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Let $f_X$ be the pdf of $X$. Let $$C = \int_{-\infty}^{\infty} f_X(x) \log F_X(x) dx $$

and define $$\mathscr{Z}_n = \left\{ x_1,\dots, x_n \in \textrm{Supp}(X) \mid \sum_{i=1}^n \log f_X(x_i) \geq n C \right\}$$

Let $\mathbf Z$ be uniform on $\mathscr{Z}_n$. If $X$ is uniform on its support $\textrm{Supp}(X)$, then $\mathscr{Z}_n =\textrm{Supp}(X)^n$ and the marginals of $\mathscr{Z}_n$ are all $X$.

Otherwise, the marginal distribution of $\mathscr{Z}_n$ has pdf proportional to

$$\int_{x_2,\dots, x_n \in \textrm{Supp}(X)} A(x_1, x_2,\dots,x_n) dx_2\dots dx_n$$ where $$A(x_1, x_2,\dots,x_n) = \begin{cases} 1 & \sum_{i=1}^n \log f_X(x_i) \geq n C \\ 0 & \sum_{i=1}^n \log f_X(x_i) < n C\end{cases}$$

which is equal to

$$f_X(x_1) \int_{x_2,\dots, x_n \in \mathbb R} B(x_1, x_2,\dots,x_n) f_X(x_2)\dots f_X(x_n) dx_2\dots dx_n$$

where $$B(x_1, x_2,\dots,x_n) = \begin{cases} e^{ n C - \sum_{i=1}^{n} \log f_X(x_i)} & \sum_{i=1}^n \log f_X(x_i) \geq n C \\ 0 & \sum_{i=1}^n \log f_X(x_i) < n C\end{cases}$$

So it suffices to check that the expectation of the random variable $B(x_1,\dots,x_n)$, with $x_2,\dots, x_n$ distributed according to $X$, is independent of $x_1 \in \textrm{Supp}(X)$. In fact, we will check that when multiplied by $\sqrt{n}$, it converges to a constant independent of $x_1$.

In other words, this is the expectation of $$\begin{cases} e^{-y} & y\geq 0 \\ 0 & y <0 \end{cases}$$ where $y = \sum_{i=1}^n \log f_X(x_i) - nC$

By the local central limit theorem, the distribution of $ \sum_{i=2}^n \log f_X(x_i) - (n-1)C$ is approximately a Gaussian with mean $0$ and variance $(n-1) v$ for $v>0$ the variance of $\log f_X(x)$. Thus, multiplied by $\sqrt{n}$, the measure of this random variable converges to a uniform measure on $\mathbb R$. Since $y$ is this random variable plus $x_1$, the expectation of $$\begin{cases} e^{-y} & y\geq 0 \\ 0 & y <0 \end{cases}$$ converges to a constant times $\int_{0}^\infty e^{-y} dy$.

Alternatively, if $\log f_X(x)$ is supported in an arithmetic progression, then the distribution of $ \sum_{i=2}^n \log f_X(x_i) - (n-1)C$ is approximately a Gaussian restricted to that arithmetic progression, so multiplying by $\sqrt{n}$ it converges to the uniform measure on the arithmetic progression, and the expectation converges to a constant times $\sum_{n=0}^{\infty} e^{-qn}$.

Let $f_X$ be the pdf of $X$. Let $$C = \int_{-\infty}^{\infty} f_X(x) \log F_X(x) dx $$

and define $$\mathscr{Z}_n = \left\{ x_1,\dots, x_n \in \textrm{Supp}(X) \mid \sum_{i=1}^n \log f_X(x_i) \geq n C \right\}$$

Note that this specializes to your examples - for a Gaussian, $\log f_X$ is quadratic, so this defines a ball, and for an exponential distribution, $\log f_X$ is linear (and supported on nonnegative reals) so this defines a simplex.

