The KLR-theorem revisited

TL;DR

This paper revisits the KLR-theorem, establishing conditions under which certain conditional expectations imply Gaussianity of variables, and introduces a new characterization involving Gaussian components.

Contribution

It provides a new characterization of Gaussian variables based on conditional expectation relations, including the first known case involving Gaussian components.

Findings

For n≥3 and ab>0, the relation holds iff variables are Gaussian.

A new characterization arises when a=1, b=-1, indicating Gaussian components.

The work extends understanding of Gaussianity through conditional expectation properties.

Abstract

For independent random variables $X_1,\ldots, X_n;Y_1,\ldots, Y_n$ with all $X_i$ identically distributed and same for $Y_j$, we study the relation \[E\{a\bar X + b\bar Y|X_1 -\bar X +Y_1 -\bar Y,\ldots,X_n -\bar X +Y_n -\bar Y\}={\rm const}\] with $a, b$ some constants. It is proved that for $n\geq 3$ and $ab>0$ the relation holds iff $X_i$ and $Y_j$ are Gaussian.\\ A new characterization arises in case of $a=1, b= -1$. In this case either $X_i$ or $Y_j$ or both have a Gaussian component. It is the first (at least known to the author) case when presence of a Gaussian component is a characteristic property.

Equations

$$E\{a\bar{X}+b\bar{Y}|X_1-\bar{X}+Y_1-\bar{Y},\dots ,X_n-\bar{X}+Y_n-\bar{Y}\}=\mathrm{const}$$

$$E\{\bar{X}|X_1-\bar{X},\dots ,X_n-\bar{X}\}=\mathrm{const}$$

$$E\{\bar{X}|X_1-\bar{X}+Y_1-\bar{Y},\dots ,X_n-\bar{X}+Y_n-\bar{Y}\}=\mathrm{const}$$

$$\sum _{j=1}^n{f}_1^{\prime }(t_j-\bar{t})\prod _{k\ne j}{f}_1(t_k-\bar{t})\prod _{j=1}^n{f}_2(t_j-\bar{t})=0,$$

$$\sum _1^n\frac{f_1^{\prime }}{f_2^{\prime }}({\tau }_j)=0,\sum _1^n{\tau }_j=0,|{\tau }_j|<ϵ.$$

$$f_1(t)=\exp c{t}^2/2,|t|<ϵ.$$

$$E\{\bar{X}|X_1-\bar{X},Y_1-\bar{Y},\dots ,X_n-\bar{X},Y_n-\bar{Y}\}=E\{\bar{X}|X_1-\bar{X},\dots ,X_n-\bar{X}\}=0.$$

$$E\{\bar{X}|X_1-\bar{X}+Y_1-\bar{Y},\dots ,X_n-\bar{X}+Y_n-\bar{Y}\}=0.$$

$$E\{a\bar{X}+b\bar{Y}|X_1-\bar{X}+Y_1-\bar{Y},\dots ,X_n-\bar{X}+Y_n-\bar{Y}\}=\mathrm{const}$$

$$a\sum _{j=1}^n{f}_1^{\prime }({\tau }_j)\prod _{k\ne j}{f}_1({\tau }_k)\prod _1^n{f}_2({\tau }_j)+b\sum _{j=1}^n{f}_2^{\prime }({\tau }_j)\prod _{k\ne j}{f}_2({\tau }_k)\prod _1^n{f}_1({\tau }_j)=0$$

$$\sum _1^{n}g({\tau }_i)=0\mathrm{if}\sum _1^n{\tau }_i=0,|{\tau }_i|<ϵ$$

$$\{f_1(t){\}}^a\{f_2(t){\}}^b=\exp c{t}^2/2,|t|<ϵ.$$

$$\prod _1^m\{f_i(t){\}}^{{\alpha }_i}=\exp c{t}^2,|t|<ϵ$$

$$E\{\bar{X}|X_1-\bar{X}+Y_1-\bar{Y},\dots ,X_n-\bar{X}+Y_n-\bar{Y}\}=$$

$$E\{\bar{Y}|X_1-\bar{X}+Y_1-\bar{Y},\dots ,X_n-\bar{X}+Y_n-\bar{Y}\}$$

$$\sum _1^{n}g({\tau }_i)=0\mathrm{if}\sum _1^n{\tau }_i=0.$$

$$f_1(t)=f_2(t)\exp c{t}^2/2$$

$$E(s_X^2|\bar{X})=\mathrm{const}$$

$$E(s_X^2|\bar{X}+\bar{Y})=E(s_Y^2|\bar{X}+\bar{Y})+c$$

$$[f_1^{\prime \prime }(t)(f_1(t){)}^{n-1}-(f_1^{\prime }(t){)}^2(f_1(t){)}^{n-2}]/(f_1(t){)}^2$$

$$=[f_2^{\prime \prime }(t)(f_2(t){)}^{n-1}-(f_2^{\prime }(t){)}^2(f_2(t){)}^{n-2}]/(f_2(t){)}^2-c$$

$$\{\log f_1(t){\}}^{\prime \prime }=\{\log f_2(t){\}}^{\prime \prime }-c$$

Full text

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with $a,b$ some constants. It is proved that for $n\geq 3$ and $ab>0$ the relation holds iff $X_{i}$ and $Y_{j}$ are Gaussian.

