This article is accepted for publishing in

Journal of Mathematical Analysis and Applications

On a characterization theorem in the space $\mathbb{R}^{n}$

G.M. Feldman

By Heyde’s theorem, the class of Gaussian distributions on the real line is
characterized by the symmetry of the conditional distribution of one linear
form of independent random variables given another. We prove an analogue of
this theorem for two independent random vectors taking values in the space
$\mathbb{R}^{n}$. The obtained class of distributions consists of convolutions
of Gaussian distributions and a distribution supported in a subspace, which is
determined by coefficients of the linear forms.

Key words and phrases: Characterization theorem, functional equation, linear
operator

Mathematical Subject Classification: 60E10, 62E10, 42B10, 39A10

##  1  Introduction

By the well-known Skitovich–Darmois theorem, Gaussian distributions on the
real are characterized by the independence of two linear forms with nonzero
coefficients of independent random variables. This result was generalized by
S.G. Ghurye and I. Olkin to independent random vectors taking values in the
space $\mathbb{R}^{n}$. Coefficients of the linear forms in this case are
invertible linear operators in $\mathbb{R}^{n}$ ([14], see also [16, §3.2]).
A theorem similar to the Skitovich–Darmois theorem was proved by C.C. Heyde.
By Heyde’s theorem, Gaussian distributions on the real line are characterized
by the symmetry of the conditional distribution of one linear form given
another. Coefficients of the linear forms are nonzero real numbers ([15],
see also [16, §13.4]). For two independent random variables Heyde’s theorem
states the following. Let $\xi_{1}$ and $\xi_{2}$ be independent random
variables with distributions $\mu_{1}$ and $\mu_{2}$. Let $a_{j}$ and $b_{j}$
be nonzero numbers such that $b_{1}a_{1}^{-1}+b_{2}a_{2}^{-1}\neq 0$. If the
conditional distribution of linear form $L_{2}=b_{1}\xi_{1}+b_{2}\xi_{2}$
given $L_{1}=a_{1}\xi_{1}+a_{2}\xi_{2}$ is symmetric, then $\mu_{j}$ are
Gaussian distributions. It is easy to see that in studying the possible
distributions $\mu_{j}$ we can assume without loss of generality that
$L_{1}=\xi_{1}+\xi_{2}$, $L_{2}=\xi_{1}+a\xi_{2}$, where $a\neq 0$. Hence the
following statement holds.

######  Theorem A

Let $\xi_{1}$ and $\xi_{2}$ be independent random variables with distributions
$\mu_{1}$ and $\mu_{2}$. Let $a\neq 0$ and $a\neq-1$. If the conditional
distribution of the linear form $L_{2}=\xi_{1}+a\xi_{2}$ given
$L_{1}=\xi_{1}+\xi_{2}$ is symmetric, then $\mu_{j}$ are Gaussian
distributions.

A number of works have been devoted to analogues of Heyde’s theorem for
various locally compact Abelian groups (see e.g. [5, 17, 6, 8, 18, 9, 11, 19,
20, 12, 13, 10]). In so doing coefficients of the linear forms are topological
automorphisms of the group. At the same time Heyde’s theorem for the space
$\mathbb{R}^{n}$ was not specially studied. The space $\mathbb{R}^{n}$ was
considered only as a special case of a locally compact Abelian group. In
particular, the following theorem was proved in [11] for arbitrary locally
compact Abelian groups containing no nonzero elements of finite order. For the
space $\mathbb{R}^{n}$ this theorem can be formulated as follows.

######  Theorem B

Let $\alpha$ be an invertible linear operator in the space $\mathbb{R}^{n}$
such that $I+\alpha$ is also an invertible operator. Let $\xi_{1}$ and
$\xi_{2}$ be independent random vectors with values in $\mathbb{R}^{n}$ and
distributions $\mu_{1}$ and $\mu_{2}$. If the conditional distribution of the
linear form $L_{1}=\xi_{1}+\xi_{2}$ given $L_{2}=\xi_{1}+\alpha\xi_{2}$ is
symmetric, then $\mu_{j}$ are Gaussian distributions.

Obviously, Theorem A is a particular case of Theorem B for $n=1$. The purpose
of this note is to prove Heyde’s theorem for the space $\mathbb{R}^{n}$
without any restrictions on an invertible linear operator $\alpha$. In other
words, we want to get a full description of the possible distributions
$\mu_{j}$ in Theorem B in the case when the operator $I+\alpha$ need not be
invertible. It turns out that if the operator $I+\alpha$ is not invertible,
then $\mu_{j}$ are convolutions of Gaussian distributions and a distribution
supported in a subspace which is determined by the operator $\alpha$.

Denote by $x=(x_{1},x_{2},\dots,x_{n})$, $x_{j}\in\mathbb{R}$, elements of the
space $\mathbb{R}^{n}$. If $x,y\in\mathbb{R}^{n}$, then put

$\langle x,y\rangle=\sum_{j=1}^{n}x_{j}y_{j}.$

Denote by $\mathbb{C}$ the complex plane. We will also use the notation
$\langle x,y\rangle$ in the case when $x,y\in\mathbb{C}^{n}$. Let $H$ be a
subspace of $\mathbb{R}^{n}$. Denote by

$A(\mathbb{R}^{n},H)=\\{x\in\mathbb{R}^{n}:\langle x,y\rangle=0\ \mbox{for
all}\ y\in H\\}$

the annihilator of $H$.

Let $\alpha$ be a linear operator in $\mathbb{R}^{n}$. Denote by
$\widetilde{\alpha}$ its adjoint operator. If a subspace $G$ of
$\mathbb{R}^{n}$ is invariant with respect to $\alpha$, then denote by
$\alpha{{}_{G}}$ the restriction of $\alpha$ to $G$. Denote by $I$ the
identity operator. Denote by $\|x\|$ a norm of a vector $x\in\mathbb{R}^{n}$
and by $\|\alpha\|$ the norm of the operator $\alpha$.

Let $P(y)$ be an arbitrary function on $\mathbb{R}^{n}$ and let
$h\in\mathbb{R}^{n}$. Denote by $\Delta_{h}$ the finite difference operator

$\Delta_{h}P(y)=P(y+h)-P(y),\quad y\in\mathbb{R}^{n}.$

Recall that a continuous function $P(y)$ is a polynomial in some neighbourhood
of zero in $\mathbb{R}^{n}$ if and only if for a nonnegative integer $m$ the
function $P(y)$ satisfies the equation

$\Delta_{h}^{m+1}P(y)=0,$ (1)

for all $y$ and $h$ in a neighborhood of zero in $\mathbb{R}^{n}$. Moreover,
the minimum $m$ at which (1) is satisfied coincides with the degree of the
poynomial $P(y)$.

Let $\mu$ be a probability distribution on $\mathbb{R}^{n}$. Denote by
$\hat{\mu}(y)$, $y\in\mathbb{R}^{n}$, the characteristic function of $\mu$.
Define the distribution $\bar{\mu}$ by the formula $\bar{\mu}(B)=\mu(-B)$ for
each Borel subset $B$ in $\mathbb{R}^{n}$. Then
$\hat{\bar{\mu}}(y)=\overline{\hat{\mu}(y)}$. Denote by $E_{x}$ the degenerate
distribution concentrated at a vector $x\in\mathbb{R}^{n}$.

##  2  Main theorem

The main result of this paper is the proof of the following theorem.

######  Theorem 2.1

Let $\alpha$ be an invertible linear operator in the space $\mathbb{R}^{n}$.
Put $K={\operatorname{Ker}}~{}(I+\alpha)$. Let $\xi_{1}$ and $\xi_{2}$ be
independent random vectors with values in $\mathbb{R}^{n}$ and distributions
$\mu_{1}$ and $\mu_{2}$. Assume that the conditional distribution of the
linear form $L_{2}=\xi_{1}+\alpha\xi_{2}$ given $L_{1}=\xi_{1}+\xi_{2}$ is
symmetric. Then there exists an $\alpha$-invariant subspace $G$ such that
$\mu_{j}$ are shifts of convolutions of symmetric Gaussian distributions
supported in $G$ and a distribution supported in $K$. Moreover, $K\cap
G=\\{0\\}$.

