Conditioning random variables

Question

I have two issues that seem to be related.

1) Suppose we have two random variables $A$ and $B$ that are conditioned by $X$. If we want to calculate $$ \frac{p(A, B | X)}{p(B | X)},$$ do we always need to use Bayes rule and get $p(A|X, B)$? Since both are conditioned on $X$, why can't we say $p(A|X)$?

2) Suppose $Y \sim \rm{Normal}(0,1)$. I think $\mathbb{E}[Y]$ is 0, but does it change the value if we calculate $\mathbb{E}[Y|Y]$, is it $\mathbb{E}[Y|Y] = Y$? I cannot understand why conditioning changes its expectation.

Answer 1

1. Starting with $$ p\left(A\middle|B\right) = \dfrac{p\left(A,B\right)}{p\left(B\right)} $$ and conditioning everything on $X$ gives $$ p\left(A\middle|B, X\right) = \dfrac{p\left(A,B\middle| X\right)}{p\left(B\middle| X\right)} $$ So $p\left(A\middle|X\right) \neq \dfrac{p\left(A,B\middle| X\right)}{p\left(B\middle| X\right)}$ in the same way that $ p\left(A\right) \neq \dfrac{p\left(A,B\right)}{p\left(B\right)} $.
2. For the particular case $\mathbb{E}\left[Y\middle|Y\right] = Y$, I like to reason about this as follows. If I gave you the value of $Y$ and asked what you expected the value of $Y$ to be, what would you answer? Well of course the expectation should be whatever the value I just gave you, which is $Y$.