In the book probabilistic robotics from Thrun there is the following definition of conditional independence ("we can condition the rule for combining probabilities of independent random variables ($p(x,y) = p(x)p(y)$) on other variables $z$"):
$$ p(x,y | z) = p(x|z)p(y|z) $$
And it is mentioned that this equation is equivalent to the two following ones:
$$ p(x|z) = p(x|z,y) $$
$$ p(y|z) = p(y|z,x) $$
Can someone please explain how the first equation is equivalent to these two equations? And which variables are independent of each other? Are all ($x$, $y$ and $z$) independent of each other or are only $x$ and $y$ independent of each other in case $z$ is given? Or are $x$ and $y$ independent of $z$?
I think what is important to remember is that the conditional probability of two independent variables $X$ and $Y$ is the following:
$$ p(x|y) = \frac{p(x) p(y)}{p(y)} = p(x) $$
In other words, if X and Y are independent, $Y$ tells us nothing about the value of $X$. However, I don't know how to infer the conditional independence equation. How is $p(x,y|z)$ equivalent to $p(x|z)p(y|z)$ in the first place?
I tried the following way:
$$ p(x,y|z) = \frac{p(x,y)p(z)}{p(z)} = p(x,y) $$
Assuming $x$ and $y$ are independent:
$$ p(x,y|z) = p(x,y) = p(x)p(y) $$
which obviously doesn't match $p(x|z)p(y|z)$.
Can someone please explain why $p(x,y|z)$ equals $p(x|z)p(y|z)$? And which variables are independent of each other, I didn't understand it from the book.
Edit
I tried a different way:
With the definition of conditional probability $p(x|y) = \frac{p(x,y)}{p(y)}$:
$$ p(x|y,z) = \frac{p(x,y,z)}{p(y,z)} = \frac{p((x,y),z)}{p(y,z)} $$
with the multiplication rule $p(x,y) = p(y|x)p(x) = p(x|y)p(y)$ follows:
$$ p(x|y,z) = \frac{p((x,y),z)}{p(y,z)} = \frac{p((x,y)|z)p(z)}{p(y|z)p(z)} = \frac{p((x,y)|z)}{p(y|z)} $$
However, I don't know how to continue or where I made a mistake.
Using the multiplication rule $P(x,y,z) = P(x,y|z)P(z)$ we first note that:
(1)
$$ P(x,y|z) = \frac{P(x,y,z)}{P(z)} = \frac{P(x,y,z)}{P(y,z)} \cdot \frac{P(y,z)}{P(z)} = P(x|y,z) \cdot P(y|z) $$
(2)
$$ P(x,y|z) = \frac{P(x,y,z)}{P(z)} = \frac{P(x,y,z)}{P(x,z)} \cdot \frac{P(x,z)}{P(z)} = P(y|x,z) \cdot P(x|z) $$
Now we assume that (2.17) holds, i.e.,
$$ P(x,y|z) = P(x|z) P(y|z) $$
Using (1) we have:
$$ P(x|z) P(y|z) = P(x,y|z) = P(x|y,z) \cdot P(y|z) \\ \Rightarrow P(x|z) = P(x|y,z) $$
which is (2.18)
Using (2) we have:
$$ P(x|z) P(y|z) = P(x,y|z) = P(y|x,z) \cdot P(x|z) \\ \Rightarrow P(y|z) = P(y|x,z) $$
which is (2.19)
Hence, (2.18) and (2.19) hold.
Now assume that (2.18) and (2.19) hold, i.e.,
$$ P(x|z) = P(x|y,z) \\ P(y|z) = P(y|x,z) $$
Then using (1) and inserting (2.18),
$$ P(x,y|z) = P(x|y,z) \cdot P(y|z)\\ = P(x|z) P(y|z) $$
Therefore (2.17) holds.
Remark: Can you prove that (2.18) and (2.19) are equivalent to each other?
$$ P(x|z) = P(x|y,z) = \frac{P(x,y,z)}{P(y,z)} = \frac{P(x,y,z)}{P(y|z)P(z)} \\ P(y|z) = P(y|x,z) = \frac{P(x,y,z)}{P(x,z)} = \frac{P(x,y,z)}{P(x|z)P(z)} $$
$$ P(x,y,z) = P(x,y,z)\\ P(x|z)P(y|z)P(z) = P(y|z)P(x|z)P(z) \\ 1 = 1 $$
$x$ and $y$ are independent of each other in case $z$ is given, or we can sya, $x$ and $y$ are independent of each other with respect to z. Basically, (2.18) tells us that $P(x|y,z)$ does not depend on value of $y$, and similarly, (2.19) tells us that $P(y|z,x)$ does not depend on value of $x$. In other words, $x$ and $y$ carry no information about each other if another variable's value $z$ is known.
For an intuitive example, let $x$ be teh height of a child, while $y$ be the number of words that the child knows. $x$ and $y$ are not independent. In genreal, if $x$ is high, $y$ is high too. But, if we introduce another variable $z$ as a child's age, then the apparent relation between $x$ and $y$ dissapears. If. for example, $z=3$ years, then all children have approximately the same heigh, while the number of words can vary and dose not depend on heigt.
Why is (2.17) correct in the first place? Again using (1),
$$ P(x,y|z) = \frac{P(x,y,z)}{P(z)} = \frac{P(x,y,z)}{P(y,z)} \cdot \frac{P(y,z)}{P(z)} = P(x|y,z) \cdot P(y|z) $$
and remembering that $x$ and $y$ are independent of each other when $z$ is given, then using the conditional probability of independent variables (2.10) $P(x|y) = \frac{P(x)P(y)}{P(y)} = P(x)$ for $P(x|y,z) = P(x|z)$ we have,
$$ P(x,y|z) = P(x|y,z) \cdot P(y|z) = P(x|z) \cdot P(y|z) $$