When Sums of Elliptical Vectors Stay Elliptical

Why Closure Under Addition Requires Care

The sum of independent random variables from a familiar scalar family often stays in that family. For elliptical vectors, independence alone does not provide the same guarantee. The shape of each summand matters: their characteristic functions must combine in a way that still depends on a single quadratic form.

This distinction is useful in distribution theory and in modeling sums of random vectors. Gaussian vectors have an especially strong closure property, while other elliptical families typically need compatible shapes or additional structure.

Definition: A random vector $X$ in $\mathbb{R}^d$ is elliptical with location $\mu$, scatter matrix $\Sigma$, and characteristic generator $\phi$ if $\Sigma$ is positive semidefinite and its characteristic function has the form $$ \varphi_X(t)=E[\exp(i t^\mathsf{T}X)] =\exp(i t^\mathsf{T}\mu)\,\phi(t^\mathsf{T}\Sigma t),\qquad t\in\mathbb{R}^d, $$ where $\phi(0)=1$. The generator must be such that this expression is a characteristic function. The matrix $\Sigma$ need not be a covariance matrix; if it is singular, the distribution may be supported on a lower-dimensional subspace.

Let $X_1$ and $X_2$ be independent elliptical vectors in the same dimension, with locations $\mu_1,\mu_2$, scatter matrices $\Sigma_1,\Sigma_2$, and generators $\phi_1,\phi_2$. Independence gives the characteristic function of their sum directly.

\varphi_{X_1+X_2}(t) =\exp\!\big(i t^\mathsf{T}(\mu_1+\mu_2)\big) \phi_1(t^\mathsf{T}\Sigma_1t)\, \phi_2(t^\mathsf{T}\Sigma_2t).

The location adds automatically. The issue is whether the product of the two generator terms can be written as one function of $t^\mathsf{T}\Sigma t$ for a single positive semidefinite matrix $\Sigma$. This gives an exact criterion, though it does not always reduce to a simple rule about the two scatter matrices.

Criterion: The sum $X_1+X_2$ is elliptical if and only if there are a positive semidefinite matrix $\Sigma$ and a function $\psi$ such that, for every $t\in\mathbb{R}^d$, $$ \phi_1(t^\mathsf{T}\Sigma_1t)\phi_2(t^\mathsf{T}\Sigma_2t) =\psi(t^\mathsf{T}\Sigma t), $$ with the resulting expression a characteristic function. In that case $\mu_1+\mu_2$ is a location for the sum.

A General Closure Condition: Proportional Scatter

A useful sufficient condition is that both summands have scatter matrices proportional to the same matrix. This includes the spherical case, where both scatter matrices are proportional to the identity.

Theorem: Suppose $X_1$ and $X_2$ are independent elliptical vectors with scatter matrices $\Sigma_1=c_1\Sigma$ and $\Sigma_2=c_2\Sigma$, where $c_1,c_2\geq 0$ and $\Sigma$ is positive semidefinite. Then $X_1+X_2$ is elliptical with location $\mu_1+\mu_2$ and scatter matrix $\Sigma$.

Proof. Set $q=t^\mathsf{T}\Sigma t$, which is nonnegative. Then $t^\mathsf{T}\Sigma_jt=c_jq$ for $j=1,2$. The characteristic function of the sum is therefore

\varphi_{X_1+X_2}(t) =\exp\!\big(i t^\mathsf{T}(\mu_1+\mu_2)\big) \phi_1(c_1q)\phi_2(c_2q).

The product depends on $t$ only through $q=t^\mathsf{T}\Sigma t$. It is the characteristic function of the sum, so it is valid by the independence assumption. Thus it has the required elliptical form with scatter $\Sigma$ and generator $\psi(q)=\phi_1(c_1q)\phi_2(c_2q)$. $\square$

Worked Example: Adding Two Spherical Vectors

Let $X_1$ and $X_2$ be independent and uniform on the unit circle in $\mathbb{R}^2$. A uniform angle $\Theta$ gives a vector $(\cos\Theta,\sin\Theta)$. Its characteristic function is

E[\exp(i t^\mathsf{T}X_1)] =\frac{1}{2\pi}\int_0^{2\pi} \exp\!\big(i(t_1\cos\theta+t_2\sin\theta)\big)\,d\theta =J_0(\|t\|),

where $J_0$ is the order-zero Bessel function and $\|t\|=\sqrt{t_1^2+t_2^2}$. The same holds for $X_2$. Independence gives

\varphi_{X_1+X_2}(t)=J_0(\|t\|)^2.

This is a function of $t^\mathsf{T}t=\|t\|^2$ alone: define $\psi(s)=J_0(\sqrt{s})^2$ for $s\geq0$. The sum is therefore spherical, and hence elliptical, with location zero and scatter $I_2$. The two summands have the same shape even though their sum is not uniform on the circle.

