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^*) ] $.


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?

  • $\begingroup$ If anyone happen to know some information of the equation $x\cdot y(x)-y'(z(x))=0$ do not hesitate to comment, thanks. $\endgroup$ – Henry.L Mar 15 '17 at 0:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.