A new characterization arises in case of $a=1,b=-1$. In this case either $X_{i}$ or $Y_{j}$ or both have a Gaussian component. It is the first (at least known to the author) case when presence of a Gaussian component is a characteristic property.

1 Introduction and an ancillary lemma

Let $X_{1},\ldots,X_{n}$ be independent identically distributed (iid) random variables with $E(|X_{i}|)<\infty,\>\bar{X}=(X_{1}+\ldots+X_{n})/n$. The KLR-theorem (see Kagan et al.(1965), also Kagan it et al.(1973), Ch. 5) claims that for $n\geq 3$ the relation

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is characteristic of Gaussian $X_{i}$.

The following lemma is a version of the KLR-theorem.

Lemma 1

Let $X_{1},\ldots,X_{n},Y_{1},\ldots,Y_{n}$ be independent with $X_{1},\ldots,X_{n}$ id with $E(|X_{i}|)<\infty$ and $Y_{1},\ldots,Y_{n}$ id. Then for $n\geq 3$ the relation

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holds iff $X_{i}$ is Gaussian.

Proof of Lemma. Replacing $X_{i}$ with $X_{i}-E(X_{i})$ the constant in (2) may be made zero. On setting $f_{1}(t)=E\{\exp itX_{j}\},\>f_{2}(t)=E\{\exp itY_{j}\}$, multiplying both parts of (2) by $\exp\sum_{1}^{n}t_{j}(X_{j}-\bar{X}+Y_{j}-\bar{Y})$ and taking the expectation results in

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where $\bar{t}=(t_{1}+\ldots+t_{n})/n$.

On setting $\tau_{j}=t_{j}-\bar{t}$ and dividing both side of (3) by $\prod_{1}^{n}f_{1}(\tau_{j})f_{2}(\tau_{j})$ for $|\tau_{j}|<\epsilon,\>\epsilon>0$ small enough where $f_{1}(\tau_{j})f_{2}(\tau_{j})\neq 0$ leads to

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This is the Cauchy classical functional equation whose solutions are linear functions, $f^{\prime}_{1}(t)/f_{1}(t)=ct$ for some $c$ and $|t|<\epsilon$ whence

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Since $f_{1}(t)$ is a characteristic function, $c<0$ and (5) holds for all $t$ proving the necessity part.

As to the sufficiency part, if $X_{i}$ is Gaussian and $Y_{i}$ arbitrary, due to independence of $(\bar{X},X_{1}-\bar{X},\ldots,X_{n}-\bar{X})$ and $(Y_{1}-\bar{Y},\ldots,Y_{n}-\bar{Y})$,one has

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Since the $\sigma-$algebra $\sigma(X_{1}-\bar{X}+Y_{1}-\bar{Y},\ldots,X_{n}-\bar{X}+Y_{n}-\bar{Y})$ is a subalgebra of $\sigma(X_{1}-\bar{X},Y_{1}-\bar{Y},\ldots,X_{n}-\bar{X},Y_{n}-\bar{Y})$, one has due to (6)

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2 The main result

Here we are studying the relation

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with the $a,b$ some constants and show that when $ab>0$ and $n\geq 3$ it is characteristic for Gaussian $X_{i}$ and $Y_{i}$ while in case of $ab<0$ a new phenomenon arises.

Theorem 1

*Let $X_{1},\ldots,X_{n},Y_{1},\ldots,Y_{n}$ be as in Lemma with

$E\{(X_{i})^{2}\}<\infty,\\ E\{(Y_{i})^{2}\}<\infty$.

For $n\geq 3$ and $ab>0$ the relation (8) holds iff $X_{i}$ and $Y_{i}$ are Gaussian.*

Proof of Theorem 1. As in Lemma, one may assume without loss in generality $E(X_{i})=E(Y_{i})=0$. Also one may always assume $a>0,b>0$.

Proceeding as in the proof of Lemma, one gets from (8) a functional equation for the characteristic functions $f_{1}(t),\>f_{2}(t)$ of $X_{i}$ and $Y_{i}$:

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holding for the $\tau_{1},\ldots,\tau_{n}$ with $\sum_{1}^{n}\tau_{i}=0,\>|\tau_{i}|<\epsilon$ with $\epsilon>0$ as in Lemma.

Dividing both sides of (9) by $\prod f_{1}(\tau_{i})f_{2}(\tau_{i})$ and setting $g(\tau)=af^{\prime}_{1}(\tau)/f_{1}(\tau)+bf^{\prime}_{2}(\tau)/f_{2}(\tau)$ results in the Cauchy functional equation

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whence for $g(t)=ct$ for some $c$ and $|t|<\epsilon$. From (10)

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From $g^{\prime}(0)=af^{\prime\prime}_{1}+bf^{\prime\prime}_{2}(0)=-aE\{(X_{i})^{2}\}-bE\{(Y_{i})^{2}\}=c$, one has $c<0$.

