Dependence of expectation operator on order properties of sample space

Question

Let $\Omega = \{R, B, G\}$ correspond to three abstract elements which aren't naturally related by any order relation in the probability space $(\Omega,2^\Omega, \mathbb{P})$ with $\mathbb{P}$ satisfying

$\mathbb{P}(\{R\}) = \frac{1}{6} \quad \mathbb{P}(\{B\}) = \frac{2}{6} \quad \mathbb{P}(\{G\}) = \frac{3}{6} $ .

Let $X: \Omega \to \mathbb{R}$ be a random variable. Since $R$, $B$, $G$ have no natural order, let's define

$X(R) = 1, \quad X(B) = 2, \quad X(G) = 3$ .

Then $\mathbb{E}[X] = \frac{13}{6}$, a value closest to the image of $B$ under $X$.

If we instead define $\quad X(R) = 3, \quad X(B) = 1, \quad X(G) = 2$ ;

then $\mathbb{E}[X] = \frac{11}{6}$, a value closest to the image of $G$ under $X$.

Clearly, any order relation on $\Omega$ induces a class of order-preserving random variables $X$ to choose from. Conversely, any choice of $X$ is equivalent to a choice of an order relation on $\Omega$ by setting $\omega_1 \leq \omega_2$ if and only if $X(\omega_1) \leq X(\omega_2)$. Any such choice produces consistent and meaningful expected values. But in absence of such a choice, in general, the result of $\mathbb{E}(X)$ seems to be arbitrary.

Does the result of $\mathbb{E}(X)$ have any meaningful interpretation in absence of an order relation specified on the elements of $\Omega$ ?

Answer 1

You overestimate the role of the order on $\Omega$. Had you defined $X$, say, like

$$X(\omega)=60 \ 000\ \text{ if }\omega=R \text{ and } 0 \text{ otherwise}$$

then the expectation would have been $10\ 000$.

So, the expectation has nothing to do with the order (if any) of the elements in $\Omega$ rather it has to do with the assignment of the $\omega s$ to the real numbers.

The best example is when we have vector valued random variables in the case of which no order can be (naturally) defined.

The assignment of the $\omega$s to reals or vectors of reals may seem to be arbitrary. But in reality, most of the time, we have (if we have) a priory/statistically given a probability distribution on the set of real numbers or of vectors, so the question does not arise the way you assumed.

Another question arises on the othe hand:

Why do we need the abstract $(\Omega,\Sigma, \mu)$ thing if the natural choice seems to be $(R^n,\Sigma, \mu)$.

The answer is that in certain cases we have many different kinds of random variables taking their values in different abstract spaces some of wich are ordered and some are not. It is very relaxing that we know that there exists an abstract $\Omega$ and we can consider our random variables as abstract functions on that $\Omega$. This is the most basic concept of the Kolmogorovian probability theory. Kolmogorov shoved that under very broad conditions such an abstract probability space exists.