It is clear that if in Theorem 2.1 $I+\alpha$ is an invertible operator, i.e.
$K=\\{0\\}$, then $\mu_{j}$ are Gaussian distributions. Thus, Theorem B
follows from Theorem 2.1.

To prove Theorem 2.1 we need a series of lemmas. The following lemma holds for
independent random variables taking values in an arbitrary locally compact
Abelian group. We formulate it for the space $\mathbb{R}^{n}$.

######  Lemma 2.2

([7, Lemma 16.1]) Let $\alpha$ be an invertible linear operator in the space
$\mathbb{R}^{n}$. Let $\xi_{1}$ and $\xi_{2}$ be independent random vectors
with values in $\mathbb{R}^{n}$ and distributions $\mu_{1}$ and $\mu_{2}$. The
conditional distribution of the linear form $L_{2}=\xi_{1}+\alpha\xi_{2}$
given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric if and only if the characteristic
functions $\hat{\mu}_{j}(y)$ satisfy the equation

$\hat{\mu}_{1}(u+v)\hat{\mu}_{2}(u+\widetilde{\alpha}v)=\hat{\mu}_{1}(u-v)\hat{\mu}_{2}(u-\widetilde{\alpha}v),\quad
u,v\in\mathbb{R}^{n}.$ (2)

Equation (2) is called the Heyde functional equation. Due to Lemma 2.2, the
proof of Theorem 2.1 reduces to the description of solutions of equation (2)
in the class of continuous normalized positive definite functions. Note that
the proof of the Ghurye–Olkin theorem, mentioned in the introduction, for two
independent random vectors reduces to solving the Skitovich–Darmois functional
equation

$\hat{\mu}_{1}(u+v)\hat{\mu}_{2}(u+\widetilde{\alpha}v)=\hat{\mu}_{1}(u)\hat{\mu}_{2}(u)\hat{\mu}_{1}(v)\hat{\mu}_{2}(\widetilde{\alpha}v),\quad
u,v\in\mathbb{R}^{n},$ (3)

in the class of continuous normalized positive definite functions. Unlike
equation (2), all solutions of equation (3) in the class of continuous
normalized positive definite functions are characteristic functions of
Gaussian distributions in the space $\mathbb{R}^{n}$.

Note that based on characterization of polynomials as the solutions set of
some functional equations, J. M. Almira in [2] proposed a new approach to
solving the Skitovich–Darmois functional equation in the space
$\mathbb{R}^{n}$ for $m\geq 2$ functions. Then, using the fact that Aichingers
equation characterizes polynomial functions (see [1] by E. Aichinger and J.
Moosbauer), J. M. Almira in [3] studied the solutions of the Skitovich–Darmois
functional equation on an arbitrary Abelian group.

We will need the following easily verified statement, which we formulate as a
lemma (see e.g. [4, Lemma 6.9]).

######  Lemma 2.3

Let $\mathbb{R}^{n}=\mathbb{R}^{p}\times\mathbb{R}^{q}$ and let $\mu$ be a
distribution on the space $\mathbb{R}^{n}$ with the characteristic function
$\hat{\mu}(s_{1},s_{2})$, $s_{1}\in\mathbb{R}^{p}$, $s_{2}\in\mathbb{R}^{q}$.
Assume that the function $\hat{\mu}(0,s_{2})$, $s_{2}\in\mathbb{R}^{q}$, is
extended to $\mathbb{C}^{q}$ as an entire function in $s_{2}$. Put
$B_{r}=\\{s_{2}=(s_{21},s_{22},\dots,s_{2q})\in\mathbb{C}^{q}:|s_{2j}|\leq r,\
j=1,2,\dots,q\\}$. Then for each fixed $s_{1}\in\mathbb{R}^{p}$ the function
$\hat{\mu}(s_{1},s_{2})$, $s_{2}\in\mathbb{R}^{q}$, is also extended to
$\mathbb{C}^{q}$ as an entire function in $s_{2}$, and for each
$s_{1}\in\mathbb{R}^{p}$ the inequality

$\max_{s_{2}\in B_{r}}|\hat{\mu}(s_{1},s_{2})|\leq\max_{s_{2}\in
B_{r}}|\hat{\mu}(0,s_{2})|$ (4)

holds.

The following lemma holds for independent random variables taking values in an
arbitrary locally compact Abelian group containing no elements of order 2. It
follows directly from Lemma 2.2. We formulate it for the space
$\mathbb{R}^{n}$.

######  Lemma 2.4

Let $\xi_{1}$ and $\xi_{2}$ be independent random vectors with values in the
space $\mathbb{R}^{n}$ and distributions $\mu_{1}$ and $\mu_{2}$. The
conditional distribution of the linear form $L_{2}=\xi_{1}-\xi_{2}$ given
$L_{1}=\xi_{1}+\xi_{2}$ is symmetric if and only if $\mu_{1}=\mu_{2}$.

The following lemma is crucial in the proof of Theorem 2.1. It describes the
possible distributions $\mu_{j}$ in Theorem B in the case when in a suitable
basis a Jordan cell with the eigenvalue $\lambda=-1$ corresponds to a linear
operator $\alpha$, i.e.

$\alpha=\alpha_{n}=\left(\begin{array}[]{ccccc}-1&1&0&\ldots&0\\\
0&-1&1&\ldots&0\\\ \ldots&\ldots&\ldots&\ldots&\ldots\\\ 0&0&0&\ldots&1\\\
0&0&0&\ldots&-1\end{array}\right).$

######  Lemma 2.5

Let $\alpha_{n}$ be the invertible linear operator in the space
$\mathbb{R}^{n}$, $n\geq 2$, of the form

$\alpha_{n}(x_{1},x_{2},\dots,x_{n})=(-x_{1}+x_{2},-x_{2}+x_{3},\dots,-x_{n-1}+x_{n},-x_{n}),\quad
x_{j}\in\mathbb{R}.$ (5)

Let $\xi_{1}$ and $\xi_{2}$ be independent random vectors with values in
$\mathbb{R}^{n}$ and distributions $\mu_{1}$ and $\mu_{2}$. Assume that the
conditional distribution of the linear form $L_{2}=\xi_{1}+\alpha_{n}\xi_{2}$
given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric. Put
$K={\operatorname{Ker}}~{}(I+\alpha_{n})$. Then we can replace the
distributions $\mu_{j}$ by their shifts $\tau_{j}$ in such a way that
$\tau_{1}=\tau_{2}$, the distribution $\tau_{j}$ is supported in the subspace
$K$, and if $\eta_{j}$ are independent identically distributed random vectors
with values in the space $\mathbb{R}^{n}$ and distribution $\tau_{j}$, then
the conditional distribution of the linear form
$M_{2}=\eta_{1}+\alpha_{n}\eta_{2}$ given $M_{1}=\eta_{1}+\eta_{2}$ is
symmetric.

Proof. Note that
$K=\\{(x_{1},0,\dots,0)\in\mathbb{R}^{n}:x_{1}\in\mathbb{R}\\}$. We divide the
proof of the lemma into three steps.

1\. By Lemma 2.2, the characteristic functions $\hat{\mu}_{j}(y)$ satisfy
equation (2). Put $\nu_{j}=\mu_{j}*\bar{\mu}_{j}$. Then
$\hat{\nu}_{j}(y)=|\hat{\mu}_{j}(y)|^{2}\geq 0$. Since $\hat{\nu}_{j}(0)=1$,
we can choose $\varepsilon>0$ in such a way that the inequalities
$\hat{\nu}_{j}(y)>0$, $j=1,2$, are valid for all $\|y\|<\varepsilon$.
Obviously, the characteristic functions $\hat{\nu}_{j}(y)$ also satisfy
equation (2). Put $P_{j}(y)=\log\hat{\nu}_{j}(y)$, where $\|y\|<\varepsilon$,
$j=1,2$. Let $\|\alpha_{n}\|=M$. It is obvious that $M\geq 1$. Put
$\delta=\varepsilon/8M$ and

$H=(I+\widetilde{\alpha}_{n})(\mathbb{R}^{n}).$

We prove that $P_{j}(y)$ are polynomials of degree at most 2 in some
neighbourhood of zero in the subspace $H$.

