6
$\begingroup$

Stein's Lemma in its standard form states that $X \sim N(0,1) \Leftrightarrow E[f'(X) - X f(X)] =0 $ for all bounded one-time differentiable functions $f$ (I am ignoring the exact conditions on $f$ for now).

A generalization of the above statement is $X \sim Q \Leftrightarrow E[(\mathcal{A}f)(X)]=0$. Here we need to find an operator $\mathcal{A}$ that acts on functions $f$ from a particular class such that it characterizes the distribution $Q$. If $Q$ has an absolutely continuous density, the operator and function class could be stated explicitly (for example, via score functions).

Are there any particularly easy to use characterizations of the operator and function class in the case of discrete $Q$ (apart from https://arxiv.org/abs/0808.2877)?

$\endgroup$
  • $\begingroup$ For any cdf $\varphi(a)=\Pr(X\le a)$, $$X\sim \varphi\iff E((\mathcal Af)(X))=0\quad\forall f$$ where $f$ ranges over all the indicator functions $1_{(-\infty,a]}$, $a\in\mathbb R$, and $$(\mathcal Af)(X)=f(X)-\varphi(\sup\{x:f(x)>0\}).$$ But I guess you don't want $\mathcal A$ to refer to $\varphi$... $\endgroup$ – Bjørn Kjos-Hanssen Dec 22 '16 at 7:00
  • $\begingroup$ try i-journals.org/ps/viewarticle.php?id=182&layout=abstract $\endgroup$ – user83457 Dec 22 '16 at 11:59
1
$\begingroup$

You may want to have a look into Hwang's Lemma which is known as a discrete (sufficient) analogy to Stein's Lemma though it is NOT a characterization. Generally speaking you can yield such a characterization when the undelying probability distribution is restricted to full rank regular exponential family because in that case you are actually characterizing the linear structure of the probability density.

[Hwang]Let $g(x)$ be a function with bounded expectation $-\infty<Eg(X)<\infty$under the probability distribution $X~\varphi$ and $-\infty<g(-1)<\infty$ Then

(a)If $X~Poisson(\lambda)$ then $E(\lambda g(X))=E(Xg(X-1))$

(b)If $X~NegBinom(r,p)$ then $E([1-p] g(X))=E(\frac{X}{r+X-1}g(X-1))$

More generally I do not think there will be a characterization in a simpler form than [Eichelsbacher&Reinert] of $X \sim Q \Leftrightarrow E[(\mathcal{A}f)(X)]=0$ as you asked for. Here is my reasoning, since in discrete probability distributions the joint density of sample function usually contains order statistics as sufficient statistics. The related empirical process will based on the order statistics[Daly et.al]. If we write the stochastic empirical process in characterization(that is what Stein did) of some differential equation like $E[(\mathcal{A}f)(X)]=0$ then it must include such a sufficient statistics. Moreover if you read [Eichelsbacher&Reinert] carefully you can see they argue that in general a Gibbs measure must contain necessary many terms to describe the sample functions.

References

[Hwang]Hwang, Jiunn Tzon. "Improving upon standard estimators in discrete exponential families with applications to Poisson and negative binomial cases." The Annals of Statistics (1982): 857-867.

[Eichelsbacher&Reinert]Eichelsbacher, P., and G. Reinert. "Stein’s method for spatial Gibbs measures." Preprint (2003).

[Daly et.al]Daly, Fraser, Claude Lefèvre, and Sergey Utev. "Stein's method and stochastic orderings." Advances in Applied Probability (2012): 343-372.

$\endgroup$

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.