Orthogonal transformation of multivariate Bernoulli-Gaussian distribution

Question

Actually, I have asked this question in https://math.stackexchange.com/questions/4330127/orthogonal-transformation-of-multivariate-bernoulli-gaussian-distribution, but I think mathoverflow might be more appropriate for it.

Recently, I studied multivariate Bernoulli-Gaussian distribution which is very useful for sparse signal processing. Suppose $X = (X_{1}, \cdots, X_{n})$ are i.i.d BG($p, \sigma^{2}$), we can know that $\mathbb{E}(X) = \mathbf{0}$ and Cov($X$) = diag($p \sigma^{2}$). Suppose $A$ is an orthogonal matrix with proper size and $Y = AX$, we can also know that $\mathbb{E}(Y) = \mathbf{0}$ and Cov($Y$) = diag($p \sigma^{2}$). Is it possible for us to know the distribution of $Y$ and is it possible for us to know $Y = (Y_{1}, \cdots, Y_{n})$ are i.i.d? Thank you.

Answer 1

Each $X_j$ has the Bernoulli-Gaussian distribution, $$P(x_j)=(1-p)\delta(x_j)+pN(x_j;0,\sigma^2).$$

To characterize the distribution of the variables $Y_i=\sum_{j}A_{ij}X_j$, for an orthogonal $n\times n$ matrix $A$, I calculate the moment generating function:

$$F(z_1,z_2,\ldots z_n)=\mathbb{E}\left[e^{\sum_{i}z_iy_i}\right]=\mathbb{E}\left[e^{\sum_{ij} z_i A_{ij} x_j}\right]=\prod_j \mathbb{E}\left[e^{\sum_i z_i A_{ij} x_j}\right]$$ $$\qquad = \prod_j \left(1-p + p \exp\left[\tfrac{1}{2} \sigma^2 \sum_{k,l} z_k z_l A_{kj}A_{lj}\right]\right).$$ For $p\neq 0,1$ this does not factorize in terms dependent only one single $z_j$, so the $Y_i$ variables are not independent.