Equidistributed sequence wrt exponential/Gaussian measure

Question

For an arbitrary probability space $(X,\mu)$, a sequence $(x_n)$ in $X$ is said to be equidistributed with respect to $\mu$ if the measures $\frac 1 n \sum_{1\le k\le n} \delta_{x_k}$ converges weakly to $\mu$ as $n \to \infty$.

The special case of a compact interval equipped with the Lebesgue measure is very well studied. One way of building equidistributed sequences is to construct an ergodic transformation preserving the measure. This post sketches such a construction for the Lebesgue measure on the real line.

Is there any known explicit construction of an equidistributed sequence for the measure associated to an exponential or Gaussian random variable, preferably that doesn’t involve applying the cumulative distribution function/its inverse?

Answer 1

Weak convergence of measures is preserved under continuous transformations: if a sequence of Borel measures $\mu_n$ converges to a Borel measure $\mu$ weakly and if $\psi$ is a continuous transformation of the appropriate topological spaces, then the sequence of the pushforward measures $\mu_n\psi^{-1}$ will converge to the pushforward measure $\mu\psi^{-1}$ weakly.

Therefore, the problem immediately reduces to transforming the uniform distribution on the interval $(0,1)$ or on the unit square $(0,1)^2$ (say) continuously to an exponential or Gaussian distribution.

So, if a sequence $(x_n)$ is equidistributed with respect to the uniform distribution on the interval $(0,1)$, then the sequence $(-\ln x_n)$ will be equidistributed with respect to the exponential distribution with mean $1$ -- because the continuous map $(0,1)\ni x\mapsto\psi(x):=-\ln x\in\mathbb R$ transforms the uniform distribution on the interval $(0,1)$ into the exponential distribution.

Similarly, if a sequence $((x_n,y_n))$ is equidistributed with respect to the uniform distribution on the unit square $(0,1)^2$, then the sequence $(\sqrt{-2\ln x_n}\cos(2\pi y_n))$ will be equidistributed with respect to the standard normal distribution, in view of the Box–Muller transform.

Answer 2

Denote by $\Phi(x)$ the cumulative distribution function of $\Gamma$, the standard normal distribution. More explicitly, $$ \Phi(x)=\int_{-\infty}^x \Gamma[dt]:=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-t^2/2} dt. $$

$\newcommand{\bR}{\mathbb{R}}$ It defines a homeomorphisms $\Phi:\bR\to (0,1)$. Its inverse $Q=\Phi^{-1}$ is the so called quantile function of $\Gamma$. If $\mu$ is the uniform distribution on $[0,1]$, then $Q_\#\mu$, the push forward of $\mu$ is the normal distribution. The push forward is defined as in Iosif Pinelis' answer,

$$Q_\#\mu[S]= \mu\big[ Q^{-1}(S)\big],$$

for any Borel subset $S\subset \bR$.

Take an ergodic transformation of $T$ of the $\big((0,1),\mu\big)$. Then $Q\circ T \circ \Phi$ is an ergodic transformation of $(\bR,\Gamma)$.