A Random Set is a set-valued RV, i.e. a map $X:\Omega\to\mathcal{C}$ from a probability space $(\Omega,\Lambda,P)$ to the family of measurable closed sets $\mathcal{C}$ on a $\sigma-$algebra $\Lambda$ built from the elementals $\omega\in\Omega$ (with $\Omega$ countable), and such RV takes set values:

$$X^{-1}(K)=\{ \omega|X(\omega)\cap K\neq\emptyset \},$$

where $K$ is a closet set called the "trap" or structuring set. For simplicity, I denoted $X^{-1}(K)\in\mathcal{X}$ by $x_i$, and $Y^{-1}(K)\in\mathcal{Y}$ by $y_j$, and $X(\omega)$ by $X$.

I'm trying to implement conditional probability distribution $Y|X$. So, for $Y=y_j$, given $X=x_i$, we can use the Bayes theorem: $$P(Y=y_j|X=x_i)=\frac{P(X=x_i|Y=y_j)P(Y=y_j)}{\sum_{y_j'\in\mathcal{Y}}P(X=x_i|Y=y_j')P(Y=y_j')}=\frac{f(x_i,y_j)}{\sum_{y_j'\in\mathcal{Y}}f(x_i,y_j')},$$ where the question is, how to define $f(x_i,y_j)$ as $X,Y$ take values on the sets of sets $\mathcal{X}=\{x_1,\dots,x_\ell\}$ and $\mathcal{Y}=\{y_1,\dots,y_{\ell'} \}$, respectively?

I've tried with this: $$(1)\hspace{10mm}f(x_i,y_j)=\sum_{(x_i',y_j')} |x_i\cap x_i' |\cdot|y_j\cap y_j'|. $$

However, I'm not sure how to verify here the usual notion of quatifying the observation of the outcome $(x_i,y_j)$ over the set of outcomes $\{ (x_i', y_j')\} $. Please, someone can help me to get the right expression for Eq. (1)?

  • 1
    $\begingroup$ Also the sentence "$X_i \cap Y_j$ provides a measure of how the attributes of both sets should be observed jointly" is very unclear. What is the "attribute" of a set? What does "observing an attribute" mean? I'm not sure whether you are a bit hurried in writing and so the sentences didn't come out right, or whether you may have some misunderstanding or confusion about the notions of random variable, event, sample space, and so on... $\endgroup$
    – pglpm
    Nov 22 '20 at 20:15
  • 1
    $\begingroup$ You say events are sets, you probably wants to say that outcomes are sets. Maybe tell us about the concrete background for your Q? $\endgroup$ Nov 23 '20 at 16:49
  • 1
    $\begingroup$ Dear @pglpm, thanks a lot for helping me to fix the terminology. Indeed, I wanted to mean that $X$ takes values on a set of sets, so what is the best way you suggest to denote such values? Now, by saying "attributes" I tried to be simple. However what I wanted to say that each value of the RV is built by elementary objects $\omega\in\Omega$ such that some $\sigma-$algebra is generated by a power set $2^\Omega$, and some topology takes place by means of finite unions. The RVs are elements of a family of closed and measurable sets, so a measure $\mu$ is present. $\endgroup$
    – Nacho
    Nov 23 '20 at 18:55
  • 1
    $\begingroup$ You are studying random sets $\endgroup$ Nov 23 '20 at 21:06
  • 1
    $\begingroup$ @pglpm I only found interesting the problem of predicting sets instead of points, although the paper point out some undesirable properties also (you are right in that the paper makes it appear that there is no difference in applying set-valued classifiers and usual ones). $\endgroup$
    – Nacho
    Nov 23 '20 at 21:53

Since the expressions for conditional probability involves a sum to marginalise the joint probabilities, you will need to specify the support of the random sets you are dealing with, which will require you to refer to classes of sets (i.e., sets of sets). Thus, the first thing that will be useful here is to use standard fonts for elements, sets, and classes. For example, you might denote a specific element value by $x$, a set of these values by $\mathcal{X}$, and a class of these sets by $\mathscr{X}$. Once you are using notation that clearly distinguishes between these things, it is fairly simple to formulate the conditional probability formula for random sets.

For simplicity, I will consider the case where your random sets are discrete random variables (i.e., both have countable support). Consider the case where we have two random sets $\mathcal{X} \in \mathscr{X}$ and $\mathcal{Y} \in \mathscr{Y}$ where the latter classes are their respective supports. Then the rule of conditional probability says that:

$$f(\mathcal{Y}| \mathcal{X}) = \frac{f(\mathcal{X}, \mathcal{Y})}{\sum_{\mathcal{Y} \in \mathscr{Y}} f(\mathcal{X}, \mathcal{Y})}.$$

Without specifying how these random sets are constructed from underlying random elements, that is really as far as you can go. If you are willing to specify how random sets are constructed from some underlying random elements then you might be able to split the term $f(\mathcal{X}, \mathcal{Y})$ into a product of probabilities for those random elements. At the moment you appear to be trying to split this term out into a sum, but you are trying to do this without giving any information on how the random sets are consructed; consequently, your attempts are all abortive.

  • $\begingroup$ Dear @Ben, I've added information on how the RVs and their possible outcomes are built. This helps you to be more specific in your answer? $\endgroup$
    – Nacho
    Nov 24 '20 at 0:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.