Let $\mathbf Z$ be uniform on $\mathscr{Z}_n$. If $X$ is uniform on its support $\textrm{Supp}(X)$, then $\mathscr{Z}_n =\textrm{Supp}(X)^n$ and the marginals of $\mathscr{Z}_n$ are all $X$.

Otherwise, the marginal distribution of $\mathscr{Z}_n$ has pdf proportional to

$$\int_{x_2,\dots, x_n \in \textrm{Supp}(X)} A(x_1, x_2,\dots,x_n) dx_2\dots dx_n$$ where $$A(x_1, x_2,\dots,x_n) = \begin{cases} 1 & \sum_{i=1}^n \log f_X(x_i) \geq n C \\ 0 & \sum_{i=1}^n \log f_X(x_i) < n C\end{cases}$$

which is equal to

$$f_X(x_1) \int_{x_2,\dots, x_n \in \mathbb R} B(x_1, x_2,\dots,x_n) f_X(x_2)\dots f_X(x_n) dx_2\dots dx_n$$

where $$B(x_1, x_2,\dots,x_n) = \begin{cases} e^{ n C - \sum_{i=1}^{n} \log f_X(x_i)} & \sum_{i=1}^n \log f_X(x_i) \geq n C \\ 0 & \sum_{i=1}^n \log f_X(x_i) < n C\end{cases}$$

So it suffices to check that the expectation of the random variable $B(x_1,\dots,x_n)$, with $x_2,\dots, x_n$ distributed according to $X$, is independent of $x_1 \in \textrm{Supp}(X)$. In fact, we will check that when multiplied by $\sqrt{n}$, it converges to a constant independent of $x_1$.

In other words, this is the expectation of $$\begin{cases} e^{-y} & y\geq 0 \\ 0 & y <0 \end{cases}$$ where $y = \sum_{i=1}^n \log f_X(x_i) - nC$

By the local central limit theorem, the distribution of $ \sum_{i=2}^n \log f_X(x_i) - (n-1)C$ is approximately a Gaussian with mean $0$ and variance $(n-1) v$ for $v>0$ the variance of $\log f_X(x)$. Thus, multiplied by $\sqrt{n}$, the measure of this random variable converges to a uniform measure on $\mathbb R$. Since $y$ is this random variable plus $x_1$, the expectation of $$\begin{cases} e^{-y} & y\geq 0 \\ 0 & y <0 \end{cases}$$ converges to a constant times $\int_{0}^\infty e^{-y} dy$.

Alternatively, if $\log f_X(x)$ is supported in an arithmetic progression, then the distribution of $ \sum_{i=2}^n \log f_X(x_i) - (n-1)C$ is approximately a Gaussian restricted to that arithmetic progression, so multiplying by $\sqrt{n}$ it converges to the uniform measure on the arithmetic progression, and the expectation converges to a constant times $\sum_{n=0}^{\infty} e^{-qn}$.

Let $f_X$ be the pdf of $X$. Let $$C = \int_{-\infty}^{\infty} f_X(x) \log F_X(x) dx $$

and define $$\mathscr{Z}_n = \left\{ x_1,\dots, x_n \in \textrm{Supp}(X) \mid \sum_{i=1}^n \log f_X(x_i) \geq n C \right\}$$

Let $\mathbf Z$ be uniform on $\mathscr{Z}_n$. If $X$ is uniform on its support $\textrm{Supp}(X)$, then $\mathscr{Z}_n =\textrm{Supp}(X)^n$ and the marginals of $\mathscr{Z}_n$ are all $X$.

