# 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?

• 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