It follows from equation (2) for the characteristic functions
$\hat{\nu}_{j}(y)$ that the functions $P_{j}(y)$ satisfy the equation

$P_{1}(u+v)+P_{2}(u+\widetilde{\alpha}_{n}v)-P_{1}(u-v)-P_{2}(u-\widetilde{\alpha}_{n}v)=0,\quad
u,v\in\mathbb{R}^{n},\ \|u\|<\delta,\ \|v\|<\delta.$ (6)

Equation (6) arises in the study of Heyde’s theorem on various locally compact
Abelian groups. To solve equation (6) we use the finite difference method.
This is a standard reasoning (see e.g. [9], [11]).

Take an arbitrary vector $k_{1}$ in the space $\mathbb{R}^{n}$ such that
$\|k_{1}\|<\delta$. Put $h_{1}=\widetilde{\alpha}_{n}k_{1}$. Replacing in (6)
$u$ by $u+h_{1}$ and $v$ by $v+k_{1}$ and subtracting equation (6) from the
obtained equation, we get

$\Delta_{l_{11}}P_{1}(u+v)+\Delta_{l_{12}}P_{2}(u+\widetilde{\alpha}_{n}v)-\Delta_{l_{13}}P_{1}(u-v)=0,\quad
u,v\in\mathbb{R}^{n},\ \|u\|<\delta,\ \|v\|<\delta,$ (7)

where $l_{11}=(I+\widetilde{\alpha}_{n})k_{1}$,
$l_{12}=2\widetilde{\alpha}_{n}k_{1}$,
$l_{13}=(\widetilde{\alpha}_{n}-I)k_{1}.$ Take an arbitrary vector $k_{2}$ in
the space $\mathbb{R}^{n}$ such that $\|k_{2}\|<\delta$. Put $h_{2}=k_{2}$.
Replacing in (7) $u$ by $u+h_{2}$ and $v$ by $v+k_{2}$ and subtracting
equation (7) from the obtained equation, we find

$\Delta_{l_{21}}\Delta_{l_{11}}P_{1}(u+v)+\Delta_{l_{22}}\Delta_{l_{12}}P_{2}(u+\widetilde{\alpha}_{n}v)=0,\quad
u,v\in\mathbb{R}^{n},\ \|u\|<\delta,\ \|v\|<\delta,$ (8)

where $l_{21}=2k_{2}$, $l_{22}=(I+\widetilde{\alpha}_{n})k_{2}.$ Take an
arbitrary vector $k_{3}$ in the space $\mathbb{R}^{n}$ such that
$\|k_{3}\|<\delta$. Put $h_{3}=-\widetilde{\alpha}_{n}k_{3}$. Replacing in (8)
$u$ by $u+h_{3}$ and $v$ by $v+k_{3}$ and subtracting equation (8) from the
obtained equation, we get

$\Delta_{l_{31}}\Delta_{l_{21}}\Delta_{l_{11}}P_{1}(u+v)=0,\ \
u,v\in\mathbb{R}^{n},\ \|u\|<\delta,\ \|v\|<\delta,$ (9)

where $l_{31}=(I-\widetilde{\alpha}_{n})k_{3}.$ Substituting $v=0$ into (9) we
find that

$\Delta_{l_{31}}\Delta_{l_{21}}\Delta_{l_{11}}P_{1}(u)=0,\quad
u\in\mathbb{R}^{n},\ \|u\|<\delta.$ (10)

Since multiplication by 2 and $I-\widetilde{\alpha}_{n}$ are invertible linear
operators in the space $\mathbb{R}^{n}$ and $k_{j}$ are arbitrary vectors in
$\mathbb{R}^{n}$ such that $\|k_{j}\|<\delta$, it follows from (10) that the
function $P_{1}(y)$ satisfies the equation

$\Delta_{h}^{3}P_{1}(y)=0,\ \ y,h\in H,\ \|y\|<\delta,\ \|h\|<\delta.$ (11)

Arguing similarly, we exclude the function $P_{1}(y)$ from equation (8) and
obtain that the function $P_{2}(y)$ also satisfies equation (11). It follows
from this that $P_{j}(y)$ are polynomials of degree at most 2 in some
neighbourhood of zero in the subspace $H$.

2\. Assume $n=2$ and prove that $\hat{\nu}_{1}(y)=\hat{\nu}_{2}(y)=1$ for all
$y\in H$. Since $\alpha_{2}(x_{1},x_{2})=(-x_{1}+x_{2},-x_{2})$,
$x_{j}\in\mathbb{R}$, we have
$\widetilde{\alpha}_{2}(y_{1},y_{2})=(-y_{1},y_{1}-y_{2})$,
$y_{j}\in\mathbb{R}$, and then
$H=\\{(0,y_{2})\in\mathbb{R}^{2}:y_{2}\in\mathbb{R}\\}$. Since $P_{j}(y)$ are
polynomials of the degree at most 2 in some neighbourhood of zero in the
subspace $H$ and $P_{j}(y)=\log\hat{\nu}_{j}(y)$ for all $\|y\|<\delta$, $y\in
H$, this implies that the characteristic functions $\hat{\nu}_{j}(0,y_{2})$
coincide with the characteristic functions of some symmetric Gaussian
distributions in some neighbourhood of zero on the real line. It is well known
that then $\hat{\nu}_{j}(0,y_{2})$, $y_{2}\in\mathbb{R}$, are the
characteristic functions of some symmetric Gaussian distributions ([16,
§1.2]). Thus, we have

$P_{j}(0,y_{2})=-\sigma_{j}y_{2}^{2},\quad y_{2}\in\mathbb{R},$ (12)

where $\sigma_{j}\geq 0$, $j=1,2$.

Since $\widetilde{\alpha}_{2}(y_{1},y_{2})=(-y_{1},y_{1}-y_{2})$,
$y_{j}\in\mathbb{R}$, equation (2) for the functions $\hat{\nu}_{j}(y)$ takes
the form

$\hat{\nu}_{1}(u_{1}+v_{1},u_{2}+v_{2})\hat{\nu}_{2}(u_{1}-v_{1},u_{2}+v_{1}-v_{2})\\\
=\hat{\nu}_{1}(u_{1}-v_{1},u_{2}-v_{2})\hat{\nu}_{2}(u_{1}+v_{1},u_{2}-v_{1}+v_{2}),\quad
u_{j},v_{j}\in\mathbb{R}.$ (13)

By Lemma 2.3, it follows from (12) that for each fixed $y_{1}\in\mathbb{R}$
the functions $\hat{\nu}_{j}(y_{1},y_{2})$ can be extended to the complex
plane $\mathbb{C}$ as entire functions in $y_{2}$. It is obvious that equation
(13) remains valid for all $u_{1},v_{1}\in\mathbb{R}$ and
$u_{2},v_{2}\in\mathbb{C}$.