Otherwise, the marginal distribution of $\mathscr{Z}_n$ has pdf proportional to

$$\int_{x_2,\dots, x_n \in \textrm{Supp}(X)} A(x_1, x_2,\dots,x_n) dx_2\dots dx_n$$ where $A(x_1, x_2,\dots,x_n) = \begin{cases} 1 & \sum_{i=1}^n \log f_X(x_i) \geq n C \ 0 & \sum_{i=1}^n \log f_X(x_i) < n C\end{cases}$$

which is equal to

$$f_X(x_1) \int_{x_2,\dots, x_n \in \mathbb R} B(x_1, x_2,\dots,x_n) f_X(x_2)\dots f_X(x_n) dx_2\dots dx_n$$

where $B(x_1, x_2,\dots,x_n) = \begin{cases} e^{ n C - \sum_{i=1}^{n} \log f_X(x_i)} & \sum_{i=1}^n \log f_X(x_i) \geq n C \ 0 & \sum_{i=1}^n \log f_X(x_i) < n C\end{cases}$$

So it suffices to check that the expectation of the random variable $B(x_1,\dots,x_n)$, with $x_2,\dots, x_n$ distributed according to $X$, is independent of $x_1 \in \textrm{Supp}(X)$. In fact, we will check that when multiplied by $\sqrt{n}$, it converges to a constant independent of $x_1$.

In other words, this is the expectation of $$\begin{cases} e^{-y} & y\geq 0 \\ 0 & y <0 \end{cases}$$ where $y = \sum_{i=1}^n \log f_X(x_i) - nC$

By the local central limit theorem, the distribution of $ \sum_{i=2}^n \log f_X(x_i) - (n-1)C$ is approximately a Gaussian with mean $0$ and variance $(n-1) v$ for $v>0$ the variance of $\log f_X(x)$. Thus, multiplied by $\sqrt{n}$, the measure of this random variable converges to a uniform measure on $\mathbb R$. Since $y$ is this random variable plus $x_1$, the expectation of $$\begin{cases} e^{-y} & y\geq 0 \\ 0 & y <0 \end{cases}$$ converges to a constant times $\int_{0}^\infty e^{-y} dy$.

Alternatively, if $\log f_X(x)$ is supported in an arithmetic progression, then the distribution of $ \sum_{i=2}^n \log f_X(x_i) - (n-1)C$ is approximately a Gaussian restricted to that arithmetic progression, so multiplying by $\sqrt{n}$ it converges to the uniform measure on the arithmetic progression, and the expectation converges to a constant times $\sum_{n=0}^{\infty} e^{-qn}$.

Let $f_X$ be the pdf of $X$. Let $$C = \int_{-\infty}^{\infty} f_X(x) \log F_X(x) dx $$

and define $$\mathscr{Z}_n = \left\{ x_1,\dots, x_n \in \textrm{Supp}(X) \mid \sum_{i=1}^n \log f_X(x_i) \geq n C \right\}$$

Let $\mathbf Z$ be uniform on $\mathscr{Z}_n$. If $X$ is uniform on its support $\textrm{Supp}(X)$, then $\mathscr{Z}_n =\textrm{Supp}(X)^n$ and the marginals of $\mathscr{Z}_n$ are all $X$.

Otherwise, the marginal distribution of $\mathscr{Z}_n$ has pdf proportional to

$$\int_{x_2,\dots, x_n \in \textrm{Supp}(X)} A(x_1, x_2,\dots,x_n) dx_2\dots dx_n$$ where $$A(x_1, x_2,\dots,x_n) = \begin{cases} 1 & \sum_{i=1}^n \log f_X(x_i) \geq n C \\ 0 & \sum_{i=1}^n \log f_X(x_i) < n C\end{cases}$$

which is equal to

$$f_X(x_1) \int_{x_2,\dots, x_n \in \mathbb R} B(x_1, x_2,\dots,x_n) f_X(x_2)\dots f_X(x_n) dx_2\dots dx_n$$

where $$B(x_1, x_2,\dots,x_n) = \begin{cases} e^{ n C - \sum_{i=1}^{n} \log f_X(x_i)} & \sum_{i=1}^n \log f_X(x_i) \geq n C \\ 0 & \sum_{i=1}^n \log f_X(x_i) < n C\end{cases}$$

So it suffices to check that the expectation of the random variable $B(x_1,\dots,x_n)$, with $x_2,\dots, x_n$ distributed according to $X$, is independent of $x_1 \in \textrm{Supp}(X)$. In fact, we will check that when multiplied by $\sqrt{n}$, it converges to a constant independent of $x_1$.

