I was trying to understand both intuitively and rigorously what the difference between $E[X\mid Y]$ vs $E[X\mid Y=y]$.

Let me tell you first the things that do make sense to me. $E[X\mid Y=y]$ makes sense to me (I think). For me it means the value that we expect $X$ to have on average given that the event $Y=y$ was observed and it has deterministic value:

$$E[X\mid Y=y] = \sum_x{xp_{X\mid Y}(x\mid y)} = \mu_{X\mid Y=y}$$

that can be computed like any other expectation (where the notation $\mu_{X\mid Y=y}$ denotes the actual real number that $E[X\mid Y=y]$ takes). $E[X\mid Y=y]$ is a function of y. Given $Y=y$, the conditional expectation will always be $\mu_{X\mid Y=y}$, which is governed by what specific y was observed.

However, $E[X\mid Y]$ makes less sense to me. I have read something similar to the following:

Notice now that Y is a random variable this time. Therefore, $E[X\mid Y]$ is a random variable.

That sort of makes sense to me because, if $Y$ is random then the expected value of $X$ has to be random too. In other words, $Y$ being random, then consequently $E[X\mid Y]$ is random too. The statement $E[X|Y]$ is a r.v. and is the exact statement I would like to understand more precisely. I feel I understand less and I would like to address it.

I feel if I really understood this concept of what $E[X\mid Y]$ really means, I should be able to answer the following questions:

1) If $E[X\mid Y]$ is random, then, what are the possible values it can take? Can it only take the values $\mu_{X\mid Y=y}$ for $y \in Y$? Say if $Y=\{1,2,3\}$ and $U_{X,Y} = \{\mu_{X\mid Y=1}= 11, \mu_{X\mid Y=2} = 22, \mu_{X\mid Y=3} = 33 \}$. Then is there any chance that $E[X\mid Y] = 123$?

2) Is the distribution of $E[X|Y]$ over $U_{X,Y}$ or over $Y$?

3) What is the probability distribution of $E[X\mid Y]$ if we have all the information we need about the distributions of $p_X(x)$ and $p_Y(y)$? Is it just the same as $p_Y(y)$? Is there a closed/specific formula for it?

4) When one is asked to find the distribution of $E[X\mid Y]$ are we asked to find $Pr[E[X\mid Y=y]]$ or $Pr[E[X\mid Y] = \mu]$? Is there a difference between the two? Is one nonsense while the other one is a valid probability distribution?

5) If we were to sketch the probability density function for $E[X \mid Y]$, would the horizontal axis be $y$ or $\mu$ ? i.e. would the probability density be a function of $y$ or of $E[X\mid Y] = \mu$? i.e. would $p_{E[X\mid Y]}(k)$ be a function of $y=Y$ or $E[X\mid Y]$?

6) Related to the above two question, it seems to me that writing an explicit formula for $E[X\mid Y=y]$ is easy, while for $E[X\mid Y]$ it is not (or at least for me). Is the formula $E[X\mid Y] = \sum_x{xp_{X\mid Y}(x\mid Y)}$? I would guess it is but for me its a very strange equation because we are conditioning on a random variable, or we are saying given $Y$, but $Y$ is random so its really not given. Therefore, I can't seem to find an expression for it that makes sense to me.

Basically, it is not clear to me what $E[X\mid Y]$ means, because I don't know what its valid outcomes are, what its distribution is (in relation to $p_X(x), p_Y(y)$ or $p_{X,Y}(x,y)$ or anything) nor can I write an explicit formula for it that makes sense to me. I can't even decide if $p_{E[X \mid Y]}(k)$ is a function of y or $\mu=E[X|Y]$.


If $\varphi(y)=E[X\mid Y=y\,]$ is the non-random function of $y$, then $E[X\mid Y\,]$ is defined to be the random variable $\varphi(Y)$.

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    $\begingroup$ The reason I down voted your answer is because,that fact is already stated in my question and I already said that fact didn't help me answer any of my other 6 doubts about what $E[X \mid Y]$ means.It might be a correct statement but does not address the main issue about my question.I am willing to change my vote if your answer address any of the real issue I raised or states more than the obvious.I apologize beforehand if the obvious does not make sense to me.If it did,this question wouldn't exist in the first place.Thanks for your time,I'm sure you meant your answer with the best intentions. $\endgroup$ – Charlie Parker Jul 20 '14 at 20:23
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    $\begingroup$ All of these considerations are well and good but they do not imply that you should downvote this quite correct and complete answer. Some strange manners are at work here... $\endgroup$ – Did Aug 4 '14 at 17:12
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    $\begingroup$ @Did The whole "point" that this question even exists is that even with that information I could not answer the six or so properties I listed.Furthermore,I don't feel its a real answer to just copy what I wrote in my own question without further explanations. With no additional insights or intuitions,its hard to tell if it was just copied from my own question.The reason I downvoted it is because I can't tell if that information was copied just from my question.Maybe its own incompetence that I can't tell the difference, but how else do you think I should have responded? $\endgroup$ – Charlie Parker Aug 5 '14 at 0:52
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    $\begingroup$ I am open for suggestions :) I Truly seek to be the best member of this community as I can :) I rather be honest with other members, than accept an answer or like an answer that I really don't understand or see how it contributes anything to answer my question. $\endgroup$ – Charlie Parker Aug 5 '14 at 0:53

