# Conditioning random variables

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.

• 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. – bob Nov 27 '18 at 4:55

## 1 Answer

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 $$Y$$.