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By Pólya’s theorem, any even real-valued function $f$ on $\mathbb R$ with $f(\infty-)=0$ which is convex on $[0,\infty)$ is positive definite. So, any such function is the (auto)covariance function of a stationary Gaussian process; see e.g. Section "Properties of the Autocovariance Function", page 2.

Now just take any two different functions, $f_1$ and $f_2$, of the Pólya class such that $f_2(t)=1-|t|=f_2(t)$ for $|t|\le1/2$. Then the corresponding stationary Gaussian processes, say $(X_{1,t})$ and $(X_{2,t})$, with the covariance functions $f_1$ and $f_2$ will have different distributions. Therefore, these two processes will be different from each other.

To be more specific, note first here that, by vertical and horizontal re-scaling, without loss of generality $a_1=a_2=1$, so that $$EX_sX_t=1-|t-s|\quad\text{if}\quad|t-s|\le u, \tag{1}$$ where $u\in(0,1)$. Let then $$Y_t:=B_{t+1}-B_t=\int_t^{t+1}dB_s,$$ where $(B_t)_{t\in\mathbb R}$ is the standard Brownian motion with $B_0=0$. Then $$EY_sY_t=1-|t-s|\quad\text{if}\quad|t-s|\le 1,$$ so that $$EY_sY_t=EX_sX_t\quad\text{if}\quad|t-s|\le u,$$ as desired.

For more examples, take any $h\in(0,1)$ and let $$U_t:=\frac1{\sqrt2}\,(Y_{(1-h)t}+Z_{(1+h)t}),$$ where $(Z_t)$ is an independent copy of the Gaussian process $(Y_t)$. Then $$EU_sU_t=1-|t-s|=EY_sY_t \quad\text{if}\quad|t-s|\le1/(1+h)$$ and hence $$EU_sU_t=EX_sX_t \quad\text{if}\quad|t-s|\le\min[u,1/(1+h)],$$ as desired.

By Pólya’s theorem, any even real-valued function $f$ on $\mathbb R$ with $f(\infty-)=0$ which is convex on $[0,\infty)$ is positive definite. So, any such function is the (auto)covariance function of a stationary Gaussian process; see e.g. Section "Properties of the Autocovariance Function", page 2.

Now just take any two different functions, $f_1$ and $f_2$, of the Pólya class such that $f_2(t)=1-|t|=f_2(t)$ for $|t|\le1/2$. Then the corresponding stationary Gaussian processes, say $(X_{1,t})$ and $(X_{2,t})$, with the covariance functions $f_1$ and $f_2$ will have different distributions. Therefore, these two processes will be different from each other.

To be more specific, note first here that, by vertical and horizontal re-scaling, without loss of generality $a_1=a_2=1$, so that $$EX_sX_t=1-|t-s|\quad\text{if}\quad|t-s|\le u, \tag{1}$$ where $u\in(0,1)$. Let then $$Y_t:=B_{t+1}-B_t=\int_t^{t+1}dB_s,$$ where $(B_t)_{t\in\mathbb R}$ is the standard Brownian motion with $B_0=0$. Then $$EY_sY_t=1-|t-s|\quad\text{if}\quad|t-s|\le 1$$ (with $EY_sY_t=0$ if $|t-s|>1$), so that $$EY_sY_t=EX_sX_t\quad\text{if}\quad|t-s|\le u,$$ as desired.

For more examples, take any $h\in(0,1)$ and let $$U_t:=\frac1{\sqrt2}\,(Y_{(1-h)t}+Z_{(1+h)t}),$$ where $(Z_t)$ is an independent copy of the Gaussian process $(Y_t)$. Then $$EU_sU_t=1-|t-s|=EY_sY_t \quad\text{if}\quad|t-s|\le1/(1+h)$$ and hence $$EU_sU_t=EX_sX_t \quad\text{if}\quad|t-s|\le\min[u,1/(1+h)],$$ as desired.

By Pólya’s theorem, any even real-valued function $f$ on $\mathbb R$ with $f(\infty-)=0$ which is convex on $[0,\infty)$ is positive definite. So, any such function is the (auto)covariance function of a stationary Gaussian process; see e.g. Section "Properties of the Autocovariance Function", page 2.

Now just take any two different functions, $f_1$ and $f_2$, of the Pólya class such that $f_2(t)=1-|t|=f_2(t)$ for $|t|\le1/2$. Then the corresponding stationary Gaussian processes, say $(X_{1,t})$ and $(X_{2,t})$, with the covariance functions $f_1$ and $f_2$ will have different distributions. Therefore, these two processes will be different from each other.

