In Example 11.6 on p. 270 of Hermans and Škulj's "Stochastic Processes" in Augustin et al.'s Introduction to Imprecise Probabilities, there is a definition of an upper transition operator as $$\overline T=I_{\cal X} \max$$

$\overline T$ is an upper prevision (upper expectation) operator, where for any function $f$ on the set of atomic outcomes $\cal X$, $\overline Tf := \overline Q (f|x)$, the upper conditional prevision of $f$ conditional on any value $x\in \cal X$ (11.8, p. 266).

I'm having trouble interpreting $I_{\cal X} \max$ in this context. I assume that $\max$ applied to $f$ is just supposed to find the maximum value of $f$ over all $x\in \cal X$. That seems reasonable.

I'm confused by seeing $I_{\cal X}$ next to $\max$, though. I take $I_{\cal X}$ to be the indicator function for $\cal X$, the entire space (see e.g. p. 260). This function returns 1 for any value at all, i.e. in this context, for any value in $\cal X$. So assuming that the range of $f$ is in $\cal X$, $I_{\cal X} (\max f)$ aways returns 1.

EDIT (revision that more accurately reflects other material in the same example):

That could be a sensible value for $\overline T$ in the example in which the definition appears, but if that interpretation is correct, then why give such a complicated definition for the function that simply returns 1 when passed any function on $\cal X$?

Further, in the subsequent equations that depend on this definition, $\max$ appears repeatedly. I don't believe that all of the instances of $\max$ come from this definition, but in no cases is the value of $\max$ ignored in such a way that 1 is always returned. This suggests that I am completely misinterpreting $\overline T=I_{\cal X} \max$.

  • $\begingroup$ I do not think ‘fuzzy-logic’ is appropriate for this question. $\endgroup$ – equaeghe Feb 18 '18 at 20:23
  • $\begingroup$ Thanks @equaeghe. I've removed the fuzzy-logic tag. It's hard to know what tags to use until there is an imprecise-probability tag available (or something with a similar meaning). $\endgroup$ – Mars Feb 19 '18 at 0:17
  • $\begingroup$ I would not add the fuzzy-logic tag as a replacement for imprecise-probability, as these are different things. It is true that both are sometimes used with similar goals. $\endgroup$ – equaeghe Feb 19 '18 at 15:31
  • $\begingroup$ @equaeghe I've removed fuzzy-logic from another IP question I asked a while ago. $\endgroup$ – Mars Feb 19 '18 at 16:13

The operator $\overline{T}$ maps functions on $\mathcal{X}$ to functions on $\mathcal{X}$. So $\overline{T}=I_{\mathcal{X}}\max$ means that $\overline{T}(f)(x)=I_{\mathcal{X}}(x)\max(f)=\max(f)$ for any function $f$ on $\mathcal{X}$ and $x$ in $\mathcal{X}$.

To understand the reasoning behind this notation, it is useful to first look at functions of scalars: Let $p:\mathbb{R}\to\mathbb{R}:x\mapsto x^2$. Now, we can multiply this function by a constant $\alpha$ in $\mathbb{R}$ to get another function, i.e., let $q=\alpha p$, meaning $q:\mathbb{R}\to\mathbb{R}:x\mapsto \alpha x^2$. You can do this because $\alpha$ belongs to the range of $p$.

Similarly, with transition operators—functions of a specific class of functions to (the same) specific class of functions—you can do the same. And also here the natural ‘constants’ belong to the range of the operator. So constants are functions in this case. (This holds because multiplication is well-defined for this space of functions; namely pointwise multiplication.) So if $\overline{T}$ is a transition operator, then $g\overline{T}$, with $g$ some function on $\mathcal{X}$, is as well (perhaps not coherent, but that is not important in this argument).

In this particular case, we are not exactly in this situation, because $\max$ is not a transition operator and the multiplication with $I_{\mathcal{X}}$ is necessary to build one from it. Nevertheless, the idea that the notion of a constant is relative, is key here to understand the notation, I think.

  • $\begingroup$ Thanks very much equaghe. This is very much appreciated. It didn't occur to me the $x$ would directly become the argument of $I_{\cal X}$. So even though $I_{\cal X}(x)$ is always 1, so in one sense it doesn't do any work, in another sense it needs to be there so that the resulting so that result of $\overline T(f)$ remains a function of $x$, and doesn't just turn into the scalar $\max(f)$. Is that the idea? Sorry to keep asking additional questions. $\endgroup$ – Mars Feb 19 '18 at 0:27
  • $\begingroup$ Also, it looks like the general rule of thumb is to read expressions like $I_{\cal X} \max$ in such a way that whatever parameters of the component functions can't be filled by one kind of argument are left for an argument of another kind. What I mean is, in this case the first argument to $I_{\cal X} \max$ is $f$, so that gets sent to $\max$, because $\max$ can accept an argument of the type $f$, but it makes no sense to apply $\max$ to a single element of $\cal X$. But $I_{\cal X}$ does apply to elements of $\cal X$, not to functions on $\cal X$, so $x$ gets sent to $I_{\cal X}$. Thank you! $\endgroup$ – Mars Feb 19 '18 at 0:31
  • $\begingroup$ Yes, see my edit. $\endgroup$ – equaeghe Feb 19 '18 at 15:53
  • $\begingroup$ Ah--that clarifies a point about notation that had contributed to my confusion: i.e. that $I_{\cal X}$ is written before $\max$, even though $\max f$ is a scalar while $I_{\cal X}$ is a function. This makes sense now. Thanks. Wonderful. $\endgroup$ – Mars Feb 19 '18 at 16:05

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