# How do you explain $f(x_4|x_3)f(x_3|x_2)f(x_2|x_1)f(x_1) = f(x_4,x_3,x_2,x_1)$?

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.

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

• 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. – hkBattousai Mar 23 '13 at 7:53

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!