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:


, 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?

  • $\begingroup$ "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. $\endgroup$ Oct 18, 2021 at 12:02
  • $\begingroup$ I said "normal-linear case". $\endgroup$ Oct 18, 2021 at 20:57
  • $\begingroup$ I overlooked that. But it raises even more questions. What does it mean, 'on normal-linear case'? $\endgroup$ Oct 18, 2021 at 21:11
  • $\begingroup$ $Y$ is linear on $X$ and they both are normal. $\endgroup$ Oct 18, 2021 at 21:48
  • $\begingroup$ What means 'Y is linear on X'? $\endgroup$ Oct 18, 2021 at 22:43

1 Answer 1


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.

    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=".")

enter image description here

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    $\begingroup$ You're sure, I thought it before but I forgot it for some reason, Good example. $\endgroup$ Oct 18, 2021 at 2:39
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    $\begingroup$ "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 $S^2 = 0.25 (X_1 - X_2)^2$. $\endgroup$ Oct 18, 2021 at 11:57
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    $\begingroup$ @SextusEmpiricus. Seems there are various interpretations of 'functinally indep'. Will ponder this. $\endgroup$
    – BruceET
    Oct 18, 2021 at 14:59
  • $\begingroup$ 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 $\endgroup$ Oct 18, 2021 at 21:51

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