Why Gaussian Sums Are More Flexible

The proportional-scatter condition is sufficient for general elliptical vectors, but Gaussian vectors allow different scatter matrices. For $X_j\sim N(\mu_j,\Sigma_j)$, the characteristic generator is $\phi_j(s)=\exp(-s/2)$. Consequently, the quadratic forms combine by addition rather than by requiring a common shape.

Theorem: If $X_1$ and $X_2$ are independent multivariate normal vectors with means $\mu_1,\mu_2$ and covariance matrices $\Sigma_1,\Sigma_2$, then $$ X_1+X_2\sim N(\mu_1+\mu_2,\Sigma_1+\Sigma_2). $$

Proof. The characteristic function of $X_j$ is $\exp(i t^\mathsf{T}\mu_j-\tfrac12t^\mathsf{T}\Sigma_jt)$. Multiplication, justified by independence, gives

\begin{aligned} \varphi_{X_1+X_2}(t) &=\exp\!\left(i t^\mathsf{T}\mu_1-\tfrac12t^\mathsf{T}\Sigma_1t\right) \exp\!\left(i t^\mathsf{T}\mu_2-\tfrac12t^\mathsf{T}\Sigma_2t\right)\\ &=\exp\!\left(i t^\mathsf{T}(\mu_1+\mu_2) -\tfrac12t^\mathsf{T}(\Sigma_1+\Sigma_2)t\right). \end{aligned}

This is the characteristic function of the stated multivariate normal distribution. Since a sum of positive semidefinite matrices is positive semidefinite, the covariance matrix is valid, including when it is singular. $\square$

Worked Example: Different Gaussian Shapes

Suppose $X_1$ and $X_2$ are independent, with $X_1\sim N((1,0)^\mathsf{T},\begin{pmatrix}1&0\\0&2\end{pmatrix})$ and $X_2\sim N((0,3)^\mathsf{T},\begin{pmatrix}4&1\\1&3\end{pmatrix})$. The second covariance matrix is positive definite: its leading diagonal entry is $4>0$ and its determinant is $4(3)-1^2=11>0$.

The means add to $(1,3)^\mathsf{T}$, and the covariance matrices add entry by entry:

\begin{pmatrix}1&0\\0&2\end{pmatrix} +\begin{pmatrix}4&1\\1&3\end{pmatrix} =\begin{pmatrix}5&1\\1&5\end{pmatrix}.

The result is $X_1+X_2\sim N((1,3)^\mathsf{T},\begin{pmatrix}5&1\\1&5\end{pmatrix})$. The covariance matrices are not proportional, but Gaussian characteristic functions still combine into the required single quadratic form.

A Stable-Family Exception

A second important closure rule applies within symmetric elliptically stable families. For $0<\alpha\leq2$, consider vectors whose characteristic functions have the form

\varphi_X(t)=\exp\!\left(i t^\mathsf{T}\mu -(t^\mathsf{T}\Sigma t)^{\alpha/2}\right).

If two independent vectors have this form with the same $\alpha$ and proportional scatter matrices $\Sigma_j=c_j\Sigma$, then the product of their characteristic functions has exponent $-\{c_1^{\alpha/2}+c_2^{\alpha/2}\}(t^\mathsf{T}\Sigma t)^{\alpha/2}$. Equivalently, the sum has the same stable form with scatter $(c_1^{\alpha/2}+c_2^{\alpha/2})^{2/\alpha}\Sigma$ and location $\mu_1+\mu_2$. The Gaussian case $\alpha=2$ has the broader closure rule just proved; for other values of $\alpha$, this proportional-shape condition is a useful guarantee, not a claim of closure for arbitrary shapes.

Worked Example: Adding Two Elliptically Stable Vectors

Let $X_1,X_2$ be independent symmetric elliptically stable vectors in $\mathbb{R}^2$, both with $\alpha=1$ and location zero. Suppose their characteristic functions are $\exp(-\sqrt{t^\mathsf{T}I_2t})$ and $\exp(-\sqrt{t^\mathsf{T}(4I_2)t})$. Their product is

\exp(-\|t\|)\exp(-2\|t\|) =\exp(-3\|t\|) =\exp\!\left(-\sqrt{t^\mathsf{T}(9I_2)t}\right).

Thus $X_1+X_2$ is in the same $\alpha=1$ family, with location zero and scatter $9I_2$. Here the scale matrices are proportional; the calculation does not establish closure for arbitrary scatter matrices.

A Non-Gaussian Sum That Is Not Elliptical

The general failure is not merely that a convenient generator is hard to find. Even independent elliptical summands with finite moments can produce a sum that cannot be elliptical. The following example uses a fourth-moment condition that every finite-fourth-moment elliptical vector must satisfy.

