Let $\vec{x} \in \mathbb{R}^n$ be a fixed vector and suppose that we are given an isotropic random vector $\vec{a} = (a_1, \dots, a_n)^T$ in $\mathbb{R}^n$ (i.e., the covariance matrix of $\vec{a}$ is the identity matrix).

Are there some general conditions on $\vec x$ and the distribution of $\vec a$ such that the following holds? There exist a (non-linear) mapping $P: \mathbb{R}^n \to \mathbb{R}^n$, depending on $\vec{x}$, statisfying the following two properties:

  1. The entries of $P(\vec a)$ and $\langle \vec a, \vec x\rangle$ are independent.
  2. The range of $Q := Id - P$ is one-dimensional, i.e., there exists some $\vec{x}' \in \mathbb{R}^n$ with $range(Q) \subset span\{\vec x'\}$.

Edit: One specification that I am particularly interested in looks as follows: Assume that the entries of $\vec a$ are deformed Gaussians, i.e., $a_i = F(g_i)$ with $g_i \sim N(0,1)$ i.i.d. and $F$ continuous and invertible.

My intention is to generalize the well-known special case of independent Gaussians, $a_i \sim N(0,1)$, where one can simply choose $P$ to be the orthogonal projection onto $\{\vec x\}^\perp$ and $\vec x' = \vec x$. This works due to the rotation invariance of isotropic Gaussian random vectors and I wonder to what extend this can be relaxed.

  • $\begingroup$ Do you want $P$ to be independent of $a$ or you don't care? $\endgroup$
    – fedja
    Commented Aug 16, 2016 at 15:41
  • $\begingroup$ It would be nice to be have a deterministic construction. But if there exists a random construction, independence of $a$ is important because I would like to apply $P$ to several independent copies of $a$. $\endgroup$
    – EmmGee
    Commented Aug 17, 2016 at 6:00

1 Answer 1


This will not work in general. Let's take $a$ as a vector with iid $\pm 1$ entries, with probability $1/2$ for each of $\pm 1$, and $x=(1,1,\ldots ,1)$ (normalize this if you prefer). Then if $\langle a, x\rangle = n$, we know for sure that $a=(1,1,\ldots , 1)$, so $P(a)$ would have to be constant almost surely if we want it independent of $\langle a, x\rangle$.

If you compare this with your example (normal distribution), then you see that this had the crucial extra property that the entries stay independent after changing coordinates by a rotation.

  • $\begingroup$ Thank you! This counterexample is already helpful for me because the above question is actually part of a more general statement, which is not true for $\pm 1$-values entries. $\endgroup$
    – EmmGee
    Commented Aug 17, 2016 at 6:16
  • $\begingroup$ But I wonder "how crucial" the extra property of rotation invariance is. Is this really necessary or can this be relaxed? I slightly reformulated the problem issue in this direction. $\endgroup$
    – EmmGee
    Commented Aug 17, 2016 at 7:21
  • $\begingroup$ @EmmGee: Your edited version runs into the exact same problem: we can take an $F$ for which $F(g_j)$ is approximately the Bernoulli RV from my example. While admittedly I haven't thought about my remark from the second paragraph much, my feeling is that it does describe what this problem is about. $\endgroup$ Commented Aug 17, 2016 at 18:32

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