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Let $X \in \mathbb{R}^d$ be a random variable with probability distribution $P$. Let $f:\mathbb{R}^d \to \mathbb{R}^d$ be an invertible function and let $P_{f}$ be the distribution of random variable $f(X)$. Suppose $P$ and $P_f$ are such that the mixture distribution $(1-\alpha)P + \alpha P_f$, for some $\alpha \in [0,1]$, is equal to the standard normal distribution $\mathcal{N}(0, I_{d\times d})$, where $I_{d\times d}$ is the $d\times d$ identity matrix.

Given $\alpha, f$, I'm interested in finding the distribution $P$ for which the above condition holds, i.e., $(1-\alpha)P + \alpha P_f$ is equal to $\mathcal{N}(0, I_{d\times d})$. (By finding a distribution what I mean is that, at any given $x \in \mathbb{R}^d$, I would like to compute the probability density of $P$ at $x$.)

Here are my questions:

1) For any given $\alpha, f$, is there a unique $P$ which satisfies this i.e., is the problem identifiable? Or can there be multiple distributions? Of course, when $\alpha = 1/2$ the problem is clearly not identifiable. Assuming $\alpha \neq 1/2$, is the problem identifiable?

2) Assuming the problem is identifiable, how can I compute the density of $P$ at any given $x \in \mathbb{R}^d$? We have the following relation between the densities of $P, P_f$ and $\mathcal{N}(0, I_{d\times d})$

$$\frac{1}{\sqrt{2\pi}} e^{-x^2/2} = (1-\alpha)p(x) + \alpha |\det{J_{f^{-1}}(x)}| p(f^{-1}(x)),$$

where $p(x)$ is the probability density of $P$ at $x$ and $\det{J_{f^{-1}}(x)}$ is the determinant of the Jacobian of $f^{-1}$ evaluated at $x$. Is there a way to compute $p(x)$ from this equation?

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$\newcommand{\al}{\alpha} \newcommand{\R}{\mathbb{R}} \renewcommand{\x}{\bar x} \newcommand{\ga}{\gamma}$ Welcome to MathOverflow! If $\al\ne1/2$ and $f$ is an involution of $\R^d$ (so that $f=f^{-1}$), then $p$ is identifiable and can be found as follows. We have \begin{equation*} \phi(x):=\frac{1}{\sqrt{2\pi}} e^{-x^2/2} = (1-\al)p(x) + \al |\det{J_{f^{-1}}(x)}| p(f^{-1}(x)) \end{equation*} for all $x\in\R^d$, which we can rewrite as \begin{equation*} (1-\al)p(x) + \al p(\x)j(x)=\phi(x), \tag{1} \end{equation*} where $j(x):=|\det{J_f(x)}|$ and $\x:=f(x)$. Replacing here $x$ by $\x$, we have \begin{equation*} (1-\al)p(\x) + \al p(x)/j(x) =\phi(\x), \tag{2} \end{equation*} since $j(\x)=1/j(x)$. Solving the system of linear equations (1)--(2) for $p(x),p(\x)$, we find \begin{equation*} p(x)=\frac{(1-\al) \phi (x)-\al j(x) \phi (\x)}{2 \al-1}. \end{equation*}

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  • $\begingroup$ Thanks. This is interesting. Can something be said when f is not an involution? $\endgroup$
    – v0511
    Jul 17 '19 at 13:52
  • $\begingroup$ One way I can think of extending your approach is by assuming that $f^{-1}$ is contractive (which also means it has a fixed point). Then one can make use of the expressions for $\phi(x), \phi(f^{-1}(x)), \phi(f^{-1}\circ f^{-1}(x)) \dots$ to at least approximately evaluate $p(x)$ $\endgroup$
    – v0511
    Jul 17 '19 at 14:13
  • $\begingroup$ More generally, if $f$ is not an involution, this suggests that the answer will depend on the structure of the orbits $\{x,f(x),f(f(x)),\dots\}$. If the orbits are finite, then I think we can reason similarly. $\endgroup$ Jul 17 '19 at 14:14

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