Finding the expectation from the cdf of a vector-valued random variable We can calculate the expectation of a real-valued random variable by integrating its cumulative distribution function. What is the analogue of this fact for a random variable that takes values in a partially ordered vector space?
Suppose that $X$ is a random variable that takes values in a partially ordered vector space $V$. Let's warm up with the case where $X$ takes only takes values in a finite set of positive vectors $U ⊂ V$. (Let $U$ contain $0$ for convenience.) In the totally ordered case, if we let $d(x) = x - y$ where $y$ is the immediate predecessor of $x$ in $U$ (and $d(0) = 0$), then we have
$$
\sum_{x ∈ U} P(X = x) · x = \sum_{x ∈ U} P(X \geq x) · d(x)
$$
But this does not make sense if $V$ is only partially ordered, since $x$ need not have a single predecessor in $U$. We could try something like
$$d(x) = x - \sum \{y ∣ \text{$y$ is an immediate predecessor of $x$ in $U$} \}$$
But this doesn't work either: we end up double-counting some values of $x$, when we have branches in $U$ that eventually join again. I've been trying out more elaborate definitions of $d(x)$ that compensate for this, and I think I might be able to get it to work, but it feels like I'm probably reinventing a wheel.
 A: I think I have an answer to the warm-up question for the case where $U$ has the structure of a finite lattice. But it's kind of complicated.
For any $x ∈ U$, let
$$\begin{aligned}
U_k^{\wedge}(x) & = \{ y_1 \wedge ⋯ \wedge y_k ∣ y_1, \dots, y_k \text{ are distinct and immediately precede $x$} \} \\
U_k^{\vee}(x) & = \{ y_1 \vee ⋯ \vee y_k ∣ y_1, \dots, y_k \text{ are distinct and immediately succeed $x$} \}
\end{aligned}$$
Then let
$$
d(x) = x + \sum_k (-1)^k \sum U_k^{\wedge}(x)
$$
Then the result holds. To show this, I'll use two lemmas.
Lemma 1. Let $E$ be a finite set of events.
$$
P(\cup E) = \sum_k (-1)^{k+1} \sum_{A_1 \neq \dots \neq A_k ∈ E} P(A_1 ∩ ⋯ ∩ A_k)
$$
This is a generalization of the simple fact that
$$P(A_1 ∪ A_2) = P(A_1) + P(A_2) - P(A_1 ∩ A_2)$$
Lemma 2. For any $x, z ∈ U$, $x ∈ U_k^{\wedge}(z)$ iff $z ∈ U_k^{\vee}(x)$.
Now consider the sum
$$
\sum_{x ∈ U} P(X \ge x)· d(x)
$$
For a given element $x ∈ U$, for which $z$ does $x$ appear as a term in $d(z)$? First, obviously $x$ appears in $d(x)$; and second, $x$ appears in any $z$ term such that $x ∈ U_k^{\wedge}(z)$, or equivalently by Lemma 2, where $z ∈ U_k^{\vee}(x)$. So the final coefficient for $x$ in the sum will be
$$
P(X \geq x) + \sum_k (-1)^k \sum_{z ∈ U_k^{\vee}(x)} P(X \geq z)
$$
Moreover, $P(X \ge x) = P(X = x) + P(X > x)$, and $X > x$ holds iff
$X \ge y_1$ or … or $X \ge y_n$, where $y_1, \dots, y_n$ are all of the immediate successors of $x$ in $U$.
So by Lemma 1,
$$
P(X > x) = \sum_k(-1)^k \sum_{z ∈ U_k^{\wedge}} P(X \ge z)
$$
So all of those terms cancel out, leaving just $P(X = x) · x$.
