Is it true that P(x|y,z)=P(x|y) if x and z be independent? Why? I know that if x and z be independent, P(xz) = P(x)P(z). 
I want to know if x and z be independent, can I cancel z from P(x|yz)? Why? 
Thanks. 
 A: $$P(x\mid y,z)=\frac{P(x,y,z)}{P(y,z)}=\frac{P(y\mid x,z)P(x)}{P(y\mid z)}\ne P(x\mid y)$$ in general unless $x\perp z\mid y$.
A: The answer is that you can't. Throw a 6-sided dice. Consider the three events:
$$X = \{1, 2\}$$
$$Y = \{1, 2, 3\}$$
$$Z = \{1, 4, 5\}$$
Then $X$ and $Z$ are independent. But :
$$\mathbb{P} (X|Y,Z) = \frac{\mathbb{P} (\{1\})}{\mathbb{P} (\{1\})} = 1,$$
$$\mathbb{P} (X|Y) = \frac{\mathbb{P} (\{1,2\})}{\mathbb{P} (\{1,2,3\})} = \frac{2}{3}.$$
A: No. Suppose you roll a (fair) $6$-sided die.
Let $x$ be the event of getting at least $5$, and let $z$ be the event of getting an odd number.  Now
$$
P(x,z) = P(\geq 5 \textrm{ and odd)} = P(\textrm{exactly }5) = 1/6 = (1/3)(1/2) = P(x)P(z),
$$
so $x$ and $z$ are independent.  (Alternatively, you could check that $P(x \mid z) = 1/3 = P(x)$, or that $P(z \mid x) = 1/2 = P(z)$.)
Now let $y$ be the event that you get at least $4$.  We see that
$$
P(\textrm{roll }\geq 5\mid \textrm{roll }\geq 4 \textrm{ and odd}) = 1,
$$
while
$$
P(\geq 5 \mid \geq 4) = 2/3
$$
A: Although other responses are useful, but here is another solution with different point of view:
$$
P(x|y,z)=\frac{P(x,y,z)}{P(y,z)}=\frac{P(x,y,z)}{\int_xP(x,y,z)dx}=\frac{P(y|x,z)P(x)P(z)}{\int_xP(y|x,z)P(x)P(z)dx}=\frac{P(y|x,z)P(x)}{\int_xP(y|x,z)P(x)dx}
$$
and the last term is obviously related to $z$, unless $y$ and $z$ be conditionally independent given $x$.
