Conditioning a variable on itself and some other variable I am having trouble understanding the implications for this kind of conditional probability:
$$E(X \mid X , Y).$$
What does the above simplify to?
 A: It is conditional expectation (not probability), and $E[X|X,Y]=X$ because $X$ is already given.
A: What may be tripping you up here is a common imprecision in notation, where people (myself included) will use the same symbol to denote both a random variable, and a particular assignment or instantiation of that variable.
I wonder if things will become clearer to you if we rewrite your expectation more precisely:
$$
E(X|X=x,Y=y)
$$
where $x$ and $y$ are the values of $X$ and $Y$ that we condition on. That is, we're calculating the expected value of $X$ given that we know that the random variable $X$ has value $x$, and $Y$ has value $y$.
Hang on, you might say, we already know the value of $X$? Exactly. So the expected value is very simple: it is the value of $X$ that we already know:
$$
E(X|X=x,Y=y)=x
$$
And obviously $Y$ becomes irrelevant - since we already know $X$ there is no information that any other variable can give us about its value.
(This may seem a little silly, because $x$ is still a placeholder for an unknown value in this equation, but at the same time it represents a "known" value of $X$. As is typical in maths, we're using variables as stand-ins for values that we could fill in. It just gets a little more gnarly when you're dealing with random variables, which are not only unknown, but do not have a definite value. $X$ here is the random variable, which is the outcome of a random phenomenon (e.g. the roll of a die). $X$ has a distribution, expected values, etc. $x$ is a particular value taken by $X$, and does not have a distribution - it just represents that particular value.)
