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.



Orthogonal transformation of multivariate Bernoulli-Gaussian distribution
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.


