# 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?

• "for example if $cor(X_1,Y)=0$ then $g_{X_1Y}(X_1,Y)=g_{X_1}(X_1)g_{Y}(Y)$, where $g_{(.)}$ " This example is a bit tricky because a correlation of zero is not the same as independence. For example, if $X \sim N(0,1)$ and $Y = f(X) = X^2$ then the variables are dependent but the correlation is zero. Oct 18, 2021 at 12:02
• I said "normal-linear case". Oct 18, 2021 at 20:57
• I overlooked that. But it raises even more questions. What does it mean, 'on normal-linear case'? Oct 18, 2021 at 21:11
• $Y$ is linear on $X$ and they both are normal. Oct 18, 2021 at 21:48
• What means 'Y is linear on X'? Oct 18, 2021 at 22:43

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)
 -0.00202886          # consistent w/ 0 correlation

s = sqrt(v)
plot(a, s, pch=".") • You're sure, I thought it before but I forgot it for some reason, Good example. Oct 18, 2021 at 2:39
• "A formal proof of the independence of $\bar{X}$ and $S^2$ can be done using a linear transformation" This proof will also show that $S^2$ is actually not really functional dependent. We can rewrite the function $S^2 = \frac{1}{n} \sum (X_i - \bar{X})^2$ such that it does not contain $\bar{X}$. For instance, in the case of $n=2$ then $S^2 = 0.25 (X_1 - X_2)^2$. Oct 18, 2021 at 11:57
• @SextusEmpiricus. Seems there are various interpretations of 'functinally indep'. Will ponder this. Oct 18, 2021 at 14:59
• I think it's impossible to formalize what that is I made a post on math stackexchange and I find out that there always be a function which links two variables (assuming some weak assumptions like not to have repeated values). I should have meant 'natural functional dependence' but your point is interesting where does come your functional dependence definition from? @SextusEmpiricus Oct 18, 2021 at 21:51