Stochastic processes with same distribution, which are no modification
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Consider stochastic processes $mathcal{X}$ and $mathcal{Y}$. How can I find an example in which $mathcal{X}$ and $mathcal{Y}$ having the same distribution but are no modification of each other? I think I understood the concept of modification. But how do I find the distributions for $mathcal{X}$ and $mathcal{Y}$? I know that there can be defined a distribution by kolmogorov's extension theorem, but I wonder how this should look like. Any hint or help will be appreciated. Thanks in advance!
probability probability-theory stochastic-processes
asked Dec 4 '18 at 0:14
user408858user408858
490111
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add a comment |
$begingroup$
Consider stochastic processes $mathcal{X}$ and $mathcal{Y}$. How can I find an example in which $mathcal{X}$ and $mathcal{Y}$ having the same distribution but are no modification of each other? I think I understood the concept of modification. But how do I find the distributions for $mathcal{X}$ and $mathcal{Y}$? I know that there can be defined a distribution by kolmogorov's extension theorem, but I wonder how this should look like. Any hint or help will be appreciated. Thanks in advance!
probability probability-theory stochastic-processes
asked Dec 4 '18 at 0:14
user408858user408858
490111
$endgroup$
$begingroup$
Let X be a stationary Gaussian process with a given power spectrum. Let Y be another stationary Gaussian process with the same power spectrum. They meet your requirement. Am I missing something?
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– herb steinberg
Dec 4 '18 at 1:03
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If $(B_t)_t$ is a Brownian motion, you can consider $X_t = B_t$ and $Y_t = - B_t$.
$endgroup$
– saz
Dec 4 '18 at 7:59
add a comment |
$begingroup$
Consider stochastic processes $mathcal{X}$ and $mathcal{Y}$. How can I find an example in which $mathcal{X}$ and $mathcal{Y}$ having the same distribution but are no modification of each other? I think I understood the concept of modification. But how do I find the distributions for $mathcal{X}$ and $mathcal{Y}$? I know that there can be defined a distribution by kolmogorov's extension theorem, but I wonder how this should look like. Any hint or help will be appreciated. Thanks in advance!
probability probability-theory stochastic-processes
asked Dec 4 '18 at 0:14
user408858user408858
490111
$endgroup$
Consider stochastic processes $mathcal{X}$ and $mathcal{Y}$. How can I find an example in which $mathcal{X}$ and $mathcal{Y}$ having the same distribution but are no modification of each other? I think I understood the concept of modification. But how do I find the distributions for $mathcal{X}$ and $mathcal{Y}$? I know that there can be defined a distribution by kolmogorov's extension theorem, but I wonder how this should look like. Any hint or help will be appreciated. Thanks in advance!
probability probability-theory stochastic-processes
probability probability-theory stochastic-processes
asked Dec 4 '18 at 0:14
user408858user408858
490111
asked Dec 4 '18 at 0:14
user408858user408858
490111
asked Dec 4 '18 at 0:14
user408858user408858
490111
asked Dec 4 '18 at 0:14
user408858user408858
490111
asked Dec 4 '18 at 0:14
user408858user408858
490111
490111
$begingroup$
Let X be a stationary Gaussian process with a given power spectrum. Let Y be another stationary Gaussian process with the same power spectrum. They meet your requirement. Am I missing something?
$endgroup$
– herb steinberg
Dec 4 '18 at 1:03
$begingroup$
If $(B_t)_t$ is a Brownian motion, you can consider $X_t = B_t$ and $Y_t = - B_t$.
$endgroup$
– saz
Dec 4 '18 at 7:59
add a comment |
$begingroup$
Let X be a stationary Gaussian process with a given power spectrum. Let Y be another stationary Gaussian process with the same power spectrum. They meet your requirement. Am I missing something?
$endgroup$
– herb steinberg
Dec 4 '18 at 1:03
$begingroup$
If $(B_t)_t$ is a Brownian motion, you can consider $X_t = B_t$ and $Y_t = - B_t$.
$endgroup$
– saz
Dec 4 '18 at 7:59
$begingroup$
Let X be a stationary Gaussian process with a given power spectrum. Let Y be another stationary Gaussian process with the same power spectrum. They meet your requirement. Am I missing something?
$endgroup$
– herb steinberg
Dec 4 '18 at 1:03
$begingroup$
Let X be a stationary Gaussian process with a given power spectrum. Let Y be another stationary Gaussian process with the same power spectrum. They meet your requirement. Am I missing something?
$endgroup$
– herb steinberg
Dec 4 '18 at 1:03
$begingroup$
If $(B_t)_t$ is a Brownian motion, you can consider $X_t = B_t$ and $Y_t = - B_t$.
$endgroup$
– saz
Dec 4 '18 at 7:59
$begingroup$
If $(B_t)_t$ is a Brownian motion, you can consider $X_t = B_t$ and $Y_t = - B_t$.
$endgroup$
– saz
Dec 4 '18 at 7:59
add a comment |
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$begingroup$
Let X be a stationary Gaussian process with a given power spectrum. Let Y be another stationary Gaussian process with the same power spectrum. They meet your requirement. Am I missing something?
$endgroup$
– herb steinberg
Dec 4 '18 at 1:03
$begingroup$
If $(B_t)_t$ is a Brownian motion, you can consider $X_t = B_t$ and $Y_t = - B_t$.
$endgroup$
– saz
Dec 4 '18 at 7:59