Is it possible to uniquely decompose a vector $\underset{d_x \times 1}{x}$ of $d_x$ random variables into dependent and independent sources of variation?
Suppose we know the distribution $P_x$ of a mean-zero vector $\underset{d_x \times 1}{x}$. Is it possible to write $x$ as the sum of two vectors of random variables $\underset{d_x \times 1}{f}$ and $\underset{d_x \times 1}{v}$, with the following conditions:
$ x = f + v $
The consitituting random variables in $v$ are independent: $p(v) = \prod_{l=0}^{d_x-1} p(v_l)$
$f$ and $v$ are independent.
$\underset{d_x \times 1}{f}$ can be written as the function of a lower-dimensional vector: $\underset{d_x \times 1}{f} = z(\underset{d_a \times 1}{a})$, with the unknown function $z: \mathbb{R}^{d_a} \rightarrow \mathbb{R}^{d_x}$ and some $d_a < d_x$.
Under these conditions, are the distributions of $f$ and $v$, $P_f$ and $P_v$, uniquely identified?
I know the above conditions imply that $$p(x) = \int_{\mathbb{R}^{d_x}} p_f(x-v) \Big( \prod_{l=0}^{d_x-1} p_{v_l}(v_l) \Big) dv_{{d_x}-1} ... dv_0 \ \ \ \ \forall x \in \mathbb{R}^L $$
However, I have not been able to show that the distributions $P_f$ and $P_v$ have to be unique.
In general, $P_f$ and $P_v$ do not have to be unique.
Simple counterexample
Suppose $a$ is a random variable and $v$ a 2-dimensional vector of random variables. $f$ and $v$ are independent. $$ \underset{2 \times 1}{x} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \underset{1 \times 1}{a} + \underset{2 \times 1}{v} $$ $$ v \sim N \begin{pmatrix} \underset{2 \times 1}{0}, & \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{pmatrix}, \ \ \ \ a \sim N(0, 1), \ \ \ \ v \perp\!\!\!\perp a $$ $$ \implies x \sim N \begin{pmatrix} \underset{2 \times 1}{0}, & \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \end{pmatrix} $$ Suppose that we know $z(.)$ is a linear function in this example: $f = \begin{pmatrix} z_0 a \\ z_1 a \end{pmatrix} = \begin{pmatrix} z_0 \\ z_1 \end{pmatrix} a$. Then we have a problem with 4 unknown parameters and 3 conditions. The first two unknown parameters are $z_0$ and $z_1$, which describe the covariance of $x_0$ and $x_1$. The latter two parameters are the variances $\sigma_{v_0}^2$ and $\sigma_{v_1}^2$, which describe the amount of idiosyncratic variation in $x$. We can rewrite the linear system:
$$ \begin{pmatrix} z_0^2 + \sigma_{v_0}^2 & z_0 z_1 \\ z_0 z_1 & z_1^2 + \sigma_{v_1}^2 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} $$
Given that $z_0^2 + \sigma_{v_0}^2 = z_1^2 + \sigma_{v_1}^2 = 2$ we know that $z_0, z_1 \leq \sqrt{2}$. Given $z_0 z_1 = 1$, this in turn implies $z_0, z_1 \geq \frac{1}{\sqrt{2}}$. All we can say about $P_f$ and $P_v$ is up to one bounded parameter: $$ z_0 = \sqrt{2 - \sigma_{v_0}^2}, \ \ \ \ z_1 = \frac{1}{\sqrt{2 - \sigma_{v_0}^2}}, \ \ \ \ \sigma_{v_1}^2 = \frac{3 - 2\sigma_{v_0}^2}{2 - \sigma_{v_0}^2}, \ \ \ \ \sigma_{v_0}^2 \in [0, 1.5] $$
Example of Uniqueness
In a linear model, $P_f$ and $P_v$ can be identified uniquely as long as the dimension $a$ is sufficiently small.
