# 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?

• 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. – Gordon Smyth Aug 21 at 6:44
• If you know X you know X...! – Xi'an Aug 21 at 7:19
• How did you come across this conditional expectation? – Sextus Empiricus Aug 21 at 7:35

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.)

• 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. – Richard Hardy Aug 22 at 8:15
• 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. – Ruben van Bergen Aug 23 at 8:51
• 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. – Artem Mavrin Aug 24 at 16:23

It is conditional expectation (not probability), and $$E[X|X,Y]=X$$ because $$X$$ is already given.

• The expectation value should give you a number. $X$ is a random variable, not a number. – probably_someone Aug 21 at 18:51
• 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 – gunes Aug 21 at 18:55
• 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? – probably_someone Aug 24 at 5:41