# Is the expectation value a function of the probability distribution

I am not very familiar with the measure theoretic approach to probability which might be why I cannot answer the following question to myself.

Say we have a discrete and finite random variable $X$ that can take $n$ values. Let $\mathcal{X}=\{1,...,n\}$ be the set of possible values that $X$ can take. Then the set $\mathcal{P}_{\mathcal{X}}$ of possible probability mass functions is the $n-1$-simplex. I.e.\ the set of functions $p:\mathcal{X} \rightarrow [0,1]$ such that $\sum_{x\in \mathcal{X}} p(x) =1$.

Now let us define a function $m:\mathcal{P}_{\mathcal{X}}\rightarrow \mathbb{R}$ that takes as an argument a probability mass function and maps it to a real number. My question is, is the expected value $\mathbb{E}[X]$ such a function?

What confuses me is that the expected value

$$\mathbb{E}[X]=\sum_{x \in \mathcal{X}} x \; p(x)$$

somehow needs access to the values that $X$ can take but it shouldn't be able to tell that by just looking at the function $p$. If we swap $\mathcal{X}$ to another set of natural numbers, e.g. $\{10,...,10+n\}$ then the expectation value changes but the set of probability mass functions remains exactly the same.

EDIT: As Batman noted in a comment, this is a bit imprecise. Let $\mathcal{Y} = \{10,...,10+n\}$ then we don't actually have $\mathcal{X}=\mathcal{Y}$ of course, we only have that there is a bijection (they are isomorphic). Then I suspect that we also don't have $$\mathcal{P}_{\mathcal{X}}=\mathcal{P}_{\mathcal{Y}}$$ but again only isomorphism. This then suggests that $m:\mathcal{P}_{\mathcal{X}} \rightarrow \mathbb{R}$ actually "sees" or "knows" $\mathcal{X}$ and the expectation value can be defined as such a function.

I now wonder how a mathematician would go about defining a function $g:\mathcal{P}_{\mathcal{X}} \rightarrow \mathbb{R}$ that doesn't "see" $\mathcal{X}$? I.e. a function that is automatically also defined for elements of $\mathcal{P}_{\mathcal{Y}}$? END EDIT.

The previous end to this question might still be relevant: This seems to suggest that the expectation value is not a function of the probability mass function. It should be a function of the random variable in some sense however.

Does this mean that the expectation value is a function on

$$\mathcal{P}_{\mathcal{X}} \times \{\mathcal{A}\subset \mathbb{Z}:|\mathcal{A}|=n\}$$

i.e. on the pairs of elements of the $n-1$-simplex and subsets of the integers of cardinality $n$?

If that is so, what is the accurate mathematical terminology for this? Do some people really define the expectation value like this?

• The probability mass function depends on $\cal X$, so the "set of probability mass functons remains exactly the same" isn't quite right; you can form a bijection between the sets of probability mass functions on the two sets , but the set of probability mass functions are different on the two sets – Batman Apr 15 '18 at 2:03
• @Batman Thank you that points out some imprecision. So you are saying in the standard approach the function $m$ would always know $\mathcal{X}$? Because at least theoretically I could also define a function that is agnostic w.r.t. $\mathcal{X}$ and only cares about the cardinality of $\mathcal{X}$. It would actually be of interest to me to know how I can define the latter function specifically and in contrast to the former. – mab Apr 15 '18 at 2:14
• You can define any function you like as long as you are precise about it, such as specifying the domain and so on. If $X$ takes values on a finite set $\mathcal{X}$ then $E[X]$ is defined as the real number $E[X]= \sum_{x \in \mathcal{X}} xP[X=x]$. It is not a function. But it certainly "depends on" the set $\mathcal{X}$ and the PMF $P[X=x]$ for $x \in \mathcal{X}$. If you want you can define a set of PMFs on $\mathcal{X}$ and then make a "function" that maps PMFs to expectations, that would be a function, each value of the function being an expectation of a certain random variable. – Michael Apr 15 '18 at 4:49
• Formally, every random variable $X$ has a cumulative distribution function $F_X(x)=P[X\leq x]$, defined for all $x \in \mathbb{R}$, and $E[X]$ is completely determined by $F_X(x)$ for all $x \in \mathbb{R}$. So, it depends entirely on the probability distribution function. – Michael Apr 15 '18 at 4:54
• @Michael Interesting point. This seems to suggest that I can define the expectation value as a function of the cumulative distribution function (CDF). The CDF in some sense contains more information than the PMF because it identifies the support $\mathcal{S}(X) \subset \mathcal{X}$. On the other hand we cannot uniquely identify $\mathcal{X}$ from a given CDF is the support is not full. For the expectation value this is of course irrelevant. – mab Apr 16 '18 at 4:52