Necessary condition: If a centered elliptical vector $W$ has positive definite covariance $C$ and finite fourth moments, then for some scalar $\kappa$, its fourth cumulants satisfy $$ \operatorname{cum}(W_i,W_j,W_k,W_l) =\kappa(C_{ij}C_{kl}+C_{ik}C_{jl}+C_{il}C_{jk}). $$

To see why, write a centered elliptical vector as $W=AZ$, where $Z$ is spherical. Spherical symmetry and finite fourth moments imply $E[Z_iZ_jZ_kZ_l]=m(\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})$ for a scalar $m$; here $\delta_{ij}$ is 1 if $i=j$ and 0 otherwise. Also $\operatorname{Cov}(Z)=vI$ for some $v>0$. Subtracting the three covariance products in the definition of the fourth cumulant gives the same paired-index expression with coefficient $m-v^2$. Applying the linear map $A$ transforms each paired identity matrix into $AA^\mathsf{T}$. Since $C=vAA^\mathsf{T}$, the result has the stated form, with $\kappa=(m-v^2)/v^2$. This condition is necessary, not sufficient.

Worked Example: Independent Elliptical Summands With a Non-Elliptical Sum

Let $U=(\cos\Theta,\sin\Theta)^\mathsf{T}$, where $\Theta$ is uniform on $[0,2\pi)$, and let $G\sim N(0,\begin{pmatrix}1&0\\0&2\end{pmatrix})$, independent of $U$. The vector $U$ is spherical, so it is elliptical; $G$ is Gaussian and therefore elliptical. We show that $W=U+G$ is not elliptical.

By symmetry, $E[U]=0$, $\operatorname{Var}(U_1)=\operatorname{Var}(U_2)=1/2$, and $\operatorname{Cov}(U_1,U_2)=0$. Independence and addition of covariance matrices give

\operatorname{Cov}(W)= \begin{pmatrix}1/2&0\\0&1/2\end{pmatrix} +\begin{pmatrix}1&0\\0&2\end{pmatrix} =\begin{pmatrix}3/2&0\\0&5/2\end{pmatrix}.

For $U$, $E[U_1^4]=E[U_2^4]=3/8$ and $E[U_1^2U_2^2]=1/8$, as follows by integrating the corresponding powers of cosine and sine over a full period. Its fourth cumulants are therefore $\operatorname{cum}(U_1,U_1,U_1,U_1)=3/8-3(1/2)^2=-3/8$ and $\operatorname{cum}(U_1,U_1,U_2,U_2)=1/8-(1/2)(1/2)=-1/8$. A Gaussian vector has zero fourth cumulants, and cumulants of independent sums add, so these are also the corresponding fourth cumulants of $W$.

If $W$ were elliptical, the necessary condition would give, for its diagonal covariance matrix, $\operatorname{cum}(W_1,W_1,W_2,W_2)=\kappa(3/2)(5/2)$. Since this cumulant is $-1/8$, it would force $\kappa=-1/30$. The same condition would then require $\operatorname{cum}(W_1,W_1,W_1,W_1)=3\kappa(3/2)^2=-9/40$. But the actual value is $-3/8$, and $-9/40\ne-3/8$. This contradiction proves that $W$ is not elliptical.

Common Mistake and Practical Implications

Incorrect reasoning: “Both summands are elliptical, so their independent sum must be elliptical.” Independence only lets us multiply the characteristic functions. That product need not depend on a single quadratic form, as the counterexample demonstrates.

Correction: Check how the quadratic forms enter the product. Proportional scatter gives a general closure guarantee; Gaussian characteristic functions allow arbitrary covariance matrices; and symmetric elliptically stable laws have their own closure rule when the stability index and shape are compatible. Outside such conditions, elliptical closure needs to be established rather than assumed.

This matters when choosing a model for an aggregate, prediction error, or other sum of multivariate contributions. A Gaussian approximation may be appropriate in some settings, but the fact that each contribution is elliptical does not by itself justify an elliptical model for the total.

Check Your Understanding

Use the characteristic-function criterion and closure conditions to answer the following.

Write the characteristic function of a sum of two independent elliptical vectors and identify the part that determines whether the sum is elliptical.
Why do proportional scatter matrices guarantee an elliptical sum, even when the two generators differ?
What feature of the Gaussian characteristic function permits closure when the covariance matrices are not proportional?
For two symmetric elliptically stable vectors with common index $\alpha$ and scatters $c_1\Sigma$ and $c_2\Sigma$, what scalar multiplies $\Sigma$ in the sum's scatter matrix?
In the non-elliptical counterexample, which two fourth cumulants contradict the necessary elliptical condition?