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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.

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    $\begingroup$ For #2, you may know that E[Y] has mean zero based on the distribution, but conditioning will change the expectation. If you know that $Y=c$ for some c, then of course, $E[Y|Y=c]=E[c]=c$. In a sense, it is trivial. Knowing that $Y=c$, you wouldn't expect Y to be zero any more. $\endgroup$ – bob Nov 27 '18 at 4:55
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  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 $Y$.
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