# How can I show that “almost all function” have property P?

The following is cross-posted from https://math.stackexchange.com/questions/1391293/is-almost-all-function-a-well-defined-concept since I didn't (yet) get an answer there. (I hope that's okay?)

Anyways, I am working on a problem which has well defined properties for the vast majority of all PDFs. I would like to make a quantitative statement along the lines of "for almost all distributions, P holds". I've found this to be relevant Is there a natural measures on the space of measurable functions? but unfortunately it'd take me days to really get familiar with the proposed survey and would greately appreciate some concrete help on this topic. (This is - by far - not a very central point of my research and I'd like to avoid making a huge theory out of it. If there's some simple argument, that would be perfect, otherwise I'll just restrict myself to class of functions that I know to have property P.)

So here's the concrete question: Consider all CDF $G:\Sigma\times X\to[0,1]$ over $\Sigma\subseteq\mathbb{R}$, indexed over parameter taken from an interval $X\subseteq \mathbb{R}$, that admit a differentiable density $g(\sigma\mid x)$ and satisfy the strict monotone likelihood ratio property, i.e. $$\frac{g(\sigma'\mid x)}{g(\sigma\mid x)}<\frac{g(\sigma'\mid x')}{g(\sigma\mid x')}$$ for all $x<x'$ and $\sigma<\sigma'$. Let's call the set of these functions $\mathcal{G}$.

Now, consider the subset $\mathcal{G_0}$ containing only CDFs that have a very specific functional form over an interval (i.e. don't have property P). In particular, a CDF is in $\mathcal{G}_0$ if and only if there exist $\sigma\leq\sigma'$, real numbers $A,B,C,D\in\mathbb{R}$ and a parameter subinterval $[\underline x,\overline x]\subset X$ such that $$G(\sigma'\mid x)=1-\frac{A-B G(\sigma\mid x)}{C-D G(\sigma\mid x)} \qquad\forall x\in[\underline x,\overline x].$$ It seems to me that such a parametric restriction of the function over an entire interval drastically reduces the 'degrees of freedom' of the function, and as such I'm hoping that one could make a statement about how rare this is. Does $\mathcal{G_0}$ have measure zero among the functional space $\mathcal{G}$? Why? And what 'measure' are we talking about?

• Just a wee suggestion. What probaialists call cdf's are known as probability measures to measure theoreticians. The space of such measures on a polish space (complete, separable metric space) is itself polish under its natural topology and in such spaces one has a concept of almost everywhere which might be of use to you. For the analogies and differences between topological and measure theoretic notions of "almost everywhere", you could consult the classic "Measure and Category" by Oxtoby (readily available online). – priel Sep 11 '15 at 10:11

No, there is no such measure, but there are some substitutes for "almost every". In particular, you might look at the notions of prevalent and shy sets defined by Hunt, Sauer and Yorke.

• Thank you for that comment – that's what I'm trying to do. Unfortunately, I get stuck at the very beginning, because the definition starts out with a vector space, and I'm looking at CDFs which do not (to my understanding) satisfy this notion (e.g. scalar multiples are no longer a CDF)... – mimuller Aug 13 '15 at 19:49