To be more specific, note first here that, by vertical and horizontal re-scaling, without loss of generality $a_1=a_2=1$, so that $$EX_sX_t=1-|t-s|\quad\text{if}\quad|t-s|\le u, \tag{1}$$ where $u\in(0,1)$. Let then $$Y_t:=B_{t+1}-B_t=\int_t^{t+1}dB_s,$$ where $(B_t)_{t\in\mathbb R}$ is the standard Brownian motion with $B_0=0$. Then $$EY_sY_t=1-|t-s|\quad\text{if}\quad|t-s|\le 1, \tag{2}$$ as desired.

By Pólya’s theorem, any even real-valued function $f$ on $\mathbb R$ with $f(\infty-)=0$ which is convex on $[0,\infty)$ is positive definite. So, any such function is the (auto)covariance function of a stationary Gaussian process; see e.g. Section "Properties of the Autocovariance Function", page 2.

Now just take any two different functions, $f_1$ and $f_2$, of the Pólya class such that $f_2(t)=1-|t|=f_2(t)$ for $|t|\le1/2$. Then the corresponding stationary Gaussian processes, say $(X_{1,t})$ and $(X_{2,t})$, with the covariance functions $f_1$ and $f_2$ will have different distributions. Therefore, these two processes will be different from each other.

To be more specific, note first here that, by vertical and horizontal re-scaling, without loss of generality $a_1=a_2=1$, so that $$EX_sX_t=1-|t-s|\quad\text{if}\quad|t-s|\le u, \tag{1}$$ where $u\in(0,1)$. Let then $$Y_t:=B_{t+1}-B_t=\int_t^{t+1}dB_s,$$ where $(B_t)_{t\in\mathbb R}$ is the standard Brownian motion with $B_0=0$. Then $$EY_sY_t=1-|t-s|\quad\text{if}\quad|t-s|\le 1,$$ so that $$EY_sY_t=EX_sX_t\quad\text{if}\quad|t-s|\le u,$$ as desired.

For more examples, take any $h\in(0,1)$ and let $$U_t:=\frac1{\sqrt2}\,(Y_{(1-h)t}+Z_{(1+h)t}),$$ where $(Z_t)$ is an independent copy of the Gaussian process $(Y_t)$. Then $$EU_sU_t=1-|t-s|=EY_sY_t \quad\text{if}\quad|t-s|\le1/(1+h)$$ and hence $$EU_sU_t=EX_sX_t \quad\text{if}\quad|t-s|\le\min[u,1/(1+h)],$$ as desired.

By Pólya’s theorem, any even real-valued function $f$ on $\mathbb R$ with $f(\infty-)=0$ which is convex on $[0,\infty)$ is positive definite. So, any such function is the (auto)covariance function of a stationary Gaussian process; see e.g. Section "Properties of the Autocovariance Function", page 2.

Now just take any two different functions, $f_1$ and $f_2$, of the Pólya class such that $f_2(t)=1-|t|=f_2(t)$ for $|t|\le1/2$. Then the corresponding stationary Gaussian processes, say $(X_{1,t})$ and $(X_{2,t})$, with the covariance functions $f_1$ and $f_2$ will have different distributions. Therefore, these two processes will be different from each other.

By Pólya’s theorem, any even real-valued function $f$ on $\mathbb R$ with $f(\infty-)=0$ which is convex on $[0,\infty)$ is positive definite. So, any such function is the (auto)covariance function of a stationary Gaussian process; see e.g. Section "Properties of the Autocovariance Function", page 2.

Now just take any two different functions, $f_1$ and $f_2$, of the Pólya class such that $f_2(t)=1-|t|=f_2(t)$ for $|t|\le1/2$. Then the corresponding stationary Gaussian processes, say $(X_{1,t})$ and $(X_{2,t})$, with the covariance functions $f_1$ and $f_2$ will have different distributions. Therefore, these two processes will be different from each other.

To be more specific, note first here that, by vertical and horizontal re-scaling, without loss of generality $a_1=a_2=1$, so that $$EX_sX_t=1-|t-s|\quad\text{if}\quad|t-s|\le u, \tag{1}$$ where $u\in(0,1)$. Let then $$Y_t:=B_{t+1}-B_t=\int_t^{t+1}dB_s,$$ where $(B_t)_{t\in\mathbb R}$ is the standard Brownian motion with $B_0=0$. Then $$EY_sY_t=1-|t-s|\quad\text{if}\quad|t-s|\le 1, \tag{2}$$ as desired.