Markov property for mixed joint densities
$begingroup$
Let $X rightarrow Y rightarrow Z$ be a Markov chain in that order, $X$ and $Y$ be jointly Gaussian, and $Z$ be a discrete random variable with finite alphabet $mathcal{Z}$.
Denote their mixed joint density as $f(X,Y,Z)$. Is it true that, due to the Markov chain assumption, $f(X,Y,Z) = f(X)f(Y|X)f(Z|Y)$? That is, $f(Z|X,Y) = f(Z|Y)$?
I am following the notation here: https://en.wikipedia.org/wiki/Joint_probability_distribution
I know this is true when all the random variables are discrete but since here I have a mixture, I am not sure weather the above make sense.
Thanks for reading me!
probability probability-theory probability-distributions markov-chains markov-process
asked Dec 6 '18 at 5:26
OliVerOliVer
617
$endgroup$
add a comment |
$begingroup$
Let $X rightarrow Y rightarrow Z$ be a Markov chain in that order, $X$ and $Y$ be jointly Gaussian, and $Z$ be a discrete random variable with finite alphabet $mathcal{Z}$.
Denote their mixed joint density as $f(X,Y,Z)$. Is it true that, due to the Markov chain assumption, $f(X,Y,Z) = f(X)f(Y|X)f(Z|Y)$? That is, $f(Z|X,Y) = f(Z|Y)$?
I am following the notation here: https://en.wikipedia.org/wiki/Joint_probability_distribution
I know this is true when all the random variables are discrete but since here I have a mixture, I am not sure weather the above make sense.
Thanks for reading me!
probability probability-theory probability-distributions markov-chains markov-process
asked Dec 6 '18 at 5:26
OliVerOliVer
617
$endgroup$
add a comment |
$begingroup$
Let $X rightarrow Y rightarrow Z$ be a Markov chain in that order, $X$ and $Y$ be jointly Gaussian, and $Z$ be a discrete random variable with finite alphabet $mathcal{Z}$.
Denote their mixed joint density as $f(X,Y,Z)$. Is it true that, due to the Markov chain assumption, $f(X,Y,Z) = f(X)f(Y|X)f(Z|Y)$? That is, $f(Z|X,Y) = f(Z|Y)$?
I am following the notation here: https://en.wikipedia.org/wiki/Joint_probability_distribution
I know this is true when all the random variables are discrete but since here I have a mixture, I am not sure weather the above make sense.
Thanks for reading me!
probability probability-theory probability-distributions markov-chains markov-process
asked Dec 6 '18 at 5:26
OliVerOliVer
617
$endgroup$
Let $X rightarrow Y rightarrow Z$ be a Markov chain in that order, $X$ and $Y$ be jointly Gaussian, and $Z$ be a discrete random variable with finite alphabet $mathcal{Z}$.
Denote their mixed joint density as $f(X,Y,Z)$. Is it true that, due to the Markov chain assumption, $f(X,Y,Z) = f(X)f(Y|X)f(Z|Y)$? That is, $f(Z|X,Y) = f(Z|Y)$?
I am following the notation here: https://en.wikipedia.org/wiki/Joint_probability_distribution
I know this is true when all the random variables are discrete but since here I have a mixture, I am not sure weather the above make sense.
Thanks for reading me!
probability probability-theory probability-distributions markov-chains markov-process
probability probability-theory probability-distributions markov-chains markov-process
asked Dec 6 '18 at 5:26
OliVerOliVer
617
asked Dec 6 '18 at 5:26
OliVerOliVer
617
edited Dec 6 '18 at 5:46
OliVer
asked Dec 6 '18 at 5:26
OliVerOliVer
617
asked Dec 6 '18 at 5:26
OliVerOliVer
617
asked Dec 6 '18 at 5:26
OliVerOliVer
617
617
add a comment |
add a comment |
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