I am very new to Bayesian statistics and I wanted to know how could I show that Uniform($0,A$) ,as $A \to \infty$, is an improper density?

I mean the Uniform($0,A$) ,as $A \to \infty$ density is given by $ \lim_{A \to \infty}f(x)=\frac{1}{A} \mathbb{1}_{[0,A]}=0$ and hence it doesnt integrate to 1 and therefore improper based on the definition I found here Does "improper" posterior or prior refer to a density function that does not integrate to 1 or to one that does not integrate to a finite value?

Am I wrong?


1 Answer 1


Basically, yes.

More formally, what you want to know is: what is the value of

$$ \int_{-∞}^∞ \lim_{A \to ∞} \tfrac{1}{A}1_{[0, A]}(x) \; dx \;\;\; ?$$

For all $x$, the limit converges to 0. To see this, notice that for any $x$ and $A$, $\tfrac{1}{A}1_{[0, A]}(x)$ is either $0$ or $\frac{1}{A}$, the latter of which is less than any $ε > 0$ for all $A > \frac{1}{ε}$. Then we have just $\int_{-∞}^∞ 0 \; dx$, which is $0$.

Because the function's integral is $0$, it isn't a proper density.

  • $\begingroup$ Hi if you read from the link I attached , I see that any finite number is considered to be proper but somehow i gather that this finite number should be non zero for it to be proper since we can scale it to make it equal to 1 . Right ? $\endgroup$ Jun 24, 2017 at 15:40
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    $\begingroup$ If the density approaches 0 everywhere it can't integrate to 1. That is why it is improper. $\endgroup$ Jun 24, 2017 at 16:23
  • $\begingroup$ @user3503589 That's right. There can't be a normalizing constant when the integral is 0. $\endgroup$ Jun 24, 2017 at 16:56
  • $\begingroup$ @Kodiologist But does it even make sense for a density to be zero everywhere? I was reading a paper by Prior distribution for variance parameter in hierarchical models by Andrew Gelman, where they talk about approximating the improper distribution $U(0, \infty)$ as a limit of proper distributions i.e $U(0,A), A \to \infty$ $\endgroup$ Jun 26, 2017 at 19:59
  • $\begingroup$ @user3503589 No, it does not make sense, because it means that the function has integral 0. The improper prior $U(0, ∞)$ indeed has no corresponding random variable or probability density function. $\endgroup$ Jun 26, 2017 at 20:10

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