1

Assume that $X\sim \mathcal{N}(0,\sigma_X^2)$ and $\epsilon \sim \mathcal{N}(0,\sigma_\epsilon^2)$ are independent and $k\in \mathbb{N}$.

Define $Y:= \sum_{i=1}^k \beta_i X^i + \epsilon$, where $\beta_1,...,\beta_k$ are real numbers.

  1. Is it possible to derive $E(X|Y)$ for general $\beta_1,...,\beta_k\in \mathbb{R}$?

  2. Do you know any large function classes $\mathcal{F} \subset \{f:\mathbb{R}\to \mathbb{R}\}$, such that we are able to derive $E(X|f(X)+\epsilon)$ when $f\in \mathcal{F}$?

John
  • 1,775
  • 14
  • 27
  • What are your thoughts, what have you tried so far? – Pontus Hultkrantz Nov 02 '20 at 14:19
  • 1
    @PontusHultkrantz I think that in my current formulation 1) is not possible. Even with second-degree polynomials and brute force density transformation one ends up with reversed conditional density $f_{X|Y}$ that contains some very nasty integrals that only simplify to something computable by hand if $\beta_1=0$. Hence, I might be better of concentrating on 2) and try to find other "large" function classes which actually allows for the derivation of the conditional expectation. I have only shown that it is possible for 2.degree polynomials with no linear term, so not very large or interesting. – John Nov 02 '20 at 19:13
  • pretty much same result for me. With $\sigma_X=\sigma_{\epsilon}=1$, brute force yields $E[X|Y=y] \propto \int_{-\infty}^{\infty} x \phi(y-f(x)) \phi(x) dx$, where $\phi$ is the standard normal pdf. Please let us know if you progress, as I am curious about the answer. – Pontus Hultkrantz Nov 03 '20 at 22:11

0 Answers0