For example, let's say a term $A(x,y)$, a function of two random variables $x$ and $y$, is the argument of an expectation over $y$. The resulting term is no longer a function of $y$. Is there a mathematical symbol that explicitly says this?

I know there are workarounds--in this arbitrary case, bar notation and dropping $y$ from the parentheses--but I'm wondering if there's a more concise and explicit way to write this. I have nothing against doing it in words instead of symbols; I'm just curious.

  • $\begingroup$ I haven't seen a specific notation for this. In this case, I'd say "There is a function $f$ such that $A(x, y) = f(\mathbb{E}y)$. $\endgroup$ – xyzzyz Sep 20 '16 at 14:47
  • $\begingroup$ This whole question depends on the concept of "function of a variable", which is troublesome to start with. $\endgroup$ – Git Gud Sep 20 '16 at 14:48
  • $\begingroup$ I think the problem is that you can regard any function as a function of any collection of additional variables as well. For example, what about constant functions? Just because a variable isn't explicitly needed to compute the values of a function doesn't mean the function isn't a function of that variable. It isn't the same thing as, say, independence. $\endgroup$ – MPW Sep 20 '16 at 14:58
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    $\begingroup$ @pondrthis Not really. Even among mathematicians that terminology is common, albeit with a grain of salt. And it's this grain of salt which hinders the formalization of this concept. If it's any help, in case the partials of $A$ exist, you can obtain the concept by claiming that $\partial _2A(x,y)=0$ for all $(x,y)$ in the domain of $A$. $\endgroup$ – Git Gud Sep 20 '16 at 15:33
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    $\begingroup$ You could say that $\exists f: X\rightarrow \mathbb{R} \ni \forall x \in X \text{ and }\forall y \in Y, f(x)=A(x,y)$, which I believe is equivalent. That is, there's a function f which is equivalent to A when their x-inputs are equal and f doesn't take a y-input. $\endgroup$ – Neil Sep 20 '16 at 19:20

You want to express that the function is independent of a certain $y$, so it doesn't change when $y$ changes. I think, in your case derivative would be useful to state this independence:

$\frac{\partial A}{\partial y} = 0$


As Git Gud commented, there is a problem with the concept of a "function of variable". However, let $A$ be the set of all possible values of y (that you want to consider), and $B$ be the target set, such that $f$ would be a function $A\rightarrow B$. The set of all functions of a set $X$ into a set $Y$ is commonly denoted as $Y^X$. Therefore, you could write your statement as $f\not\in Y^X$.

If, however, the correct interpretation of "$f$ is not a function of $y$" is the one described by Eike Schulte in the comments on this answer, then I think that what you want to write is $$A(x,y_1) = A(x,y_2) \;\;\;\forall y_1,y_2\in B.$$

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    $\begingroup$ This does not really answer the question: “Not being a function of $y$” expresses independence of $y$, i.e. that the function is constant “in the $y$ direction”. But constant functions are still functions. $\endgroup$ – Eike Schulte Sep 20 '16 at 17:41
  • $\begingroup$ @Eike Schulte Does it answer the question now? $\endgroup$ – confusedStudent Sep 20 '16 at 19:09

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