# Estimating the probability density of a component of a mixture distribution

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?

$$\newcommand{\al}{\alpha} \newcommand{\R}{\mathbb{R}} \renewcommand{\x}{\bar x} \newcommand{\ga}{\gamma}$$ 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 $$\begin{equation*} p(x)=\frac{(1-\al) \phi (x)-\al j(x) \phi (\x)}{2 \al-1}. \end{equation*}$$
• 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 $p(x)$ Jul 17 '19 at 14:13
• 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. Jul 17 '19 at 14:14