Given a joint probability density function over the random vector $=(1,2,3,4)$ that factorizes as $(1,2,3,4)=(1,4|2)(2,3|1)$
show that the following two statements hold
- $X_{1} \bot X_{2}$
- $X_{3} \bot X_{4} | X_{1}, X_{2}$
so far i have come up with the following.
1: \begin{equation} p(x_{1},x_{2}) = p(x_{1}) * p(x_{2}) \tag{1} \end{equation}
\begin{equation} p(x_{1} | x_{2}) = p(x_{1}) \tag{2} \end{equation} \begin{equation} p(x_{2} | x_{1}) = p(x_{2}) \tag{3} \end{equation}
\begin{equation} \begin{split} p(x_{1},x_{2}) & = \sum_{x_{3}}\sum_{x_{4}}p(x_{1},x_{2},x_{3},x_{4}) \\ & = \sum_{x_{3}}\sum_{x_{4}}p(x_{1},x_{4} | x_{2}) p(x_{2},x_{3} | x_{1}) \\ & = \sum_{x_{4}}p(x_{1},x_{4} | x_{2}) \sum_{x_{3}}p(x_{2},x_{3} | x_{1}) \\ & = p(x_{1} | x_{2}) p(x_{2} | x_{1}) \\ \\ p(x_{1},x_{2}) & = p(x_{1} | x_{2}) * p(x_{2}) = p(x_{2} | x_{1}) * p(x_{1}) \end{split} \end{equation} I am unsure if this is correct, and if it is how i should continue.
2. \begin{equation} \begin{split} P(x_{3},x_{4}|x_{1},x_{2}) &= \frac{P(x_3,x_4,x_1,x_2)}{P(x_1,x_2)}\\ & = \frac{P(x_1, x_4 | x_2) P(x_2, x_3 | x_1)}{P(x_1 | x_2)P(x_2 | x_1)}\\ \end{split} \end{equation}
Unsure if this correct and how i should continue.
In general i would like to know if i am on the right track with this exercise, and how to finish it. I am also unsure if i need to prove (1),(2) and (3) or if just one of them suffices.
For the first part, the last line in the derivation should be $p\left(x_1|x_2\right)p\left(x_2|x_1\right)$ since you have integrated $x_4$ out of the expression. So, what you have derived is
\begin{eqnarray*} p\left(x_1,x_2\right) &=& p\left(x_1|x_2\right)p\left(x_2|x_1\right)\\ &=& \frac{p\left(x_1,x_2\right)^2}{p\left(x_2\right)p\left(x_1\right)} \end{eqnarray*}
whence $p\left(x_1,x_2\right) = p\left(x_2\right)p\left(x_1\right)$ as required.
For the second part, observe that
\begin{eqnarray*} p\left(x_1,x_4|x_2\right) &=& p\left(x_4|x_1,x_2\right)p\left(x_1|x_2\right)\\ p\left(x_2,x_3|x_1\right) &=& p\left(x_3|x_1,x_2\right)p\left(x_2|x_1\right) \end{eqnarray*}
which together with the previous result will give you what you need.