I want to know the effect of differentiation on the independence of random variables. For a random variable $X$, when are $f^{(n)}(X)$ and $f^{(n+k)}(X)$ independent?, $\forall n\geq0\;, k\geq 1$.
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1$\begingroup$ One guess is when $X$ is a degenerate random variable, but you should check to verify. $\endgroup$– GalenCommented Mar 17, 2023 at 20:41
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4$\begingroup$ There's nothing special about two derivatives, so you're really asking about when $(U,V)=(f(X),g(X))$ are independent for sufficiently differentiable functions $f$ and $g.$ Equivalently, you are asking when a 2D variable $(U,V)$ that is restricted to a connected subset of the image of a differentiable curve be independent. That answers itself when you draw a picture of that image: it must be either a horizontal or vertical line segment. That is, one of $U$ or $V$ (or both) are constant. This does not imply $X$ is degenerate, but @Galen has the right general idea. $\endgroup$– whuber ♦Commented Mar 17, 2023 at 21:04
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1$\begingroup$ It also works for $k$ large enough when $f$ is a polynomial. $\endgroup$– Xi'anCommented Mar 18, 2023 at 5:27
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$\begingroup$ Perhaps there are also cases involving limits, such as $k \rightarrow \infty$ for a given $n$. $\endgroup$– GalenCommented Mar 18, 2023 at 5:29
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1$\begingroup$ As another example of the complications, let $X$ have a uniform distribution on $\{0,1,2,3\}.$ It is straightforward to construct a function $f$ for which the graph of $f^{(n)}$ passes through the points $(0,0),$ $(1,1),$ $(2,1),$ and $(3,0),$ while $f^{(n+k)}$ passes through $(0,0),$ $(1,0),$ $(2,1),$ and $(3,1),$ so that $(f^{(n)}(X),f^{(n+k)}(X))$ has the uniform distribution on $\{(0,0),(1,0),(0,1),(1,1)\}.$ Clearly $f^{(n)}(X)$ and $f^{(n+k)}(X)$ are independent. $\endgroup$– whuber ♦Commented Mar 19, 2023 at 14:03
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A more modest property, namely a lack of correlation, holds in a series of cases. Consider the case when $f$ is invertible and let the density of $X$ write as $p(f(x))$ (wlog). Further assume (wlog) that $\mathbb E[f(X)]=0$. Then, by a change of variables, \begin{align}\text{cov}(f(X),f'(X))&=\int f(x)f'(x)p(f(x))\text dx\\&=\int f(x)\frac{\text df(x)}{\text dx} p(f(x))\text dx\\&= \int f(x) p(f(x)){\text df(x)}\\ &= \mathbb E_p[F]\\ &= 0 \end{align}
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$\begingroup$ This looks like a good idea, but it's unclear how your "$f$" might be related to the "$f$" of the question. $\endgroup$– whuber ♦Commented Mar 18, 2023 at 11:08
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$\begingroup$ That's not at all evident, because the question concerns the nth and (n+k)th derivatives of $f,$ neither of which are mentioned here. $\endgroup$– whuber ♦Commented Mar 19, 2023 at 13:27