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(Philosophically, the following question is of a similar flavour to A stochastic process that is 1st and 2nd order (strictly) stationary, but not 3rd order stationaryA stochastic process that is 1st and 2nd order (strictly) stationary, but not 3rd order stationary, but more "advanced".)

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, let $(E,\mathcal{E})$ be a measurable space, and let $X:[0,\infty) \times \Omega \to E$ be a measurable function with the property that for any $n \in \mathbb{N}$, for all $\tau,t_1,\ldots,t_n \geq 0$ the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1,\omega),\ldots,X(t_n,\omega) \right)$

is equal to the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1+\tau,\omega),\ldots,X(t_n+\tau,\omega) \right)$.

Is it necessarily the case that for any $n \in \mathbb{N}$, for all $A \in \mathcal{E}^{\otimes n}$ and all $s,\tau,t_1,\ldots,t_n \geq 0$, the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [0,s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

is equal to the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [\tau,\tau+s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

(where $\textrm{Leb}$ denotes the Lebesgue measure)?

How about in the special case that $(E,\mathcal{E})=(\mathbb{R},\mathcal{B}(\mathbb{R}))$?

Now I've asked the above question in quite a general form, but what I ultimately want is: if $(X_t)_{t \geq 0}$ is a measurable stationary stochastic process in continuous time, then is the discrete-time stochastic process $\left( \int_n^{n+1} g(X_t) \, dt \right)_{n \geq 0}$ necessarily also stationary (where $g$ is a bounded measurable function)??

(Philosophically, the following question is of a similar flavour to A stochastic process that is 1st and 2nd order (strictly) stationary, but not 3rd order stationary, but more "advanced".)

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, let $(E,\mathcal{E})$ be a measurable space, and let $X:[0,\infty) \times \Omega \to E$ be a measurable function with the property that for any $n \in \mathbb{N}$, for all $\tau,t_1,\ldots,t_n \geq 0$ the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1,\omega),\ldots,X(t_n,\omega) \right)$

is equal to the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1+\tau,\omega),\ldots,X(t_n+\tau,\omega) \right)$.

Is it necessarily the case that for any $n \in \mathbb{N}$, for all $A \in \mathcal{E}^{\otimes n}$ and all $s,\tau,t_1,\ldots,t_n \geq 0$, the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [0,s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

is equal to the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [\tau,\tau+s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

(where $\textrm{Leb}$ denotes the Lebesgue measure)?

How about in the special case that $(E,\mathcal{E})=(\mathbb{R},\mathcal{B}(\mathbb{R}))$?

Now I've asked the above question in quite a general form, but what I ultimately want is: if $(X_t)_{t \geq 0}$ is a measurable stationary stochastic process in continuous time, then is the discrete-time stochastic process $\left( \int_n^{n+1} g(X_t) \, dt \right)_{n \geq 0}$ necessarily also stationary (where $g$ is a bounded measurable function)??

(Philosophically, the following question is of a similar flavour to A stochastic process that is 1st and 2nd order (strictly) stationary, but not 3rd order stationary, but more "advanced".)

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, let $(E,\mathcal{E})$ be a measurable space, and let $X:[0,\infty) \times \Omega \to E$ be a measurable function with the property that for any $n \in \mathbb{N}$, for all $\tau,t_1,\ldots,t_n \geq 0$ the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1,\omega),\ldots,X(t_n,\omega) \right)$

is equal to the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1+\tau,\omega),\ldots,X(t_n+\tau,\omega) \right)$.

Is it necessarily the case that for any $n \in \mathbb{N}$, for all $A \in \mathcal{E}^{\otimes n}$ and all $s,\tau,t_1,\ldots,t_n \geq 0$, the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [0,s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

is equal to the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [\tau,\tau+s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

(where $\textrm{Leb}$ denotes the Lebesgue measure)?

How about in the special case that $(E,\mathcal{E})=(\mathbb{R},\mathcal{B}(\mathbb{R}))$?

Now I've asked the above question in quite a general form, but what I ultimately want is: if $(X_t)_{t \geq 0}$ is a measurable stationary stochastic process in continuous time, then is the discrete-time stochastic process $\left( \int_n^{n+1} g(X_t) \, dt \right)_{n \geq 0}$ necessarily also stationary (where $g$ is a bounded measurable function)??

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(Philosophically, the following question is of a similar flavour to A stochastic process that is 1st and 2nd order (strictly) stationary, but not 3rd order stationary, but more "advanced".)

