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I am having trouble understanding the implications for this kind of conditional probability:

$$E(X \mid X , Y).$$

What does the above simplify to?

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    $\begingroup$ If you condition a random variable on itself then it becomes deterministic. Think of it like this: if I tell you what the value of X is then it becomes a known value and there is no more uncertainty. Hence any conditional probability or conditional expectation becomes trivial. $\endgroup$ – Gordon Smyth Aug 21 at 6:44
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    $\begingroup$ If you know X you know X...! $\endgroup$ – Xi'an Aug 21 at 7:19
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    $\begingroup$ How did you come across this conditional expectation? $\endgroup$ – Sextus Empiricus Aug 21 at 7:35
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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.)

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    $\begingroup$ Does random process mean random phenomenon / random experiment here? I am used to random process meaning a sequence of random variables such as an AR(1) process. $\endgroup$ – Richard Hardy Aug 22 at 8:15
  • $\begingroup$ Yes, thanks for pointing that out. I did not intend to invoke that particular meaning of random process, but rather the more colloquial meaning of "a process that is random". I will change it to "random phenomenon" to avoid any confusion. $\endgroup$ – Ruben van Bergen Aug 23 at 8:51
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    $\begingroup$ It's not imprecise to write $E(X\mid X, Y)$; this is not using $X$ to mean both a random variable and an instantiation of that variable, it is conditioning on $X$ as a random variable, which is perfectly well-defined (in this case $E(X\mid X, Y)$ is itself a random variable, not a number). In fact, that's the definition that usually comes first in formal treatments of the theory of conditioning, followed by the version where you condition on $X = x$, which is more theoretically subtle. $\endgroup$ – Artem Mavrin Aug 24 at 16:23
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It is conditional expectation (not probability), and $E[X|X,Y]=X$ because $X$ is already given.

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    $\begingroup$ The expectation value should give you a number. $X$ is a random variable, not a number. $\endgroup$ – probably_someone Aug 21 at 18:51
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    $\begingroup$ no, expectation gives you a function of the RVs on the given side. If there is nothing on the given side, then it gives you a constant @probably_someone $\endgroup$ – gunes Aug 21 at 18:55
  • $\begingroup$ So $X$ is a constant on the right-hand side, and a random variable on the left-hand side before the bar, and a constant on the left-hand side after the bar, correct? $\endgroup$ – probably_someone Aug 24 at 5:41

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