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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^*) ] $.

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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?

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  • $\begingroup$ 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

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