In other words, this is the expectation of $$\begin{cases} e^{-y} & y\geq 0 \\ 0 & y <0 \end{cases}$$ where $y = \sum_{i=1}^n \log f_X(x_i) - nC$

By the local central limit theorem, the distribution of $ \sum_{i=2}^n \log f_X(x_i) - (n-1)C$ is approximately a Gaussian with mean $0$ and variance $(n-1) v$ for $v>0$ the variance of $\log f_X(x)$. Thus, multiplied by $\sqrt{n}$, the measure of this random variable converges to a uniform measure on $\mathbb R$. Since $y$ is this random variable plus $x_1$, the expectation of $$\begin{cases} e^{-y} & y\geq 0 \\ 0 & y <0 \end{cases}$$ converges to a constant times $\int_{0}^\infty e^{-y} dy$.

Alternatively, if $\log f_X(x)$ is supported in an arithmetic progression, then the distribution of $ \sum_{i=2}^n \log f_X(x_i) - (n-1)C$ is approximately a Gaussian restricted to that arithmetic progression, so multiplying by $\sqrt{n}$ it converges to the uniform measure on the arithmetic progression, and the expectation converges to a constant times $\sum_{n=0}^{\infty} e^{-qn}$.

Let $$C = \int_{-\infty}^{\infty} F_X(x) \log F_X(x) dx $$

and define $$\mathscr{Z}_n = \left\{ x_1,\dots, x_n \in \textrm{Supp}(X) \mid \sum_{i=1}^n \log F_X(x_i) \geq n C \right\}$$

Let $\mathbf Z$ be uniform on $\mathscr{Z}_n$. If $X$ is uniform on its support $\textrm{Supp}(X)$, then $\mathscr{Z}_n =\textrm{Supp}(X)^n$ and the marginals of $\mathscr{Z}_n$ are all $X$.

Otherwise, the marginal distribution of $\mathscr{Z}_n$ has pdf proportional to

$$\int_{x_2,\dots, x_n \in \textrm{Supp}(X)} A(x_1, x_2,\dots,x_n) dx_2\dots dx_n$$ where $A(x_1, x_2,\dots,x_n) = \begin{cases} 1 & \sum_{i=1}^n \log F_X(x_i) \geq n C \ 0 & \sum_{i=1}^n \log F_X(x_i) < n C\end{cases}$$

which is equal to

$$F_X(x_1) \int_{x_2,\dots, x_n \in \mathbb R} B(x_1, x_2,\dots,x_n) F_X(x_2)\dots F_X(x_n) dx_2\dots dx_n$$

where $B(x_1, x_2,\dots,x_n) = \begin{cases} e^{ n C - \sum_{i=1}^{n} \log F_x(x_i)} & \sum_{i=1}^n \log F_X(x_i) \geq n C \ 0 & \sum_{i=1}^n \log F_X(x_i) < n C\end{cases}$$

So it suffices to check that the expectation of the random variable $B(x_1,\dots,x_n)$, with $x_2,\dots, x_n$ distributed according to $X$, is independent of $x_1 \in \textrm{Supp}(X)$. In fact, we will check that when multiplied by $\sqrt{n}$, it converges to a constant independent of $x_1$.

In other words, this is the expectation of $$\begin{cases} e^{-y} & y\geq 0 \\ 0 & y <0 \end{cases}$$ where $y = \sum_{i=1}^n \log F_X(x_i) - nC$

By the local central limit theorem, the distribution of $ \sum_{i=2}^n \log F_X(x_i) - (n-1)C$ is approximately a Gaussian with mean $0$ and variance $(n-1) v$ for $v>0$ the variance of $\log F_X(x)$. Thus, multiplied by $\sqrt{n}$, the measure of this random variable converges to a uniform measure on $\mathbb R$. Since $y$ is this random variable plus $x_1$, the expectation of $$\begin{cases} e^{-y} & y\geq 0 \\ 0 & y <0 \end{cases}$$ converges to a constant times $\int_{0}^\infty e^{-y} dy$.

