# What does "version" mean here?

"Assumption 1. There is a version of $$f(x)$$ that is twice continuously differentiable"

Note that $$f(x)=E(Y|x)$$ is an unknown function to be estimated with kernel regression using data $$\{Y_{i},X_{i}\}_{i=1}^{n}$$.

My question is what does "a version" of a function being continuous mean? How does this assumption differ from the following statement:

"$$f(x)$$ is twice continuously differentiable"

$$E(Y\mid x)$$ is a conditional expectation, which means it is a random variable.

The distributional properties of random variables are not changed when changes are made to those variables that obviously do not change any probabilities associated with them. In particular, when $$X$$ and $$X^\prime$$ are random variables defined on the same probability space $$\Omega$$ and the chance that they differ is zero, we say each is a "version" of the other.

It is also said that two versions of a random variable are almost surely equal.

Formally, $$X^\prime$$ is a version of $$X$$ when there is an event $$\mathcal{E}\subset\Omega$$ for which

1. $$\mathbb{P}(\mathcal E) = 0$$ ($$\mathcal E$$ is a set of measure zero) and

2. $$X\mid_{\Omega\setminus\mathcal E} = X^\prime\mid_{\Omega\setminus\mathcal E}$$ ($$X$$ and $$X^\prime$$ are the same function on the complement of $$\mathcal E$$).

It is an immediate (and obvious) consequence of probability axioms and definitions that $$X$$ and $$X^\prime$$ have the same distribution function and that $$\mathbb{P}(X\ne X^\prime)=0.$$

For instance, let $$X$$ be a random variable with a standard Normal distribution. Define $$X^\prime = X$$ for all $$\omega\in\Omega$$ except for which $$X(\omega)=0,$$ where we may define $$X^\prime = 1$$ (say). Because a Normal distribution has zero chance of equalling any particular value like $$0,$$ this does not change any of the distributional properties of $$X:$$ $$X^\prime$$ still has a standard Normal distribution.

The same concepts hold in any measure space. It is possible, depending on the context, that the $$f$$ in the question is being considered as a function $$f:\mathbb{R}\to\mathbb{R}.$$ In such cases the sense of "version" means with respect to Lebesgue measure rather than a probability measure. The usual terminology is that two "versions" of a function are equal almost everywhere. Thus, the assertion in the question could mean there exists a function $$g$$ and a set $$\mathcal{E}\subset\mathbb{R}$$ of Lebesgue measure zero for which property $$(2)$$ above holds: that is, apart from their values on $$\mathcal E,$$ $$f$$ and $$g$$ are the same function.

Why is this important and interesting? Because, for instance, the function $$g$$ defined by

$$g(x) = x^2$$

is everywhere smooth, but the function $$f$$ defined to equal $$g$$ on the irrational numbers and otherwise is (say) equal to $$-1$$ on all rational numbers has no derivative anywhere. Nevertheless, because the rational numbers have Lebesgue measure zero, $$g$$ is a version of $$f.$$ In this way we have stripped away "trivial" aspects of $$f$$ to focus on its basic underlying "structure" as revealed by $$g.$$

• Thank you so much! This is extremely helpful, and is the clearest explanation I can imagine. Apr 8, 2020 at 17:37