This is a follow-up of this question: : Compute $\phi_X(t)=E(e^{it^\top X})$ if $X\stackrel{d}{=}\mu + \xi AU$ with $AA^\top=\Sigma$
Let $X\stackrel{d}{=}\mu + \xi AU$ be an elliptical distributed random vector, with $\xi \in \mathbb{R}$ and $U\in \mathbb{R^n}$ independent, $\mu$ an $n$-dimensional vector; A a $n \times n$ matrix with $A A^T=\Sigma$;
Then the following holds:
- The characteristic function $\phi(t)=\mathbf{E}(e^{it^\top X})$ of $X$ is of the form: $$\phi_X(t)=e^{it^\top \mu}\psi(t^\top \Sigma t)$$
- Any linear combination of elliptically distributed vectors is elliptical.
- Marginal distributions of elliptically distributed vectors are elliptical.
Proof: 1. For the characteristic function of $X$ we have: \begin{eqnarray*} \phi_X(t) &=& \mathbb{E}\left( \exp(i t^\top(\mu + \xi AU) \right) =\exp(i t^\top \mu) \mathbb{E}\left(\exp(i t^\top\xi AU)\right)=\exp(i t^\top \mu)\mathbb{E}\left( \mathbb{E}\left( \exp(i t^\top\xi AU) | \xi \right) \right) \\ &=& \exp(i t^\top \mu) \mathbb{E}\left( \phi_{U}\left(A^\top \xi t \right) \right)=\exp(i t^\top \mu) \mathbb{E}\left( \psi_{U}\left(t^\top \xi A A^\top \xi t \right) \right) \\ &=& \exp(i t^\top \mu) \mathbb{E}\left( \psi_{U}\left(\xi^2 t^\top \Sigma t \right) \right), \end{eqnarray*} where we used the law of total expectation and $\phi_U(t)=\psi(t^Tt)$. Now defining $\psi(s)=\mathbb{E}\left( \psi_{U}\left(\xi^2 s \right) \right)$, we get the desired result. \ In fact it is also possible to prove the other direction. A random variable is elliptically distributed if and only if the characteristic function is of the form $\phi_X(t)=e^{it^\top \mu}\psi(t^\top \Sigma t)$. For a proof, see [Frahm, 2004].
- We only prove it for $X$, $Y$ being independent. \ Let $X$ and $Y \in \mathbb{R}^n$ be two elliptically distributed vectors, independent of each other. Since $\phi_{cX}(t)=\exp(i t^\top c \mu) \mathbb{E}\left(\exp(i t^\top c \xi AU)\right)=\exp(i t^\top c \mu)\psi(t^\top c \Sigma c t)=\exp(i t^\top c \mu)\psi(c^2 t^\top \Sigma t)$, $\psi$ is also only dependent on $t^\top \Sigma t$ and for this reason with the comment in (1), $cX$ and $bY$ are also elliptical $\forall b,c \in \mathbb{R}$.
We then have: \begin{eqnarray*} \phi_{aX+bY}(t) &=& \phi_{aX}(t) \cdot \phi_{bY}(t)=\exp(i t^\top (\mu_1+\mu_2))\psi_{cX}(t^\top \Sigma t)\psi_{bY}(t^\top \Sigma t), \end{eqnarray*} where $\mu_1$ and $\mu_2$ describes the mean of $cX$, respectively $bY$ and $\psi_{cX}$ respectively $\psi_{bY}$ describes the functions for describing the characteristic function of $cX$ and $bY$. \ Defining $\psi(.)=\psi_{cX}(.)\psi_{bY}(.)$ yields the desired result.
- We prove that $BX$ is elliptical for all matrices $B \in \mathbb{R}^{m \times n}$. We have: \begin{eqnarray*} \phi_{BX}(t) &=& \mathbb{E}\left( \exp(i t^\top BX \right)=\phi_{X}(B^\top t) =\exp(i t^\top B \mu)\psi_X(t^\top B \Sigma B^\top t). \end{eqnarray*}
So first of all: Is everything I wrote so far correct? But now with 3. I don't know how to continue;
Saying that the characteristic function consists of a function $\psi$, dependent only on $t^T \Sigma t$ means that the function $\psi$ is only dependent on the 1-dimensional product of $t^T \Sigma t$, which is clearly not given with $t^\top B \Sigma B^\top t$;
I got the hint to prove it like that on this page but it seems to not work somehow - or am I doing a mistake?
I'm very glad about some help;
The advices I got are can also be found here: Compute $\phi_X(t)=E(e^{it^\top X})$ if $X\stackrel{d}{=}\mu + \xi AU$ with $AA^\top=\Sigma$