Independent distributions determined by projections Let $X_1 \sim \mathbb P_1, \dots, X_n \sim \mathbb P_n$ be independent random variables. Let $b_1, \dots b_n$ be a basis of $\mathbb R^n$.
Question: Once I know the distributions of the random variables
\begin{align*}
         \langle b_i , X \rangle
\end{align*}
for all $i= 1, \dots, n$, does this uniquely determine my distributions $\mathbb P_1, \dots, \mathbb P_n$ ?
 A: No, it doesn't. At least, not for any basis $b$.
Here is a Gaussian counterexample: let $X_1$ and $X_2$ be standard Gaussians with mean $0$ and variance $\sigma_i^2, i=1,2$, and let $b_1 := e_1 + e_2$, $b_2 := e_1 - e_2$ be the basis vectors, where $e_i$ are the standard basis vectors. Now both $\langle X,b_1 \rangle = X_1 + X_2$ and $\langle X,b_2 \rangle = X_1 - X_2$ are standard Gaussians with variance $\sigma_1^2 + \sigma_2^2$, so from their distributions you will only know $\sigma_1^2 + \sigma_2^2$, but not the individual $\sigma_i^2$.
By using symmetric stable laws this counterexample can be generalized to bases for which there is an $\alpha \in (0,2]$, such that the matrix $( |b_{ij}|^\alpha )$ is degenerate.
On the other hand, if for every natural $m$ the matrix $(b_{ij}^m)$ is nondegenerate (e.g. this is the case for Vandermonde matrices) then we can compute cumulants of $X_i$ from cumulants of $\langle X, b_i \rangle$ - if they exist, of course. So under this nondegeneracy assumption on $b$ + existence of moments + uniqueness of solution to moment problem, we can restore $X_i$.
So the interesting question here is whether there exist nontrivial $b$'s for which the law is determined by projections under no additional restrictions on $X$.
