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Start with the simple identity: $$(f(x) - a)(g(x) - b) + a(g(x) - b) + b(f(x) - a) = f(x)g(x) - ab.$$

If $a$ and $b$ are the respective values of $f$ and $g$ at some point, then, after dividing both sides by $\Delta x$ and letting $\Delta x\to 0$ the first term vanishes and we get a proof of the product rule.

But if $a$ and $b$ are the respective average values of $f$ and $g$, then after averaging the whole expression, the first term is the only one on the left that does not vanish and we get the familiar identity $\mathrm{cov}(f,g) = E(fg) - E(f)E(g)$, where $\mathrm{cov}(f,g)$ is by definition the average of the first term.

I don't know that I've ever seen it stated that the product rule and this identity on covariances are the local and global versions of something.

Is this just one instance of some broader pattern?

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2 Answers 2

I do not know about the broader pattern in general but it reminds me of the "probability" and "free probability" settings.

You take an algebra with a trace $\tau$ on it, i.e. a (circular) linear form on it (for technical details, you have to add the good topology on it) and you define some notion of independence (the usual one or the free one). Then you can evaluate non-linear quantities such as $$\tau(a^n b^k a^m \ldots b^p)$$ (where $a$ and $b$ are independent elements of the algebra and $n$, $k$, $m$ and $p$ are integers) as a combination by sums of products of the only quantities $$\tau(a^i), \quad \tau(b^j)$$

See http://www.mast.queensu.ca/~speicher/survey.html for various survey articles by Speicher.

It seems to me that you are following the same kind of procedures here and you just change the algebra and/or the trace on it...

Damien.

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I understand that you consider a "parametrized" identity on a space of differentiable functions $$ \mathcal{L}_{a,b}(f,g)=\[f-a\]\[g-b\]+a\[g-b\]+b\[f-a\]-fg+ab=0 $$ and two different linear forms acting on that space, namely $\phi : f\mapsto f'(x_0)$ and $\psi : f\mapsto\mathbb{E}(f)$, which induce the identities $\phi \[\mathcal{L}_{x_0,\;x_0}(f,g)\]=0$ (the product rule) and $\psi\[\mathcal{L}_{\mathbb{E}(f),\;\mathbb{E}(g)}(f,g)\]=0$ (the definition of the covariance).

Of course you can evaluate this identity along all the (linear) maps you want playing with the parameters $a$ and $b$, but maybe your question was about a potential more subtle relation between "derivative" and "variance" of functions ? In this case, I have the impression that these two identities are different in nature, since the parameters $a$ and $b$ have to be chosen in a very different way for each form.

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