Lemma 2

Let $f_{1}t),\ldots,f_{m}(t)$ be characteristic functions and $\alpha_{1},\ldots,\alpha_{m}$ positive numbers. The relation

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holds for some $c<0,\epsilon>0$ iff $f_{1}(t),\ldots,f_{m}(t)$ are characteristic functions of Gaussian distributions.

For a proof of Lemma 2 generalizing Cramer classical theorem on the components of a Gaussian distribution see Zinger and Linnik (1955) (see also Kagan et al.(1973, Ch. 1).

Applying Lemma 2 to (11) completes the proof of Theorem 1.

Turn now to the case when in (8) $a=1,b=-1$. Before stating the result, remind that a random variable $X$ is said having a Gaussian component if $X$ is equidistributed with $U+\xi$, denoted $X\cong U+\xi$, where $U$ and $\xi$ are independent and $\xi$ is (non-degenerate) Gaussian.

Theorem 2

*Let $X_{1},\ldots,X_{n},Y_{1},\ldots,Y_{n}$ be as in Theorem 1. Assume that the characteristic functions $f_{1}(t),\>f_{2}(t)$ of $X_{i}$ and $Y_{i}$ do not vanish.

The relation*

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holds for $n\geq 3$ iff either (i) $X_{i}$ and $Y_{i}$ have the same (arbitrary) distribution or (ii) $Y_{i}$ is arbitrary (with $f_{2}(t)\neq 0$) while $X_{i}\cong Y_{i}+\xi$ with $Y_{i}$ and Gaussian $\xi$ independent or (iii) $X_{i}$ is arbitrary while $Y_{i}\cong X_{i}+\eta$ with $X_{i}$ and Gaussian $\eta$ independent. In other words, unless $X_{i}$ and $Y_{i}$ are identically distributed, at least one of them has a Gaussian component.

Proof of Theorem 2. Proceeding as in proof of Theorem 1 and taking into account the condition $\prod_{1}^{n}f_{1}(\tau_{i})f_{2}(\tau_{i})\neq 0$ one gets the following Cauchy equation for $g(t)=f^{\prime}_{1}(t)/f_{1}(t)-f^{\prime}_{2}(t)/f_{2}$:

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From (14),

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with $c=f^{\prime\prime}_{1}(0)-f^{\prime\prime}_{2}(0)=E\{(Y_{i})^{2}\}-E\{(X_{i})^{2}\}$.

If $c=0$, is $X_{i}$ and $Y_{i}$ are identically distributed. If $c<0$, $X_{i}$ has a Gaussian component $\xi$ with variance $c$, $X_{i}\cong Y_{i}+\xi$ while $Y_{i}$ is arbitrary. If $c>0$, $Y_{i}$ has a Gaussian component $\eta$ with variance $-c$, $Y_{i}\cong X_{i}+\eta$ while $X_{i}$ is arbitrary. This proves Theorem 2.

In Kagan and Klebanov (2010) were described some analytic properties characterizing random variables with Gaussian components. It would be interesting to find a statistical property characteristic for distributions with Gaussian components. Here is such an example.

Set $s_{X}^{2}=(1/n-1)\sum_{1}^{n}(X_{i}-\bar{X})^{2},s_{Y}^{2}=(1/n-1)\sum_{1}^{n}(Y_{i}-\bar{Y})^{2}$. In Lukacs (1942) was shown that if $n\geq 2$ the relation

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holds iff $X_{i}$ is Gaussian. A version of (16) in spirit of (13) is

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where $c=var(X_{i})-var(Y_{i})$.

Theorem 3

Under the conditions of Theorem 2, the relation (17) holds for $n\geq 2$ if (i) in case of $c>0$ $X_{i}\cong Y_{i}+\xi$ where $\xi$ is independent of $Y_{i}$ Gaussian random variable with variance $c$, (ii) in case of $c<0$ $Y_{i}\cong X_{i}+\eta$ where $\eta$ is independent of $X_{i}$ Gaussian random variable with variance $-c$, (iii) in case of $c=0$ $X_{i}\cong Y_{i}$ (remind that we assumed $E(X_{i})=E(Y_{i})=0$).

Proof of Theorem 3. Multiplying both sides of (17) by $\exp\{it(\bar{X}+\bar{Y})\}$, taking the expectation and then dividing both sides by $(f_{1}(t))^{n}(f_{2}(t))^{n}$ results in a differential equation for the characteristic functions $f_{1}(t),\>f_{2}(t)$ of $X_{i}$ and $Y_{i}$:

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equivalent to

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whence $f_{1}(t)=f_{2}(t)\exp(-ct^{2}/2)$ proving Theorem 3.