We will verify that if $\hat{\nu}_{j}(y_{1},0)>0$ for all $|y_{1}|<\delta$,
$j=1,2$, then the functions $\hat{\nu}_{j}(y_{1},y_{2})$ do not vanish for all
$|y_{1}|<\delta$ and $y_{2}\in\mathbb{C}$. Assume the contrary, let
$\hat{\nu}_{1}(\widetilde{y}_{1},\widetilde{y}_{2})=0$, where
$|\widetilde{y}_{1}|<\delta$, $\widetilde{y}_{2}\in\mathbb{C}$. Take arbitrary
numbers $u_{1}\in\mathbb{R}$ and $u_{2}\in\mathbb{C}$. Put
$v_{1}=\widetilde{y}_{1}-u_{1}$, $v_{2}=\widetilde{y}_{2}-u_{2}$ and
substitute $u_{j}$ and $v_{j}$ into equation (13). Then the left-hand side of
equation (13) is equal to zero. Hence the equality

$\hat{\nu}_{1}(2u_{1}-\widetilde{y}_{1},2u_{2}-\widetilde{y}_{2})\hat{\nu}_{2}(\widetilde{y}_{1},\widetilde{y}_{2}-\widetilde{y}_{1}+u_{1})=0$
(14)

is satisfied for all $u_{j}$. Since $\hat{\nu}_{2}(\widetilde{y}_{1},y_{2})$
is not identically zero an entire function in $y_{2}$, take
$u_{1}=\widetilde{u}_{1}$ in such a way that
$\hat{\nu}_{2}(\widetilde{y}_{1},\widetilde{y}_{2}-\widetilde{y}_{1}+\widetilde{u}_{1})\neq
0$ and $|2\widetilde{u}_{1}-\widetilde{y}_{1}|<\delta$. Obviously, we can do
it. Then it follows from (14) that
$\hat{\nu}_{1}(2\widetilde{u}_{1}-\widetilde{y}_{1},2u_{2}-\widetilde{y}_{2})=0$
for all $u_{2}\in\mathbb{C}$, that contradicts the fact that
$\hat{\nu}_{1}(2\widetilde{u}_{1}-\widetilde{y}_{1},y_{2})$ is a not
identically zero entire function in $y_{2}$. Arguing similarly it is easy to
make sure that the function $\hat{\nu}_{2}(y_{1},y_{2})$ does not vanish for
all $|y_{1}|<\delta$, $y_{2}\in\mathbb{C}$.

Taking into account (12), by Lemma 2.3, it follows from inequality (4) that
$\hat{\nu}_{j}(y_{1},y_{2})$ are entire functions in $y_{2}$ of order at most
2 for all $|y_{1}|<\delta$. Since the functions $\hat{\nu}_{j}(y_{1},y_{2})$
for all $|y_{1}|<\delta$, $y_{2}\in\mathbb{C}$ do not vanish, Hadamard’s
theorem on the representation of entire functions of finite order implies that
the representation

$\hat{\nu}_{j}(y_{1},y_{2})=\exp\\{a_{j}(y_{1})y_{2}^{2}+b_{j}(y_{1})y_{2}+c_{j}(y_{1})\\},\quad|y_{1}|<\delta,\
y_{2}\in\mathbb{C},$ (15)

is valid, where $a_{j}(y_{1}),b_{j}(y_{1}),c_{j}(y_{1})\in\mathbb{C}$,
$j=1,2$. Substituting $u_{1}=v_{1}=0$, $u_{2}=v_{2}=y_{2}/2$ into (13), we get

$\hat{\nu}_{1}(0,y_{2})=\hat{\nu}_{2}(0,y_{2}),\ \ y_{2}\in\mathbb{R}.$ (16)

It follows from (12) and (16) that $\sigma_{1}=\sigma_{2}=\sigma$. Thus, (12)
and (15) imply that $a_{1}(0)=a_{2}(0)=-\sigma$, $b_{1}(0)=b_{2}(0)=0$.
Obviously, without loss of generality, we can assume that
$c_{1}(0)=c_{2}(0)=0$.

Substituting $u_{1}=v_{1}=y_{1}/2$ into (13) and taking into account (15), we
obtain

$a_{1}(y_{1})(u_{2}+v_{2})^{2}+b_{1}(y_{1})(u_{2}+v_{2})+c_{1}(y_{1})-\sigma(u_{2}+y_{1}/2-v_{2})^{2}\\\
=-\sigma(u_{2}-v_{2})^{2}+a_{2}(y_{1})(u_{2}-y_{1}/2+v_{2})^{2}+b_{2}(y_{1})(u_{2}-y_{1}/2+v_{2})\\\
+c_{2}(y_{1})+2\pi in(y_{1}),\ \ |y_{1}|<\delta,\ u_{2},v_{2}\in\mathbb{C},$
(17)

where the function $n(y_{1})$ takes integer values. Considering the left-hand
side and the right-hand side of (17) as polynomials in $u_{2}$ and $v_{2}$ and
equating in (17) the coefficients of $u_{2}$ and $v_{2}$, we get

$b_{1}(y_{1})-\sigma y_{1}=-a_{2}(y_{1})y_{1}+b_{2}(y_{1}),\quad
b_{1}(y_{1})+\sigma
y_{1}=-a_{2}(y_{1})y_{1}+b_{2}(y_{1}),\quad|y_{1}|<\delta.$

This implies that $\sigma=0$. Hence
$\hat{\nu}_{1}(0,y_{2})=\hat{\nu}_{2}(0,y_{2})=1$ for all
$y_{2}\in\mathbb{R}$.

3\. We will prove the lemma by induction. Let $n=2$. We have,

$H=(I+\widetilde{\alpha}_{2})(\mathbb{R}^{2})=\\{(0,y_{2})\in\mathbb{R}^{2}:y_{2}\in\mathbb{R}\\}.$

Substituting $u_{1}=v_{1}=0$, $u_{2}=v_{2}=y_{2}/2$ into equation (13) for the
functions $\hat{\mu}_{j}(y_{1},y_{2})$, we get that
$\hat{\mu}_{1}(0,y_{2})=\hat{\mu}_{2}(0,y_{2})$ for all $y_{2}\in\mathbb{R}$.
It follows from what was proved in step 2 that
$|\hat{\mu}_{1}(0,y_{2})|=|\hat{\mu}_{2}(0,y_{2})|=1$ for all
$y_{2}\in\mathbb{R}$. Hence there is a real number $x$ such that
$\hat{\mu}_{j}(0,y_{2})=\exp\\{ixy_{2}\\}$ for all $y_{2}\in\mathbb{R}$,
$j=1,2$. Put $t_{1}=(x,-x)$, $t_{2}=(0,-x)$. Consider the distributions
$\tau_{j}=\mu_{j}*E_{t_{j}}$. It is obvious that $\hat{\tau}_{j}(0,y_{2})=1$
for all $y_{2}\in\mathbb{R}$, $j=1,2$. It follows from this that the
distributions $\tau_{j}$ are supported in the annihilator
$A(\mathbb{R}^{2},H)$. It is obvious that $A(\mathbb{R}^{2},H)=K$. Since
$t_{1}+\alpha_{2}t_{2}=0$, the characteristic functions $\hat{\tau}_{j}(y)$
satisfy equation (2). By Lemma 2.2, this implies that if $\eta_{j}$ are
independent random vectors with values in the space $\mathbb{R}^{2}$ and
distributions $\tau_{j}$, then the conditional distribution of the linear form
$M_{2}=\eta_{1}+\alpha_{2}\eta_{2}$ given $M_{1}=\eta_{1}+\eta_{2}$ is
symmetric. Since $K={\operatorname{Ker}}~{}(I+\alpha_{2})$, the restriction of
the operator $\alpha_{2}$ to the subspace $K$ coincides with $-I$. It means
that if we consider $\eta_{j}$ as independent random vectors with values in
$K$, then the conditional distribution of the linear form
$M_{2}=\eta_{1}-\eta_{2}$ given $M_{1}=\eta_{1}+\eta_{2}$ is symmetric. It
follows from Lemma 2.4, applying to the subspace $K$, that
$\tau_{1}=\tau_{2}$. Thus, when $n=2$ the lemma is proved.

