Multivariate Zero-Bias Transform The zero-bias transform for a univariate random variable $W$ is defined as a random variable $W^*$ satisfying 
\begin{align}
\mathbb{E} [ W \cdot f(W )] = \mathbb{E} [ f' (W^*)]
\end{align}
for any differentiable function $f$ satisfying some regularity conditions. The probability density function of $W^*$ is given by $f(t ) = \mathbb{E} [ W \cdot \mathbf{1} \{ W \geq t \} ]$.
For the multivariate case, zero-bias transformation is only known for multivariate Gaussian, whose zero-bias transform is itself. Let $X \sim N(0, I_d) $, then we have 
$
\mathbb{E} [ X \cdot f(X) ] = \mathbb{E} [ \nabla f(X)].
$
Is it possible to define multivariate zero-bias transform for general random vectors? We could first think $X$ has i.i.d. entries, then for each $j \in [d]$, 
we can apply the univariate result to obtain 
\begin{align}
\mathbb{E} [ X_j \cdot f(X_1, X_2, \ldots, X_d) ]  = \mathbb{E} [ \partial _jf( X_1, \ldots, X_j^*, X_{j+1}, \ldots, X_n)].
\end{align}
In this case, I could not find a random vector $X^* \in \mathbb{R}^d$ such that 
\begin{align}
\mathbb{E} [ X_j \cdot f(X_1, X_2, \ldots, X_d) ] = \mathbb{E} [ \partial_j f (X^*) ] , 
\end{align}
or equivalently, $\mathbb{E} [ X  \cdot f(X ) ] = \mathbb{E} [ \nabla f(X^*) ] $.
 A: Steve.
Consider the following 
$$\mathbb{E} [ X_j \cdot f(X_1, X_2, \ldots, X_d)-\partial _jf( X_1, \ldots, X_j^*, X_{j+1}, \ldots, X_n) ]  = 0$$ 
We have to assume that $f$ has a pointwise nonvanishing Jacobian in following derivations, which is not the same as nonvanishing Fisher information in the usual regularity conditions.
If $X_j \cdot f(X_1, X_2, \ldots, X_d)-\partial _jf( X_1, \ldots, X_j^*, X_{j+1}, \ldots, X_n)$ is a complete family $(X_j,X_j^{*})\mid X_1\cdots X_{j-1},X_{j+1},\cdots X_n$, then the expectation equation reduces to $$X_j \cdot f(X_1, X_2, \ldots, X_d)-\partial _jf( X_1, \ldots, X_j^*, X_{j+1}, \ldots, X_n)=0$$ which is a differential equation in form of $x\cdot y(x)-y'(z(x))=0$ where $y(\bullet )$ is known and $z(x)$ is to be solved. I am not sure such a variational equation has a solution...
On the contrary if the above $(X_j,X_j^{*})\mid X_1\cdots X_{j-1},X_{j+1},\cdots X_n$ is not complete for some $j$, then there will be more than one solutions since expectation(integral) equation usually have Fredholm structure.
Could you please state your motivation to the problem in OP too?