Alternatively, if $\log F_X(x)$ is supported in an arithmetic progression, then the distribution of $ \sum_{i=2}^n \log F_X(x_i) - (n-1)C$ is approximately a Gaussian restricted to that arithmetic progression, so multiplying by $\sqrt{n}$ it converges to the uniform measure on the arithmetic progression, and the expectation converges to a constant times $\sum_{n=0}^{\infty} e^{-qn}$.

Let $f_X$ be the pdf of $X$. Let $$C = \int_{-\infty}^{\infty} f_X(x) \log F_X(x) dx $$

and define $$\mathscr{Z}_n = \left\{ x_1,\dots, x_n \in \textrm{Supp}(X) \mid \sum_{i=1}^n \log f_X(x_i) \geq n C \right\}$$

Let $\mathbf Z$ be uniform on $\mathscr{Z}_n$. If $X$ is uniform on its support $\textrm{Supp}(X)$, then $\mathscr{Z}_n =\textrm{Supp}(X)^n$ and the marginals of $\mathscr{Z}_n$ are all $X$.

Otherwise, the marginal distribution of $\mathscr{Z}_n$ has pdf proportional to

$$\int_{x_2,\dots, x_n \in \textrm{Supp}(X)} A(x_1, x_2,\dots,x_n) dx_2\dots dx_n$$ where $A(x_1, x_2,\dots,x_n) = \begin{cases} 1 & \sum_{i=1}^n \log f_X(x_i) \geq n C \ 0 & \sum_{i=1}^n \log f_X(x_i) < n C\end{cases}$$

which is equal to

$$f_X(x_1) \int_{x_2,\dots, x_n \in \mathbb R} B(x_1, x_2,\dots,x_n) f_X(x_2)\dots f_X(x_n) dx_2\dots dx_n$$

where $B(x_1, x_2,\dots,x_n) = \begin{cases} e^{ n C - \sum_{i=1}^{n} \log f_X(x_i)} & \sum_{i=1}^n \log f_X(x_i) \geq n C \ 0 & \sum_{i=1}^n \log f_X(x_i) < n C\end{cases}$$

So it suffices to check that the expectation of the random variable $B(x_1,\dots,x_n)$, with $x_2,\dots, x_n$ distributed according to $X$, is independent of $x_1 \in \textrm{Supp}(X)$. In fact, we will check that when multiplied by $\sqrt{n}$, it converges to a constant independent of $x_1$.

In other words, this is the expectation of $$\begin{cases} e^{-y} & y\geq 0 \\ 0 & y <0 \end{cases}$$ where $y = \sum_{i=1}^n \log f_X(x_i) - nC$

By the local central limit theorem, the distribution of $ \sum_{i=2}^n \log f_X(x_i) - (n-1)C$ is approximately a Gaussian with mean $0$ and variance $(n-1) v$ for $v>0$ the variance of $\log f_X(x)$. Thus, multiplied by $\sqrt{n}$, the measure of this random variable converges to a uniform measure on $\mathbb R$. Since $y$ is this random variable plus $x_1$, the expectation of $$\begin{cases} e^{-y} & y\geq 0 \\ 0 & y <0 \end{cases}$$ converges to a constant times $\int_{0}^\infty e^{-y} dy$.

Alternatively, if $\log f_X(x)$ is supported in an arithmetic progression, then the distribution of $ \sum_{i=2}^n \log f_X(x_i) - (n-1)C$ is approximately a Gaussian restricted to that arithmetic progression, so multiplying by $\sqrt{n}$ it converges to the uniform measure on the arithmetic progression, and the expectation converges to a constant times $\sum_{n=0}^{\infty} e^{-qn}$.