Let $n>2$. We note that
$\widetilde{\alpha}_{n}(y_{1},y_{2},\dots,y_{n})=(-y_{1},y_{1}-y_{2},\dots,y_{n-1}-y_{n})$.
Hence equation (2) for the functions $\hat{\nu}_{j}(y)$ takes the form

$\hat{\nu}_{1}(u_{1}+v_{1},u_{2}+v_{2},\dots,u_{n}+v_{n})\hat{\nu}_{2}(u_{1}-v_{1},u_{2}+v_{1}-v_{2},\dots,u_{n}+v_{n-1}-v_{n})\\\
=\hat{\nu}_{1}(u_{1}-v_{1},u_{2}-v_{2},\dots,u_{n}-v_{n})\hat{\nu}_{2}(u_{1}+v_{1},u_{2}-v_{1}+v_{2},\dots,u_{n}-v_{n-1}+v_{n}),\
\ u_{j},v_{j}\in\mathbb{R}.$ (18)

Substituting $u_{1}=u_{2}=\dots=u_{n-2}=0$, $v_{1}=v_{2}=\dots=v_{n-2}=0$ into
(18) we obtain

$\hat{\nu}_{1}(0,\dots,0,u_{n-1}+v_{n-1},u_{n}+v_{n})\hat{\nu}_{2}(0,\dots,0,u_{n-1}-v_{n-1},u_{n}+v_{n-1}-v_{n})\\\
=\hat{\nu}_{1}(0,\dots,0,u_{n-1}-v_{n-1},u_{n}-v_{n})\hat{\nu}_{2}(0,\dots,0,u_{n-1}+v_{n-1},u_{n}-v_{n-1}+v_{n}),\
\ u_{j},v_{j}\in\mathbb{R}.$ (19)

We see that equation (19), up to the notation, coincides with equation (13).
Therefore, as proven in step 2, we have
$\hat{\nu}_{1}(0,\dots,0,y_{n})=\hat{\nu}_{2}(0,\dots,0,y_{n})=1$ for all
$y_{n}\in\mathbb{R}$. Hence
$|\hat{\mu}_{1}(0,\dots,0,y_{n})|=|\hat{\mu}_{2}(0,\dots,0,y_{n})|=1$ for all
$y_{n}\in\mathbb{R}$. Put
$L=\\{(0,\dots,0,y_{n})\in\mathbb{R}^{n}:y_{n}\in\mathbb{R}\\}$. Arguing as in
the case when $n=2$ we can replace the distributions $\mu_{j}$ by their shifts
$\tau_{j}$ in such a way that the distributions $\tau_{j}$ are supported in
the annihilator $A(\mathbb{R}^{n},L)=\mathbb{R}^{n-1}$. Moreover, if
$\eta_{j}$ are independent random vectors with values in the space
$\mathbb{R}^{n}$ and distributions $\tau_{j}$, then the conditional
distribution of the linear form $M_{2}=\eta_{1}+\alpha_{n}\eta_{2}$ given
$M_{1}=\eta_{1}+\eta_{2}$ is symmetric. We note that
$\alpha_{n}(\mathbb{R}^{n-1})=\mathbb{R}^{n-1}$ and the restriction of the
operator $\alpha_{n}$ to the subspace $\mathbb{R}^{n-1}$ coincides with the
operator $\alpha_{n-1}$. It means that if we consider $\eta_{j}$ as
independent random vectors with values in $\mathbb{R}^{n-1}$, then the
conditional distribution of the linear form
$M_{2}=\eta_{1}+\alpha_{n-1}\eta_{2}$ given $M_{1}=\eta_{1}+\eta_{2}$ is
symmetric. The lemma is proved by induction.

The statement of the lemma can not be strengthened. Indeed, let $\omega$ be an
arbitrary distribution supported in the subspace
$K={\operatorname{Ker}}~{}(I+\alpha_{n})$. Let $t_{1}$ and $t_{2}$ be some
vectors in $\mathbb{R}^{n}$ such that $t_{1}+\alpha_{n}t_{2}=0$. Put
$\mu_{j}=\omega*E_{t_{j}}$, $j=1,2$. Taking into account that the restriction
of the operator $\alpha_{n}$ to the subspace $K$ coincides with $-I$, Lemma
2.2 and Lemma 2.4 imply that if $\xi_{1}$ and $\xi_{2}$ are independent random
vectors with values in the space $\mathbb{R}^{n}$ and distributions $\mu_{j}$,
then the conditional distribution of the linear form
$L_{2}=\xi_{1}+\alpha_{n}\xi_{2}$ given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric.
$\Box$

The following statement implies from Lemma 2.5.

######  Corollary 2.6

Let $\alpha$ be an invertible linear operator in the space $\mathbb{R}^{n}$.
Put $K={\operatorname{Ker}}~{}(I+\alpha)$ and suppose $K\neq\\{0\\}$. Let
$\xi_{1}$ and $\xi_{2}$ be independent random vectors with values in
$\mathbb{R}^{n}$ and distributions $\mu_{1}$ and $\mu_{2}$. Assume that the
conditional distribution of the linear form $L_{2}=\xi_{1}+\alpha\xi_{2}$
given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric. Then there exists an
$\alpha$-invariant subspace $G$ satisfying the condition $K\cap G=\\{0\\}$ and
such that some shifts $\tau_{j}$ of the distributions $\mu_{j}$ are supported
in the subspace $K\times G$. Moreover, if $\eta_{j}$ are independent random
vectors with values in the space $\mathbb{R}^{n}$ and distributions
$\tau_{j}$, then the conditional distribution of the linear form
$M_{2}=\eta_{1}+\alpha\eta_{2}$ given $M_{1}=\eta_{1}+\eta_{2}$ is symmetric,
and the restriction of the operator $\alpha$ to the subspace $K\times G$ is of
the form $(-I,\alpha_{G})$.

Proof. Represent the space $\mathbb{R}^{n}$ as a direct sum of two
$\alpha$-invariant subspaces $\mathbb{R}^{n}=F\times G$, where $F$ is the root
subspace corresponding to the eigenvalue $\lambda=-1$ of the operator
$\alpha$, and the operator $I+\alpha$ is invertible in the subspace $G$. The
operator $\alpha$ can be written in the form $\alpha=(\alpha_{F},\alpha_{G})$.
In order not to complicate the notation we assume that $F=\mathbb{R}^{p}$,
$G=\mathbb{R}^{q}$, a basis in the space $\mathbb{R}^{p}$ is chosen in such a
way that
$\mathbb{R}^{p}=\mathbb{R}^{n_{1}}\times\mathbb{R}^{n_{2}}\times\dots\times\mathbb{R}^{n_{k}}$,
and the operator $\alpha_{F}$ is of the form
$\alpha_{F}=(\alpha_{n_{1}},\alpha_{n_{2}},\dots,\alpha_{n_{k}})$, where
$\alpha_{n_{j}}=-I$, if $n_{j}=1$, and $\alpha_{n_{j}}$ is an invertible
linear operator in the space $\mathbb{R}^{n_{j}}$ which is given by formula
$(\ref{16.05.1})$, if $n_{j}\geq 2$. By Lemma 2.2, the characteristic
functions $\hat{\mu}_{j}(y)$ satisfy equation (2) which takes the form

$\hat{\mu}_{1}(u_{1}+v_{1},u_{2}+v_{2})\hat{\mu}_{2}(u_{1}+\widetilde{\alpha}_{F}v_{1},u_{2}+\widetilde{\alpha}_{G}v_{2})\\\
=\hat{\mu}_{1}(u_{1}-v_{1},u_{2}-v_{2})\hat{\mu}_{2}(u_{1}-\widetilde{\alpha}_{F}v_{1},u_{2}-\widetilde{\alpha}_{G}v_{2}),\quad
u_{1},v_{1}\in\mathbb{R}^{p},\ u_{2},v_{2}\in\mathbb{R}^{q}.$ (20)

Substitute $u_{2}=v_{2}=0$ into (20). Taking into account Lemma 2.2 and
applying successively Lemma 2.5 to each of the subspaces $\mathbb{R}^{n_{j}}$,
where $n_{j}\geq 2$, we find as a result from the obtained equation that there
exist vectors $t_{j}\in\mathbb{R}^{p}$ such that the distributions
$\tau_{j}=\mu_{j}*E_{t_{j}}$ are supported in the subspace $K\times G$.
Moreover, if $\eta_{j}$ are independent random vectors with values in $K\times
G$ and distributions $\tau_{j}$, then the conditional distribution of the
linear form $M_{2}=\eta_{1}+\alpha\eta_{2}$ given $M_{1}=\eta_{1}+\eta_{2}$ is
symmetric. Obviously, the restriction of the operator $\alpha$ to the subspace
$K\times G$ is of the form $(-I,\alpha_{G})$. $\Box$

######  Remark 2.7

Assume that under the conditions of Theorem 2.1 $K\neq\\{0\\}$, i.e.
$\lambda=-1$ is an eigenvalue of the operator $\alpha$. Assume also that the
root subspace corresponding to the eigenvalue $\lambda=-1$ does not coincide
with the eigenspace. Then, as was proved in Corollary 2.6, some shifts of the
distributions $\mu_{j}$ are supported in a proper subspace of the space
$\mathbb{R}^{n}$. Hence $\mu_{j}$ are singular distributions.

