theta\mid x) = \frac{\displaystyle f_{\Theta}(\theta) f_{X\mid\Theta}(x\mid\theta)}{\displaystyle\int f_{\Theta}(\theta ') f_{X\mid\Theta}(x\mid\theta ') \, d\theta '}$

We know: $$f_{\Theta} (\theta)= \left\{ \begin{array}{l} 1, ~~~~if~0 \leq \theta \leq 1\\ 0, ~~~~otherwise\\ \end{array}\right.$$

and in the interested interval:

$f_{X\mid\Theta}(x\mid\theta) = 1 / \theta^n$

How do they finally achieve the answer boxed in red below? What happened to the term $f_{X\mid\Theta}(x\mid\theta)$ which seems missing?

Answer 1

They've not gotten rid of $f_{∣Θ}(∣)$ they've evaluated $_Θ()$ as $1$ within the given range. Then substituted in their value of $f_{∣Θ}(∣) = \frac{1}{}$ within that same range.

What they've not made clear is that they've adapted the range that they're working over as $x<<1$ which allows them to do this. Hope that helps!