This is getting no attention, so I'll try this here:

• The Robbins lemma, named after Herbert Robbins, says that if $X\sim\operatorname{Poisson}(\lambda)$ and $g$ is a function for which $\operatorname{E}(|X g(X)|) < \infty,$ then $$\operatorname{E}(Xg(X)) = \lambda \operatorname{E}(g(X+1)).$$
• "Gaussian integration by parts" is an identity that says that under suitable assumptions about the function $g$, if $X\sim N(0,\sigma^2),$ then $$ \operatorname{E}(Xg(X)) = \sigma^2\operatorname{E} (g\,'(X)). $$

Both of these propositions are used in empirical Bayes methods.

Both of these are of the form $$ \operatorname{E}(Xg(X)) = \operatorname{var}(X) \cdot \operatorname{E} ((Tg)(X)) $$ where $T$ is a linear operator on functions $g$.

QUESTION: Might there be, for each linear operator $T$, some probability distribution for which this holds? And might all of these be useful in empirical Bayes methods?

This is getting no attention, so I'll try this here:

• The Robbins lemma, named after Herbert Robbins, says that if $X\sim\operatorname{Poisson}(\lambda)$ and $g$ is a function for which $\operatorname{E}(|X g(X)|) < \infty,$ then $$\operatorname{E}(Xg(X)) = \lambda \operatorname{E}(g(X+1)).$$
• "Gaussian integration by parts" is an identity that says that under suitable assumptions about the function $g$, if $X\sim N(0,\sigma^2),$ then $$ \operatorname{E}(Xg(X)) = \sigma^2\operatorname{E} (g\,'(X)). $$

Both of these propositions are used in empirical Bayes methods.

Both of these are of the form $$ \operatorname{E}(Xg(X)) = \operatorname{var}(X) \cdot \operatorname{E} ((Tg)(X)) $$ where $T$ is a linear operator on functions $g$.

QUESTION: Might there be, for each linear operator $T$, some probability distribution for which this holds? And might all of these be useful in empirical Bayes methods?

(P.S.) BETTER BUT LESS LOGICALLY PRECISE VERSION: Are both of these instances of some more general fact of interest?