Proof of Theorem 2.1. By Theorem B, if $K=\\{0\\}$, i.e. $\lambda=-1$ is not
an eigenvalue of the operator $\alpha$, then $\mu_{j}$ are Gaussian
distributions. Therefore, we assume $K\neq\\{0\\}$, i.e. $\lambda=-1$ is an
eigenvalue of the operator $\alpha$. Let $F$ be the root subspace
corresponding to the eigenvalue $\lambda=-1$ of the operator $\alpha$.
Corollary 2.6 allows us to prove the theorem, assuming that $\alpha_{F}=-I$.
In other words, the root subspace corresponding to the eigenvalue $\lambda=-1$
of the operator $\alpha$ is the eigenspace. Then equation (20) takes the form

$\hat{\mu}_{1}(u_{1}+v_{1},u_{2}+v_{2})\hat{\mu}_{2}(u_{1}-v_{1},u_{2}+\widetilde{\alpha}_{G}v_{2})\\\
=\hat{\mu}_{1}(u_{1}-v_{1},u_{2}-v_{2})\hat{\mu}_{2}(u_{1}+v_{1},u_{2}-\widetilde{\alpha}_{G}v_{2}),\quad
u_{1},v_{1}\in\mathbb{R}^{p},\ u_{2},v_{2}\in\mathbb{R}^{q}.$ (21)

Substitute $u_{1}=v_{1}=0$ into (21). Taking into account Lemma 2.2, it
follows from Theorem B applying to the space $\mathbb{R}^{q}$ that
$\hat{\mu}_{j}(0,y_{2})$ are the characteristic functions of Gaussian
distributions, i.e.

$\hat{\mu}_{j}(0,y_{2})=\exp\\{-\langle A_{j}y_{2},y_{2}\rangle+i\langle
b_{j},y_{2}\rangle\\},\quad y_{2}\in\mathbb{R}^{q},$ (22)

where $A_{j}$ is a symmetric positive semidefinite $q\times q$ matrix,
$b_{j}\in\mathbb{R}^{q}$, $j=1,2$. Substituting $u_{2}=v_{2}=0$ into (21) and
taking into account (22) we obtain that $b_{1}+\alpha_{G}b_{2}=0$. Hence we
can replace the distributions $\mu_{j}$ by their shifts
$\tau_{j}=\mu_{j}*E_{-b_{j}}$ and suppose that $b_{1}=b_{2}=0$ in (22), i.e.

$\hat{\mu}_{j}(0,y_{2})=\exp\\{-\langle A_{j}y_{2},y_{2}\rangle\\},\quad
y_{2}\in\mathbb{R}^{q},\ j=1,2.$ (23)

By Lemma 2.3, it follows from (23) that for each fixed
$y_{1}\in\mathbb{R}^{p}$ the function $\hat{\mu}_{j}(y_{1},y_{2})$ can be
extended to $\mathbb{C}^{q}$ as an entire function in $y_{2}$. It is obvious
that equation (21) remains valid for all $u_{1},v_{1}\in\mathbb{R}^{p}$,
$u_{2},v_{2}\in\mathbb{C}^{q}$.

Substituting $u_{1}=v_{1}=y_{1}/2$, $u_{2}=v_{2}=0$ into (21) we see that
$\hat{\mu}_{1}(y_{1},0)=\hat{\mu}_{2}(y_{1},0)$ for all
$y_{1}\in\mathbb{R}^{p}$. Make sure that if for a fixed
$\widetilde{y}_{1}\in\mathbb{R}^{p}$ the inequalities

$\hat{\mu}_{j}(\widetilde{y}_{1},0)\neq 0,\quad j=1,2,$ (24)

hold, then $\hat{\mu}_{j}(\widetilde{y}_{1},y_{2})\neq 0$ for all
$y_{2}\in\mathbb{C}^{q}$, $j=1,2$. Suppose the contrary, let
$\hat{\mu}_{1}(\widetilde{y}_{1},\widetilde{y}_{2})=0$ for some
$\widetilde{y}_{2}\in\mathbb{C}^{q}$. Substitute
$u_{1}=v_{1}=\widetilde{y}_{1}/2$,
$u_{2}=(I+\widetilde{\alpha}_{G})^{-1}\widetilde{\alpha}_{G}\widetilde{y}_{2}$,
$v_{2}=(I+\widetilde{\alpha}_{G})^{-1}\widetilde{y}_{2}$ into (21). Then the
left-hand side of equation (21) is equal to zero. In view of (23) and (24),
the right-hand side of equation (21) is nonzero. If
$\hat{\mu}_{2}(\widetilde{y}_{1},\widetilde{y}_{2})=0$ for some
$\widetilde{y}_{2}\in\mathbb{C}^{q}$, then substituting
$u_{1}=-v_{1}=\widetilde{y}_{1}/2$,
$u_{2}=v_{2}=(I+\widetilde{\alpha}_{G})^{-1}\widetilde{y}_{2}$ into (21), we
get the contradiction, because the left-hand side of equation (21) is equal to
zero and the right-hand side is not. Thus, if inequalities (24) hold, then the
function $\hat{\mu}_{2}(\widetilde{y}_{1},y_{2})$ is also nonzero for all
$y_{2}\in\mathbb{C}^{q}$.

So, we proved that if $\hat{\mu}_{j}(y_{1},0)\neq 0$, $j=1,2$, for some
$y_{1}\in\mathbb{R}^{p}$, then the functions $\hat{\mu}_{j}(y_{1},y_{2})$ can
be extended to $\mathbb{C}^{q}$ as entire functions in $y_{2}$ without zeros.
Hence we have the representations

$\hat{\mu}_{j}(y_{1},y_{2})=\exp\\{Q_{j}(y_{1},y_{2})\\},\quad
y_{2}\in\mathbb{C}^{q},\ j=1,2,$

where $Q_{j}(y_{1},y_{2})$ are entire functions in $y_{2}$ in
$\mathbb{C}^{q}$.

By Lemma 2.3, it follows from inequality (4) and (23) that by Hadamard’s
theorem on the representation of entire functions of finite order, the
restriction of the functions $Q_{j}(y_{1},y_{2})$ to each complex plane in
$\mathbb{C}^{q}$ passing through zero are polynomials of degree at most 2.
Hence the functions $Q_{j}(y_{1},y_{2})$ are polynomials of degree at most 2
in $y_{2}$. Thus, we have a representation

$\hat{\mu}_{j}(y_{1},y_{2})=\exp\\{\langle
A_{j}(y_{1})y_{2},y_{2}\rangle+\langle
b_{j}(y_{1}),y_{2}\rangle+c_{j}(y_{1})\\},\quad y_{2}\in\mathbb{C}^{q},$ (25)

where $A_{j}(y_{1})$ are symmetric complex $q\times q$ matrices,
$b_{j}(y_{1})\in\mathbb{C}^{q}$, $c_{j}(y_{1})\in\mathbb{C}$.

