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It is not possible to write a function representing the uniform distribution, say $\mathcal{U}$, on the real line, $\mathbb{R}$ (in a Bayesian context, this is an improper prior). My question is: is it possible to write a generalized function (aka, a distribution, but I avoid this term as it may cause confusion with the "distribution" used in probability terminology) corresponding to $\mathcal{U}(\mathbb{R}) $?


Additional text to give more context to the question. Just as Dirac's Delta can be represented as a limiting process of gaussians, i.e., $$ \delta(t) = \lim_{\sigma^2 \rightarrow 0} \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{1}{2 \sigma^2} t^2 } $$ in principle, $\mathcal{U}(\mathbb{R})$ could be represented as $$ \mathcal{K}(t) = \lim_{ \sigma^2 \rightarrow +\infty} \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{1}{2 \sigma^2} t^2 } $$ in some sense of the limit. Clearly, it must hold that $$ \int^{+\infty}_{-\infty} \mathcal{K}(t) dt = 1 $$ just as with Dirac's Delta. I do not know if there are technical difficulties that arise in measure theory for this case.

Related question: link. But I would like a bit more details as an answer.

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  • $\begingroup$ If you do a calculation with $\sigma$ and then let $\sigma$ tend to $\infty$ then you get a result that which you can interpret as if it belonged to the all-universal distribution. However, if you try with another distribution with a parameter playing a similar role then the result may be different... $\endgroup$
    – zoli
    Commented Jan 31, 2018 at 14:07
  • $\begingroup$ The problem, in that case, may lie within the properties of the other distribution (e.g., it may not be everywhere smooth or absolutely integrable). Once you formally define $\delta(t)$, you use $\delta(t)$ right from the start of the calculation. Same thing for $\mathcal{K}(t)$. There is no limiting process involved anymore. After all, $\delta(t)$ is not defined (strictly speaking) that way, so we should expect the same for $\mathcal{K}(t)$. If you have a specific example in mind, you can post it as an answer. $\endgroup$ Commented Jan 31, 2018 at 14:43
  • $\begingroup$ Years ago, I did such a calculation. I will try to recall the details. $\endgroup$
    – zoli
    Commented Jan 31, 2018 at 17:00
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    $\begingroup$ Would it make sense to use a Césaro summation type of definition $$\mathbf E[f(\mathcal U)]:=:(\mathcal U,f):=\lim_{N\to\infty}\frac{1}{2N}\int_{-N}^Nf(y)dy.$$ It looks like a positive preserving bounded linear functional on $C_b(\mathbb R)$ with norm $1$. This would give you existence of a finitely additive probability measure. Is it wrong or a totally trivial object? ($C_b(\mathbb R):=$ bounded real-valued continuous functions on $\mathbb R$). (would it be just the 0 measure $C_0(\mathbb R)$ (functions vanishing at infinity)?) $\endgroup$
    – Rgkpdx
    Commented Feb 16, 2018 at 12:44
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    $\begingroup$ @MikeHawk That isn't a probability density function because it doesn't integrate to 1 over the real line. Your proposed PDF does not correspond to any cumulative distribution function. $\endgroup$
    – tparker
    Commented Feb 18, 2018 at 0:59

1 Answer 1

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Since no one answered, I would like to post here some useful insights on the matter.

The first, and most important, insight is that the improper uniform is inherently non-local. Distributions like Dirac's Delta, $\delta(t)$, have a local behavior, which is important. This issue makes the mathematical formalization of $\mathcal{K}(t)$ harder than it may appear.

The second comment is an attempted definition is the following. Consider the so-called integral mean (or average) over an interval $[a,b]$, given by $$ \frac{1}{|a-b|} \int^b_{a} f(t) dt $$ which is well-defined as long as the function $f$ is well-behaved. We can consider the limit of a sequence of increasing symmetric intervals as the mean of the function itself, given by $$ \mathbb{E}[f] \triangleq \lim_{N \rightarrow +\infty} \frac{1}{2N} \int^N_{-N} f(t)dt $$ and use this as our definition. In other words, the improper uniform can be thought of as an operator associating a real number to a function, i.e., $\mathbb{E} : \mathcal{F} \rightarrow \mathbb{R}$, where $\mathcal{F}$ is some function space with some regular properties (e.g. Schwartz space). This approach can likely be refined significantly, but I invite the author to post his answer since, as of now, no better definition has been given (unfortunately, the bounty expired).

There is, however, one significant aspect which I think can be improved upon. Intuition tells us that $\mathcal{K}(t)$ should be an infinitesimally thin "carpet", spread over $\mathbb{R}$. So, it makes sense that it should share some properties of the zero-function (the pointwise limit of gaussians for $\sigma^2 -> +\infty$). This can be accomplished via the following definition $$ \int_{\mathbb{R}} \mathcal{K}(t) f(t) dt = \lim_{\sigma^2 \rightarrow +\infty} \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{\mathbb{R}} e^{-\frac{t^2}{2 \sigma^2}} f(t) dt $$ Let us have a deeper look at this. For $f(t)=1$, we have that the result is $1$, so that $$ \int_{\mathbb{R}} \mathcal{K}(t)dt = 1 $$ is satisfied. Consider now $f(t)=t$, for which the integral on the LHS is zero. This is, intuitively, correct, as $\mathcal{K}(t)$ should behave like the zero-function and simply make the integral vanish, but not always! We should expect some functions, like $f(t)=c$, for any $c \neq 0$, not to be annihilated. Functions with divergences also may be into this category. In other words, the integral is zero for some (but not all!) functions $f$, making $\mathcal{K}(t)$ "look like" the zero-function. To be more precise, there exists a function space $\mathcal{F}_0 \subsetneq \mathcal{F}$, where $\mathcal{F} = \mathcal{F}(\mathbb{R})$ is the space of all functions in $\mathbb{R}$ (regular enough to have an integral), such that $$ \forall f \in \mathcal{F}_0: \quad \int_{\mathbb{R}} \mathcal{K}(t) f(t)dt = 0 $$ which means that $\mathcal{K}(t)$ acts like the zero-measure in $\mathcal{F}_0$. But, and this is the key point, it acts nontrivially when other functions ($\not\in \mathcal{F}_0$) come into play.

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    $\begingroup$ It is worth noting that $\mu(A)=\int K(t) 1_A(t)dt$ does not correspond to a measure (even assuming the existence of the "function" K); indeed for any bounded set $A$, we have $\mu(A)=0$, but since R can be expressed as a countable union of such sets, the countable additivity axiom does not hold. $\endgroup$ Commented Feb 23, 2018 at 14:57
  • $\begingroup$ Can you find a function such that $\mathbb E[f]\neq \int_{\mathbb R}\mathcal K(t)f(t)dt$? $\endgroup$
    – Rgkpdx
    Commented Feb 23, 2018 at 15:31

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