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, let $(E,\mathcal{E})$ be a measurable space, and let $X:[0,\infty) \times \Omega \to E$ be a measurable function with the property that for any $n \in \mathbb{N}$, for all $\tau,t_1,\ldots,t_n \geq 0$ the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1,\omega),\ldots,X(t_n,\omega) \right)$

is equal to the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1+\tau,\omega),\ldots,X(t_n+\tau,\omega) \right)$.

Is it necessarily the case that for any $n \in \mathbb{N}$, for all $A \in \mathcal{E}^{\otimes n}$ and all $s,\tau,t_1,\ldots,t_n \geq 0$, the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [0,s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

is equal to the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [\tau,\tau+s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

(where $\textrm{Leb}$ denotes the Lebesgue measure)?

How about in the special case that $(E,\mathcal{E})=(\mathbb{R},\mathcal{B}(\mathbb{R}))$?

Now I've asked the above question in quite a general form, but what I ultimately want is: if $(X_t)_{t \geq 0}$ is a measurable stationary stochastic process in continuous time, then is the discrete-time stochastic process $\left( \int_n^{n+1} g(X_t) \, dt \right)_{n \geq 0}$ necessarily also stationary (where $g$ is a bounded measurable function)??

(Philosophically, the following question is of a similar flavour to A stochastic process that is 1st and 2nd order (strictly) stationary, but not 3rd order stationary, but more "advanced".)

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, let $(E,\mathcal{E})$ be a measurable space, and let $X:[0,\infty) \times \Omega \to E$ be a measurable function with the property that for any $n \in \mathbb{N}$, for all $\tau,t_1,\ldots,t_n \geq 0$ the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1,\omega),\ldots,X(t_n,\omega) \right)$

is equal to the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1+\tau,\omega),\ldots,X(t_n+\tau,\omega) \right)$.

Is it necessarily the case that for any $n \in \mathbb{N}$, for all $A \in \mathcal{E}^{\otimes n}$ and all $s,\tau,t_1,\ldots,t_n \geq 0$, the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [0,s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

is equal to the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [\tau,\tau+s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

(where $\textrm{Leb}$ denotes the Lebesgue measure)?

How about in the special case that $(E,\mathcal{E})=(\mathbb{R},\mathcal{B}(\mathbb{R}))$?

(Philosophically, the following question is of a similar flavour to A stochastic process that is 1st and 2nd order (strictly) stationary, but not 3rd order stationary, but more "advanced".)

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, let $(E,\mathcal{E})$ be a measurable space, and let $X:[0,\infty) \times \Omega \to E$ be a measurable function with the property that for any $n \in \mathbb{N}$, for all $\tau,t_1,\ldots,t_n \geq 0$ the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1,\omega),\ldots,X(t_n,\omega) \right)$

is equal to the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1+\tau,\omega),\ldots,X(t_n+\tau,\omega) \right)$.

Is it necessarily the case that for any $n \in \mathbb{N}$, for all $A \in \mathcal{E}^{\otimes n}$ and all $s,\tau,t_1,\ldots,t_n \geq 0$, the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [0,s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

is equal to the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [\tau,\tau+s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

(where $\textrm{Leb}$ denotes the Lebesgue measure)?

How about in the special case that $(E,\mathcal{E})=(\mathbb{R},\mathcal{B}(\mathbb{R}))$?

Now I've asked the above question in quite a general form, but what I ultimately want is: if $(X_t)_{t \geq 0}$ is a measurable stationary stochastic process in continuous time, then is the discrete-time stochastic process $\left( \int_n^{n+1} g(X_t) \, dt \right)_{n \geq 0}$ necessarily also stationary (where $g$ is a bounded measurable function)??

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Is it true that all stationary measurable stochastic processes are "measurably stationary"?

(Philosophically, the following question is of a similar flavour to A stochastic process that is 1st and 2nd order (strictly) stationary, but not 3rd order stationary, but more "advanced".)

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, let $(E,\mathcal{E})$ be a measurable space, and let $X:[0,\infty) \times \Omega \to E$ be a measurable function with the property that for any $n \in \mathbb{N}$, for all $\tau,t_1,\ldots,t_n \geq 0$ the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1,\omega),\ldots,X(t_n,\omega) \right)$

is equal to the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1+\tau,\omega),\ldots,X(t_n+\tau,\omega) \right)$.

Is it necessarily the case that for any $n \in \mathbb{N}$, for all $A \in \mathcal{E}^{\otimes n}$ and all $s,\tau,t_1,\ldots,t_n \geq 0$, the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [0,s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

is equal to the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [\tau,\tau+s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

(where $\textrm{Leb}$ denotes the Lebesgue measure)?

How about in the special case that $(E,\mathcal{E})=(\mathbb{R},\mathcal{B}(\mathbb{R}))$?