If $Z=E[X\mid Y]$ a random variable then it means $F_{Z}(z)=P(Z\leq z)=P(Y\in\{y\in\mathbb{R}:E[X\mid Y=y]\leq z\})$.

If you want to think intuitively, it is very simple: to choose a random value for $Z$, first choose $Y$ randomly according to the probability of $Y$, you obtain $Y=y$. Then the value $Z$ takes is $E[X\mid Y=y]$.

In fact, this is just a special case of the general construction: if $f$ is a real Borel measurable function, then $Z=f(Y)$ is defined to be a random variable such that $F_{Z}(z)=P(Z\leq z)=P(Y\in f^{-1}(-\infty,z])$. Here $E[X\mid Y=y]$ is a real function that take $y$ and return $E[X\mid Y=y]$.

The reason for the intuitive notion to be slightly complicated is because of the possibility of the function being not injective. For example, let's suppose $X$ and $Y$ is independent. Then $Z$ would be a value that take on the value of $E[X]$ with probability $1$, regardless of how $Y$ are distributed: this is because $E[X\mid Y=y]$ is a constant function.


There are fundamental differences between $E[X\mid Y=y]$ and $E[X\mid Y]$. You've got the fist one down, in that we are conditioning on a specific event. In the latter case, you are merely told the likelihood of various events, not which ones actually happen.

Calculating the distribution of $E[X\mid Y]$ is exactly the same as calculating the distribution of $f(X)$ for some integrable function $f$ given that you know the distribution of $Y$, since $E[X\mid Y]$ is just a special case of a function of $Y$.

Bottom line: Don't think of it as an expectation, think of it as just the distribution of the function of a random variable.

  • $\begingroup$ Viewing $E[X|Y] = g(Y)$ as a function of a random variable was helpful. Then $g(Y) = E[X|Y]$ is just a function of a random variable and its distribution is just $Pr[g(Y) = \mu] = Pr[E[X|Y] = \mu] = \sum_{y:g(y)=\mu } Pr[Y=y] = \sum_{y:E[X|Y=y]=\mu} Pr[Y=y]$?? $\endgroup$ – Charlie Parker Jul 20 '14 at 17:19
  • $\begingroup$ @CharlieParker yes, that is correct..you got it :) Note that this is exactly what Bryan Schmuland wrote as well. You are merely transferring the probabilities of various Y's to the space of possible $E[X|Y]$'s. Note that $P(X=x)$ does not occur anywhere in this definition (except via the actual calculation of $E[X|Y=y]$ so it really is a random variable derived from the distribution of Y. $\endgroup$ – user76844 Jul 21 '14 at 6:04

You have been given good answers to your main question, so I think some technical remarks may also be of some value to you. Let's say you model some random phenomena on a probability space $(\Omega,\mathscr F,\mathsf P)$ and you have random variables $X:(\Omega,\mathscr F) \to (A,\mathscr A)$ and $Y:(\Omega,\mathscr F)\to (B,\mathscr B)$. Then $\xi = \mathsf E[X|Y]$ is also a random variable, that is a map $\xi:(\Omega,Y^{-1}(\mathscr B)) \to (A,\mathscr A)$ whereas $\eta(y) = \mathsf E[X|Y =y]$ is a mapping $\eta:(B,\mathscr B) \to (A,\mathscr A)$.

You can think of $\xi = \eta(Y)$, so in fact $\eta$ may contain more information. That is, let $Y$ tell us whether the train arrives in the morning $B = \{m,e\}$, and $X$ be the delay of the train. You can specify $\eta$ as an expected delay given the period of arrival, e.g. $\eta(m) = 5$ and $\eta(e) = 2$ as in the morning the situation on the rail road is more uncertain. Now, if you have a random period of arrival $Y$ which is almost surely evening, then by computing $\xi$ you will observe that $\mathsf E[X|e] = 2$, but you'll have no clue what $\mathsf E[X|m]$ shall be.

It is also true that $\xi$ does always exist, whereas for the existence of $\eta$ you may need to make some assumptions on $A$, which are nevertheless satisfied in case $A = \Bbb R$ you are interested in.


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