Does probability independence have anything to do with functional independence (on general case)? On normal-linear case I know it holds for example if $cor(X_1,Y)=0$ then $g_{X_1Y}(X_1,Y)=g_{X_1}(X_1)g_{Y}(Y)$, where $g_{(.)}$ is the distribution of (.), and that imples to $\beta_1=0$ on:
$Y=\beta_1X_1$
, since $cor$ works as measure to evaluate how good a linear approximation between two variables is and those implications hold we have that there is no linear function such that $Y=f(X_1)$ (or there is with $\beta_1=0$ but it is not informative at all) when both $X_1$ and $Y$ are
stochastically independent and normal distribuited.
But I'm not sure if it holds for general case, that is if $g_{X_1Y}(X_1,Y)=g_{X_1}(X_1)g_{Y}(Y)$ so there is no function $f$ such that $Y=f(X_1)$ (I call this functional independence). Am able to go functional dependence as probability depencence or is it pretty wrong? and if I'm not is there any relation between they both?
 A: For random normal data, the sample mean $\bar X$ and the sample variance $S^2$ are independent random variables. But they are not 'functionally' independent because $\bar X$ is used in the definition of $S^2.$
A formal proof of the independence of $\bar X$ and $S^2$ can be done
using a linear transformation in $n$-space or by moment generating functions. [This is a consequence of Basu's Thm; see example. Also, see this Q&A; "Related" page listed in the margin.]
I show a simulation in R below that illustrates the idea (in terms of correlation), using $m$ normal
samples of size $n.$ With $m = 100\,000$ iterations the correlation
should be accurate to one or two decimal places.
    set.seed(2021)
    n = 10;  m = 10^5
    x = rnorm(m*n, 50, 7)
    DTA = matrix(x, ncol=n)  # each row a sample
    a = apply(DTA, 1, mean)  # m sample means
    v = apply(DTA, 1, var)   # m sample variances
    cor(a, v)
    [1] -0.00202886          # consistent w/ 0 correlation
    
    s = sqrt(v)
    plot(a, s, pch=".")


