If X is independent to Y and Z, does it imply that X is independent to YZ ?

You can see it as having information on $Y$ is not enough to determine $X$ an identically with information on $Z$ however if you have combined informations it could determined $X$.

An example I came up with is (may be not a good one but it helps me understand whats going on).

You have a set of peoples on with you can know the colours of the hair, given by the variable $X$, there first name given by $Y$, and there last name given by $Z$.

The independence of $X$ and $Y$ tells you that even if you know there first name that does not help you in knowing the colours of there hairs. Intuitively it says that the color of the hair does not depend on your name.

The same with the last name.

However if you know both the first name and last name you know which person it is exactly hence you know the color of its hairs.

I hope this example is not confusing you :S

Short answer: No, $X \perp Y , X \perp Z$ doesn't imply $X \perp YZ$

Let's say you do an experiment where you choose a number randomly from:

{1,2,3,4,6,7,8,9}

Let X = 1 if:

(Your chosen number is even AND less than five) OR (your chosen number is odd AND greater than 5)

and 0 otherwise

Let Y = 1 if your chosen number is even, 0 otherwise Let Z = 1 if your chosen number is greater than 5, 0 otherwise

Now we know X = 1 with probability - 0.5 (2,4,7,9) and 0 with probability 0.5 (1,3,6,8)

If we know Y = 1, X is still 1 with p = 0.5 (2,4) and 0 with p = 0.5 (6,8) If we know Y = 0, X is still 1 with p = 0.5 (7,9) and 0 with p = 0.5 (1,3)

If we know Z = 1, X is still 1 with p = 0.5 (7,9) and 0 with p = 0.5 (6,8) If we know Z = 0, X is still 1 with p = 0.5 (2,4) and 0 with p = 0.5 (1,3)

So X is independent of Y and X is independent of Z.

But knowing if a number is even AND knowing if it's greater than 5 (Y & Z), makes us know X with certainty. e.g. Y = 1, Z = 1, then YZ = 1, X has to be 0 with probability 1 (as X is 0 if the number is an even number > 5)

$P(X = 0 | Y ) = P(X = 0) = 0.5$ (same for X = 1)

$P(X = 0 | Z ) = P(X = 0) = 0.5$ (same for X = 1)

But

$P(X = 0 | YZ = 1) = 1 \neq P(X=0) $ (similar for X = 1)