Assume $\hat{\mu}_{j}(y_{1},0)\neq 0$, $j=1,2$. Substituting
$u_{1}=v_{1}=y_{1}/2$ into (21) and taking into account (25), we find from the
obtained equation

$\langle A_{1}(y_{1})(u_{2}+v_{2}),u_{2}+v_{2}\rangle+\langle
A_{2}(0)(u_{2}+\widetilde{\alpha}_{G}v_{2}),u_{2}+\widetilde{\alpha}_{G}v_{2}\rangle\\\
=\langle A_{1}(0)(u_{2}-v_{2}),u_{2}-v_{2}\rangle+\langle
A_{2}(y_{1})(u_{2}-\widetilde{\alpha}_{G}v_{2}),u_{2}-\widetilde{\alpha}_{G}v_{2}\rangle,\quad
u_{2},v_{2}\in\mathbb{C}^{q}.$ (26) $\langle
b_{1}(y_{1}),u_{2}+v_{2}\rangle+\langle
b_{2}(0),u_{2}+\widetilde{\alpha}_{G}v_{2}\rangle=\langle
b_{1}(0),u_{2}-v_{2}\rangle+\langle
b_{2}(y_{1}),u_{2}-\widetilde{\alpha}_{G}v_{2}\rangle,\quad
u_{2},v_{2}\in\mathbb{C}^{q}.$ (27)

The equalities

$A_{1}(y_{1})+A_{2}(0)=A_{1}(0)+A_{2}(y_{1})$ (28)

and

$A_{1}(y_{1})+\alpha_{G}A_{2}(0)\widetilde{\alpha}_{G}=A_{1}(0)+\alpha_{G}A_{2}(y_{1})\widetilde{\alpha}_{G}$
(29)

follow from (26). Substituting $u_{1}=y_{1}$, $v_{1}=0$ into (21) and taking
into account (25), we get from the obtained equation

$\langle A_{1}(y_{1})(u_{2}+v_{2}),u_{2}+v_{2}\rangle+\langle
A_{2}(y_{1})(u_{2}+\widetilde{\alpha}_{G}v_{2}),u_{2}+\widetilde{\alpha}_{G}v_{2}\rangle\\\
=\langle A_{1}(y_{1})(u_{2}-v_{2}),u_{2}-v_{2}\rangle+\langle
A_{2}(y_{1})(u_{2}-\widetilde{\alpha}_{G}v_{2}),u_{2}-\widetilde{\alpha}_{G}v_{2}\rangle,\quad
u_{2},v_{2}\in\mathbb{C}^{q}.$ (30)

Equation (30) implies that

$A_{1}(y_{1})+\alpha_{G}A_{2}(y_{1})=0.$ (31)

Taking into account that $I+\alpha_{G}$ is an invertible operator, (23), (29)
and (31) imply that $A_{1}(y_{1})=A_{1}(0)=-A_{1}$. Then it follows from (23)
and (28) that $A_{2}(y_{1})=A_{2}(0)=-A_{2}$. Thus, we have proved that if
$\hat{\mu}_{j}(y_{1},0)\neq 0$, then

$A_{1}(y_{1})=-A_{1},\quad A_{2}(y_{1})=-A_{2}.$ (32)

We find from (27) that

$b_{1}(y_{1})+b_{2}(0)=b_{1}(0)+b_{2}(y_{1}),\quad
b_{1}(y_{1})+\alpha_{G}b_{2}(0)=-b_{1}(0)-\alpha_{G}b_{2}(y_{1}).$ (33)

It follows from (23) and (25) that $b_{1}(0)=b_{2}(0)=0$. Given this, and
taking into account that $I+\alpha_{G}$ is an invertible operator, (33)
implies that if $\hat{\mu}_{j}(y_{1},0)\neq 0$, then
$b_{1}(y_{1})=b_{2}(y_{1})=0$ and hence in view of (25) and (32), the
representations

$\hat{\mu}_{j}(y_{1},y_{2})=\exp\\{-\langle
A_{j}y_{2},y_{2}\rangle+c_{j}(y_{1})\\},\quad y_{2}\in\mathbb{C}^{q},\quad
j=1,2,$ (34)

are valid for the functions $\hat{\mu}_{j}(y_{1},y_{2})$. Moreover, in this
case substituting $u_{1}=v_{1}=y_{1}/2$, $u_{2}=v_{2}=0$ in equation (21) and
taking into account (34), we obtain that

$\hat{\mu}_{1}(y_{1},0)=\hat{\mu}_{2}(y_{1},0)=\exp\\{c_{1}(y_{1})\\}=\exp\\{c_{2}(y_{1})\\},\
\ y_{1}\in\mathbb{R}^{p}.$ (35)

Assume $\hat{\mu}_{j}(\widetilde{y}_{1},0)=0$, $j=1,2$, for some
$\widetilde{y}_{1}\in\mathbb{R}^{p}$. We verify that then
$\hat{\mu}_{j}(\widetilde{y}_{1},y_{2})=0$ for all $y_{2}\in\mathbb{C}^{q}$,
$j=1,2$. Put $u_{1}=v_{1}=\widetilde{y}_{1}/2$, $u_{2}=-v_{2}=y_{2}$ in
equation (21). Then the left-hand side of equation (21) is equal to zero. In
view of (23), we have
$\hat{\mu}_{2}(\widetilde{y}_{1},(I+\widetilde{\alpha}_{G})y_{2})=0$ for all
$y_{2}\in\mathbb{C}^{q}$. Since $(I+\widetilde{\alpha}_{G})$ is an invertible
operator, we have $\hat{\mu}_{2}(\widetilde{y}_{1},y_{2})=0$ for all
$y_{2}\in\mathbb{C}^{q}$. Substituting $u_{1}=-v_{1}=\widetilde{y}_{1}/2$,
$u_{2}=\widetilde{\alpha}_{G}y_{2}$, $v_{2}=-y_{2}$ into equation (21), we
make sure that $\hat{\mu}_{1}(\widetilde{y}_{1},y_{2})=0$ for all
$y_{2}\in\mathbb{C}^{q}$.

Denote by $\gamma_{j}$ the symmetric Gaussian distribution in the space
$\mathbb{R}^{q}$ with the characteristic function

$\hat{\gamma}_{j}(y_{2})=\exp\\{-\langle A_{j}y_{2},y_{2}\rangle\\},\quad
y_{2}\in\mathbb{R}^{q},\ j=1,2.$ (36)

In view of $\hat{\mu}_{1}(y_{1},0)=\hat{\mu}_{2}(y_{1},0)$ for all
$y_{1}\in\mathbb{R}^{p}$, denote by $\omega$ a distribution in the space
$\mathbb{R}^{p}$ with the characteristic function

$\hat{\omega}(y_{1})=\hat{\mu}_{1}(y_{1},0)=\hat{\mu}_{2}(y_{1},0),\quad
y_{1}\in\mathbb{R}^{p}.$ (37)

Taking into account that $\hat{\omega}(y_{1})=0$ if and only if
$\hat{\mu}_{2}(y_{1},y_{2})=0$ for all $y_{2}\in\mathbb{R}^{q}$, (34)–(37)
imply the representations

$\hat{\mu}_{j}(y_{1},y_{2})=\exp\\{-\langle
A_{j}y_{2},y_{2}\rangle\\}\hat{\omega}(y_{1}),\quad y_{1}\in\mathbb{R}^{p},\
y_{2}\in\mathbb{R}^{q},\ j=1,2.$

This implies the statement of the theorem.

Note that we also proved that the intersection of each of the supports of the
symmetric Gaussian distributions $\gamma_{j}$ with the root subspace
corresponding to the eigenvalue $\lambda=-1$ of the operator $\alpha$ is equal
to zero.

The statement of the theorem can not be strengthened. Indeed, consider an
invertible linear operator $\alpha$ in the space $\mathbb{R}^{n}$. Let $G$ be
an $\alpha$-invariant subspace in $\mathbb{R}^{n}$ such that the operator
$I+\alpha$ is invertible in $G$. Let $G$ be isomorphic to $\mathbb{R}^{q}$.
Denote by $\gamma_{j}$ a Gaussian distribution in the space $\mathbb{R}^{q}$
with the characteristic function (36). Moreover, we will also assume that the
equality

$A_{1}+A_{2}\widetilde{\alpha}_{G}=0$ (38)

holds. Put $K={\operatorname{Ker}}~{}(I+\alpha)$. Let $\omega$ be a
distribution on $K$. Let $x_{1}$ and $x_{2}$ be some vectors in
$\mathbb{R}^{n}$ such that

$x_{1}+\alpha x_{2}=0.$ (39)

Put $\mu_{j}=\gamma_{j}*\omega*E_{x_{j}}$, $j=1,2$. Let $\xi_{1}$ and
$\xi_{2}$ be independent random vectors with values in $\mathbb{R}^{n}$ and
distributions $\mu_{j}$. It is easy to see that (38) implies that the
characteristic functions $\hat{\gamma}_{j}(y)$ satisfy equation (2). It
follows from Lemma 2.2 and Lemma 2.4, applying to the subspace $K$, that the
characteristic function $\hat{\omega}(y)$ satisfies equation (2). Moreover,
(39) implies that the functions $(x_{j},y)$ satisfy equation (2). Hence the
characteristic functions $\hat{\mu}_{j}(y)$ satisfy equation (2). By Lemma
2.2, the conditional distribution of the linear form
$L_{2}=\xi_{1}+\alpha\xi_{2}$ given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric.
$\Box$

We complement Theorem 2.1 by a more detailed description of possible
distributions $\mu_{j}$ in Theorem 2.1 for the space $\mathbb{R}^{2}$
depending on the spectrum of the linear operator $\alpha$. If $\lambda$ is an
eigenvalue of the operator $\alpha$, denote by $L_{\lambda}$ the corresponding
eigenspace. In particular, if
$K={\operatorname{Ker}}~{}(I+\alpha)\neq\\{0\\}$, then $K=L_{-1}$. We consider
two cases: $\lambda=-1$ is an eigenvalue of the operator $\alpha$ or not.

