# Construction of a Transformation of a Random Vector that Preserves Independence

Let $X_1, \dots, X_n$ be $n$ independent random variables, not necessarily normal.

Let $Y_1 = \sum_{i=1}^{n}\alpha_i X_i$ a given linear combination of the random variables.

Is there a known, "canonical" way of constructing a set of functions $f_2,\dots,f_n$ such that

1. All $Y_i, i \in \{1,\dots,n\}$ are mutually independent (where $Y_i = f_i(X_1,\dots,X_n), i > 1$);

2. The set defines a transformation of variables, in the sense that $X_1,\dots,X_n$ are fully determined given $Y_1 = y_1,\dots,Y_n = y_n$.

Note: $f_i$ must be non-linear, otherwise all $X_i$ would be normal (see When linear combinations of independent random variables are still independent?).

It is clear that functions that transform only one variable do not affect independence.

If we may assume continuous variables then there exist nondecreasing real functions that transform them to standard normal variables (still independent but now also identically distributed).

After this transformation apply arbitrary rotations in n-space. This results in another set of independent standard normal variables.

Then apply arbitrary functions to the individual resulting variables.

Now if $f_1$ is already given as a linear combination of the $X_1,\ldots,X_n$ then this may not work out literally. Here is an idea, I will not blame you if you do not accept it as a full answer.

Without loss of generality assume that all coefficients $\alpha_i$ are nonzero and consider the line $L_t$ defined by $f_1=t$ and $X_3=x_3, X_4=x_4, \ldots=X_n=x_n$ having fixed values. There almost certainly exists a function $g_t:L_t\to L_t$ which is nondecreasing (in the sense of the $x_1$ component of points on $L_t$) and such that the variable $g_t(X_1),$ restricted to $L_t,$ is standard "normal" around the point of $L_t$ that has minimum distance from the origin. Let $Y_1=f_1(X_1,\ldots,X_n)$ be the coordinate in the direction of the gradient of $f_1.$ Then one should be able to demonstrate that $g_{Y_1}(X_1)$ is independent of $f_1, X_3,\ldots,X_n.$

• Thanks for your prompt answer. The problem is that $Y_1$ is a given linear combination of the original variables, not necessarily equal to some rotation of the "normalized" variables. – Ferdinand.kraft Dec 15 '15 at 20:17