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Let $x_1=x(n_1)$, $x_2=x(n_2)$, $x_3=x(n_3)$ and $x_4=x(n_4)$ be random Markov processes $(n_1 < n_2 < n_3 < n_4)$.

I don't understand the identity given below on their probability density functions.
Where does it come from. How is it derived? Please explain it clearly.

$$ \begin{array}{rcl} f(x_4|x_3)f(x_3|x_2)f(x_2|x_1)f(x_1) & = & f(x_4|x_3)f(x_3|x_2)f(x_2,x_1) \\ & = & f(x_4|x_3)f(x_3,x_2,x_1) \\ & = & f(x_4,x_3,x_2,x_1) \\ \end{array} $$

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  • $\begingroup$ I added the detail that the processes are Markov. I'm sorry for not mentioning it earlier. I didn't know that they system being Markov is essential for this identity. $\endgroup$ – hkBattousai Mar 23 '13 at 7:53
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This is to address your revised question -- the Markov Chain, where $x_4$ depends only on $x_3$, $x_3$ on $x_2$...

In general, $f(x_4,x_3,x_2,x_1) = f(x_4|x_3,x_2,x_1)f(x_3,x_2,x_1)$, which means $x_4$ depends on all the $x_3,x_2,x_1$. Obviously, one can decompose $f(x_3,x_2,x_1)$ recursively until $x_1$ and finally one arrives at full decomposition (in a lengthy form).

However, in case of Markov Chain, we have $x_4$ depends only on $x_3$ (instead of all the $x_3,x_2,x_1$). This is to say $f(x_4|x_3,x_2,x_1) = f(x_4|x_3)$. Noting this, decompose recursively and you get precisely the identity in your question!

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This equation, which in essence a way to decompose the Density Function, does not always hold! As it implies, the variables here are actually conditionally dependent such that their relationships can be captured by a Directed Acyclic Graph. You can refer to any textbook on probabilistic Graphic Model (e.g., the so-called PRML)

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  • $\begingroup$ BTW, according to the form of your equation, I believe it's a Markov Chain. $\endgroup$ – pengsun.thu Mar 23 '13 at 7:44
  • $\begingroup$ The first identity in the series of three always holds, actually, but none of the two latter ones holds in general, as you explain. $\endgroup$ – Did Mar 23 '13 at 7:45
  • $\begingroup$ @pengsun.thu Yes it was a Markov Chain. I edited my question. Please read it again. $\endgroup$ – hkBattousai Mar 23 '13 at 7:56

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