$1$. $\lambda=-1$ is an eigenvalue of the operator $\alpha$. Then the
characteristic equation of the operator $\alpha$ has only real roots. Two
cases are possible.

1A. The operator $\alpha$ has another eigenvalue $\lambda=\lambda_{0}$, where
$\lambda_{0}\neq-1$. In this case in a basis consisting of eigenvectors of
$\alpha$, the diagonal matrix $\alpha={\rm diag}\\{-1,\lambda_{0}\\}$
corresponds to the operator $\alpha$. Theorem 2.1 easily implies the following
alternative. If $\lambda_{0}>0$, then $\mu_{j}=\omega*E_{x_{j}}$, where
$\omega$ is a distribution supported in $K$, $j=1,2$. In so doing $K$ is a
one-dimensional subspace of $\mathbb{R}^{2}$. If $\lambda_{0}<0$, then
$\mu_{j}=\gamma_{j}*\omega$, where $\gamma_{j}$ are Gaussian distributions
supported in $L_{\lambda_{0}}$, and $\omega$ is a distribution supported in
$K$.

1B. The operator $\alpha$ has the only eigenvalue $\lambda=-1$. If the root
subspace corresponding to the eigenvalue $\lambda=-1$ does not coincide with
$K$, then the Jordan cell

$\alpha=\left(\begin{array}[]{cc}-1&1\\\ 0&-1\end{array}\right)$

in a suitable basis corresponds to the operator $\alpha$. In this case by
Lemma 2.5, $\mu_{j}=\omega*E_{x_{j}}$, where $\omega$ is a distribution
supported in $K$, $j=1,2$. In so doing $K$ is a one-dimensional subspace of
$\mathbb{R}^{2}$. If the root subspace corresponding to the eigenvalue
$\lambda=-1$ coincides with $K$, then $\alpha=-I$, and by Lemma 2.4,
$\mu_{1}=\mu_{2}=\mu$, where $\mu$ is an arbitrary distribution on
$\mathbb{R}^{2}$.

$2$. $\lambda=-1$ is not an eigenvalue of the operator $\alpha$. By Theorem B,
in this case $\mu_{j}$ are Gaussian distributions. Find out what can be said
about the supports of the Gaussian distributions $\mu_{j}$, in particular,
when $\mu_{j}$ are degenerate distributions.

The characteristic functions $\hat{\mu}_{j}(y_{1},y_{2})$ are represented in
the form

$\hat{\mu}_{j}(y_{1},y_{2})=\exp\\{-\langle
A_{j}(y_{1},y_{2}),(y_{1},y_{2})\rangle+i\langle
b_{j},(y_{1},y_{2})\rangle\\},\quad(y_{1},y_{2})\in\mathbb{R}^{2},$ (40)

where $A_{j}$ is a symmetric positive semidefinite $2\times 2$ matrix,
$b_{j}\in\mathbb{R}^{2}$, $j=1,2$. By Lemma 2.2, the characteristic functions
$\hat{\mu}_{j}(y_{1},y_{2})$ satisfy equation (2). Depending on the spectrum
of the operator $\alpha$, a matrix of a rather simple form in a suitable basis
corresponds to the operator $\alpha$. Substituting (40) into (2) we get that
the matrices $A_{j}$ satisfy the equation

$A_{1}+A_{2}\widetilde{\alpha}=0.$ (41)

The description of solutions of equation (41) in the class of symmetric
positive semidefinite $2\times 2$ matrices implies the description of the
supports of the Gaussian distributions $\mu_{j}$. Two cases are possible.

2A. The characteristic equation of the operator $\alpha$ has only real roots.

(i). The operator $\alpha$ has two different eigenvalues $\lambda=\lambda_{1}$
and $\lambda=\lambda_{2}$. In this case in a basis consisting of eigenvectors
of $\alpha$, the diagonal matrix $\alpha={\rm
diag}\\{\lambda_{1},\lambda_{2}\\}$ corresponds to the operator $\alpha$. Then
the following statements are valid. If $\lambda_{1}>0$ and $\lambda_{2}>0$,
then $\mu_{j}$ are degenerate distributions. If $\lambda_{1}\lambda_{2}<0$,
then $\mu_{j}=\gamma_{j}*E_{x_{j}}$, where $\gamma_{j}$ are symmetric Gaussian
distributions supported in $L_{\lambda_{j}}$, where $\lambda_{j}<0$. If
$\lambda_{1}<0$ and $\lambda_{2}<0$, then $\mu_{j}=\gamma_{j}*E_{x_{j}}$,
where $\gamma_{j}$ are symmetric Gaussian distributions. In so doing either
$\gamma_{j}$ are degenerate distributions concentrated at zero, or the
supports of $\gamma_{j}$ are one of subspaces $L_{\lambda_{j}}$, or the
supports of $\gamma_{j}$ are the space $\mathbb{R}^{2}$.

(ii). The operator $\alpha$ has the only eigenvalue $\lambda=\lambda_{0}$. If
the root space corresponding to the eigenvalue $\lambda=\lambda_{0}$ does not
coincide with $L_{\lambda_{0}}$, then the Jordan cell

$\alpha=\left(\begin{array}[]{cc}\lambda_{0}&1\\\
0&\lambda_{0}\end{array}\right)$

in a suitable basis corresponds to the operator $\alpha$. Then, if
$\lambda_{0}>0$, then $\mu_{j}$ are degenerate distributions. If
$\lambda_{0}<0$, then $\mu_{j}=\gamma_{j}*E_{x_{j}}$, where $\gamma_{j}$ are
symmetric Gaussian distributions supported in $L_{\lambda_{0}}$. If the root
space corresponding to the eigenvalue $\lambda=\lambda_{0}$ coincides with
$L_{\lambda_{0}}$, then $\alpha=\lambda_{0}I$. In this case, if
$\lambda_{0}>0$, then $\mu_{j}$ are degenerate distributions. If
$\lambda_{0}<0$, then $\mu_{j}=\gamma_{j}*E_{x_{j}}$, where $\gamma_{j}$ are
symmetric Gaussian distributions. In so doing either $\gamma_{j}$ are
degenerate distributions concentrated at zero, or the supports of $\gamma_{j}$
are the same one-dimensional subspace of $\mathbb{R}^{2}$, or the supports of
$\gamma_{j}$ are the space $\mathbb{R}^{2}$.

2B. The roots of the characteristic equation of the operator $\alpha$ are of
the form $\lambda_{1}=x_{0}+iy_{0}$, $\lambda_{1}=x_{0}-iy_{0}$, where
$y_{0}\neq 0$. Then the matrix

$\alpha=\left(\begin{array}[]{cc}x_{0}&y_{0}\\\
-y_{0}&x_{0}\end{array}\right)$

in a suitable basis corresponds to the operator $\alpha$. In this case
$\mu_{j}$ are degenerate distributions.

Acknowledgements

I thank the reviewer for getting my attention to articles [1]–[3].

This article was written during my stay at the Department of Mathematics
University of Toronto as a Visiting Professor. I am very grateful to Ilia
Binder for his invitation and support.

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