Why is the bottom limit of the conditional probability $x$ in Bayesian Statistics? I am learning bayesian statistics and was stuck when trying to understand the following example:
Romeo and Juliet start dating, but Juliet will be late on any date by a random amount X, uniformly distributed over the interval [0, $\theta$]. The parameter $\theta$ is unknown and is modelled as the value of a random variable $\Theta$, uniformly distributed between zero and one hour. Assuming that Juliet was late by an amount $x$ on their first date, how should Romeo use this information to update the distribution of $\Theta$?
The sample solution is as follows:
$f_\Theta(\theta)$ = 1 if $0 \leq \theta \leq 1$, 0 otherwise
$f_{X|\Theta}(x|\theta) = \frac{1}{\theta}$ if $0 \leq x \leq \theta$, 0 otherwise
The posterior pdf is:
$$
f_{\Theta|X} = \frac{f_{\Theta}(\theta)f_{X|\Theta}(x|\theta)}{\int_0^1{f_\Theta(\theta')f_{X|\Theta}(x|\theta')d\theta'}}
$$
The following step is where I have a problem:
$$
\frac{1/\theta}{\int_x^1{1/\theta'}d\theta'}
$$
How did the limits for the integeral go from (0, 1) to (x, 1). I cannot find the justification for this step or why the limits is changing. Thank you for your help. 
 A: As you correctly stated, 
$f_{X|\Theta}(x|\theta) = \frac{1}{\theta}$ if $0 \leq x \leq \theta$, 0 otherwise
the pdf is zero for $\theta<x$. So the lower limit of the integral over $\theta'$ can be replaced by $x$.
A: Romeo now knows that Juliet can be $x$ or more late, i.e $\theta \ge x$ and so that $x \le \theta \le 1$.  
In likelihood terms, the likelihood for $\theta$ given an observation of $x$ is proportional to $\frac{1}{\theta}$ when $\theta \ge x$ and is $0$ when $\theta \lt x$, which we can combine, writing with an indicator function, as $\frac1\theta I_{\theta \ge x}$
With your prior density for $\theta$ of $1$ when $0 \le \theta \le 1$ and $0$ otherwise, 
the posterior density for $\theta$ is $\dfrac{\frac1\theta I_{\theta \ge x}}{\int\limits_0^1  \frac1{\theta'} I_{\theta' \ge x}\, d\theta'} $ when $0 \le \theta \le 1$ and $0$ otherwise, 
and you can simplify the numerator to $\frac1\theta$ when $x \le \theta \le 1$ and $0$ otherwise and similarly the integrand of the the denominator, making the whole expression for the posterior  $$\dfrac{\frac1\theta}{\int\limits_x^1  \frac1{\theta'} \, d\theta'} =-\dfrac{1 }{\theta\log_e(x)}$$ when $x \le \theta \le 1$ and $0$ otherwise
