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.


2 Answers 2


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

  • $\begingroup$ Oh ok. That makes sense. Thanks a lot :D $\endgroup$ Commented Apr 30, 2019 at 14:12
  • $\begingroup$ So, what is the probability of Juliet coming on time? Is it $\ln(0)=\infty$? $\endgroup$
    – Yola
    Commented Jun 4, 2022 at 12:39
  • $\begingroup$ @ Yola The probability of Juliet coming on time if you know the value of $\theta$ is $\int_{x=0}^0 \frac1\theta \,dx= 0$. If you do not, then using the prior distribution for $\theta$, it is $\int_{x=0}^0\int_{\theta=0}^1\frac{1 }{\theta} \, d\theta \,dx = 0$, while using the posterior distribution for $\theta$ having previously observed $x_0$, it is $\int_{x=0}^0\int_{\theta=x}^1 -\frac{1 }{\theta^2\log_e(x_0)} \, d\theta \,dx = 0$. This is what happens with continuous distributions: the probability of any particular value is $0$. $\endgroup$
    – Henry
    Commented Jun 4, 2022 at 17:42

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 $x$.


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