Suppose $a$ is a random variable and $v$ a 3-dimensional vector of random variables. $f$ and $v$ are independent. $$ \underset{3 \times 1}{x} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \underset{1 \times 1}{a} + \underset{3 \times 1}{v} $$ $$ v \sim N \begin{pmatrix} \underset{3 \times 1}{0}, & \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{pmatrix}, \ \ \ \ a \sim N(0, 1), \ \ \ \ v \perp\!\!\!\perp a $$ $$ \implies x \sim N \begin{pmatrix} \underset{2 \times 1}{0}, & \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix} \end{pmatrix} $$ Suppose that we know $z(.)$ is a linear function in this example: $f = \begin{pmatrix} z_0 a \\ z_1 a \\ z_2 a \end{pmatrix} = \begin{pmatrix} z_0 \\ z_1 \\ z_2 \end{pmatrix} a = \underset{3 \times 1}{z} a$. Suppose that in addition we know that the dimension of $a$ is one. Then we have a problem with 6 unknown parameters and 6 conditions. The first three unknown parameters are in $\underset{3 \times 1}{z}$, which describe the covariances of $\underset{3 \times 1}{x}$. The latter three parameters are the variances $\sigma_{v_j}^2$ $l=0,1,2$, which describe the amount of idiosyncratic variation in $\underset{3 \times 1}{x}$. We can rewrite the linear system:
$$ \begin{pmatrix} z_0^2 + \sigma_{v_0}^2 & z_0 z_1 & z_0 z_2 \\ z_0 z_1 & z_1^2 + \sigma_{v_1}^2 & z_1 z_2 \\ z_0 z_2 & z_1 z_2 & z_2^2 + \sigma_{v_2}^2 \end{pmatrix} = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix} $$ Given that $a$ is one-dimensional by assumption, we know that this system has a unique solution: $$ z_0 = z_1 = z_2 = 1, \ \ \ \ \sigma_{v_0}^2 = \sigma_{v_1}^2 = \sigma_{v_2}^2 = 1 $$
General identifiability of the integral equation for linear systems
In both examples linearity was key to argue for identifiability based on the number of parameters. When $\underset{d_x}{x}$ has any number of dimensions $x$, we can find the number of dimensions $d_a$ that $\underset{d_a}{a}$ can have at most. We find that number such that the number of unknown parameters is smaller than the number of observed parameters (in $x$'s covariance matrix).
$$ \underbrace{\frac{1}{2}(d_a - 1) d_a}_{\text{covariances of } a} + \underbrace{d_x d_a}_{\text{linear effect of } a \text{ on } x} \leq \underbrace{\frac{1}{2} (d_x - 1) d_x}_{\text{(co-)variances of } x} $$ $$ d_a \leq \frac{1}{2} - d_x + \sqrt{2d_x^2 - 2d_x + 1} $$ $$ \underset{d_x \rightarrow \infty}{\text{lim}}\Big(\frac{d_a}{d_x}\Big) \leq \sqrt{2} - 1 \approx 0.414 $$ As $d_x$ increases, the ratio of the dimension of factors $a$ compared to the dimension $d_x$ of observed $x$, $\frac{d_a}{d_x}$, has to be smaller than $0.414$. For any combination of $k \leq d_a$ factors we consider, there must be more than $2.414k$ different components of $\underset{d_x \times 1}{x}$ associated with those $k$ factors to be able to uniquely identify the linear model parameters in $z$ and the covariance between the $k$ factors. When $d_x$ is small that ratio will be slightly different.
General identifiability of the integral equation for nonlinear systems
Two questions remain:
The problem can be interpreted as a multidimensional integral equation with an unknown difference kernel. If we let $P_v$ be the difference kernel, we could apply the Fourier transform to $P_v$ and $P_x$ to find $P_f$ (Polyanin, A.D. and Manzhirov, A.V., 2008. Handbook of integral equations. CRC press, p. 586).
Unfortunately the problem here is not as straightforward. Instead of knowing a kernel, we have two unknown multidimensional distributions, on which significant dimension and independence restrictions are imposed. In a linear system I have provided an example for identifiability. I conjecture that just like in linear systems the dimension of $a$ ($d_a$) plays a key role in their nonlinear counterpart. It would be fantastic to hear the opinion on nonlinear identifiability from an expert in integral equations.
Summary
In general, $P_f$ and $P_v$ do not have to be unique. In linear systems $P_f$ and $P_v$ are uniquely identified if the dimension of $a$, $d_a$ is sufficiently small. In nonlinear systems we need to solve an atypical integral equation where instead of a known kernel dimension and independence restrictions are imposed on two unknown multidimensional distributions. I do not know the implications for identifiability $P_f$ and $P_v$ in a nonlinear system, but conjecture that the dimension of $a$, $d_a$, again plays a crucial role.