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Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space and $X: \Omega \rightarrow \mathcal{X}$ and $Y: \Omega \rightarrow \mathcal{Y}$ be random variables.

I have two questions comparing the conditional expectations $\mathbb{E}[Y|X]$ and $\mathbb{E}[Y|X=x]$.

1.) What is the difference between $\mathbb{E}[Y|X=x]$ and $\mathbb{E}[Y|X]$ in terms of their mathematical properties? I.e., from which space to which space do they map, which variables are they functions of and if they are measurable?

2.) If $X$ and $Y$ are absolutely continuous real valued random variables and have a joint density $f_{X, Y}$ with respect to the Lebesgue-measure $\lambda \otimes \lambda =\lambda^2$, Then: $$\mathbb{E}[Y|X=x] = \int y \, f_{Y|X=x}(y) \, \lambda(dy) = \int y \, \frac{f_{X, Y}(x,y)}{f_X(x)}\lambda(dy).$$

My question is now, how would that equation look like for $\mathbb{E}[Y|X]$? Does it make sense to write something like $f_{Y|X=X}(y)$ and $f_{X,Y}(X,y)$ and do these quantity even exist? I.e., should I write $$\mathbb{E}[Y|X] = \int y \, f_{Y|X=X} \lambda(dy) = \int y \, \frac{f_{X, Y}(X,y)}{f_X(X)}\lambda(dy)$$?

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    $\begingroup$ $\mathbb{E}[Y\mid X=x]$ is a value which depends on $x$, while $\mathbb{E}[Y\mid X]$ is a function of $X$ and so a random variable. If you define $g(x) = \mathbb{E}[Y\mid X=x]$, then you can say $\mathbb{E}[Y\mid X]= g(X)$ $\endgroup$
    – Henry
    Commented Oct 25, 2021 at 16:57
  • $\begingroup$ So but then $\mathbb{E}[Y|X=x]$ is also a function of $x$, no? $\endgroup$
    – guest1
    Commented Oct 26, 2021 at 13:52
  • $\begingroup$ $g(x)$ is a function of $x$, though for given $x$ it is just a value. But it is not a random variable or a function of a random variable $\endgroup$
    – Henry
    Commented Oct 26, 2021 at 13:58
  • $\begingroup$ So what I am wondering is how I should interprete $\mathbb{E}[Y|X=x]$. So is that a function $g(\cdot)$ or is it a function value with a specific fixed $x$, i.e., is it the function $g$ evaluated at $x$, $g(x)$? $\endgroup$
    – guest1
    Commented Oct 26, 2021 at 14:15
  • $\begingroup$ It is certainly the later, since it is the conditional expectation of $Y$ given that $X$ takes the specific value $x$. $\endgroup$
    – Henry
    Commented Oct 26, 2021 at 14:18

2 Answers 2

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In measure-theoretic probability, it is important to understand that while $E[Y|X]$ is a carefully defined mathematical object (as a Radon-Nikodym derivative), "$E[Y|X = x]$" is merely a shorthand notation that is based on $E[Y|X]$. Specifically, since (a fixed version of) $E[Y|X]$ is $\sigma(X)$-measurable, there exists (cf. Theorem 20.1 in Probability and Measure (3rd ed.) by Patrick Billingsley; Lemma of Theorem 9.1.2 in A Course in Probability Theory by Kai Lai Chung) a Borel measurable function $\varphi$ such that $E[Y|X] = \varphi(X)$. "$E[Y|X = x]$" is then defined as \begin{align*} E[Y|X = x] = \varphi(x), x \in \mathbb{R}. \tag{1}\label{1} \end{align*}

This convention is further clarified by Chung as follows (also see remarks by me at the end of this answer):

As a consequence of the theorem, the function $E[Y|X]$ of $\omega$ is constant a.e. on each set on which $X(\omega)$ is constant. By an abuse of notation, the $\varphi(x)$ above is sometimes written as $E[Y|X = x]$.

...

Generalization to a finite number of $X$'s is straightforward. Thus one version of $E[Y|X_1, \ldots, X_n]$ is $\varphi(X_1, \ldots, X_n)$, where $\varphi$ is an $n$-dimensional Borel measurable function, and by $E[Y|X = x_1, \ldots, X_n = x_n]$ is meant $\varphi(x_1, \ldots, x_n)$.

To summarize, in measure-theoretic probability theory, there is only one unambiguous definition of the concept conditional expectation, which is $E[Y|X]$. In contrast, the widely used notation "$E[Y|X = x]$" does not possess its own specific definition, but is merely a derivative of the fundamental notion $E[Y|X]$. In other words, there is a chronological order between these two objects: one cannot define $E[Y|X = x]$ independently without first settling down the definition of $E[Y|X]$. For this reason, "$E[Y|X = x]$" is not a universally accepted notation in rigorous probability texts -- for example, I have never seen Billingsley used $E[Y|X = x]$ for a single time throughout his classical text Probability and Measure.

For your second question, what you stated is correct. The same expression of $E[Y|X]$ in this case is also given in Problem 34.2 in Probability and Measure. On the other hand, you seemed to imply that this expression is a corollary of the expression $E[Y|X = x]$ which you laid out first. As I commented above, this is not the case (as you reversed the correct chronological order): you can verify that \begin{align*} g(X) := \frac{\int_{-\infty}^\infty f(X, y)ydy}{\int_{-\infty}^\infty f(X, y)dy} \end{align*} is a version of $E[Y|X]$ directly by checking $g(X)$ satisfies (with the help of Fubini's theorem) the two defining properties of the conditional expectation. After completing this step, it is then customary to write $g(x) = \frac{\int_{-\infty}^\infty f(x, y)ydy}{\int_{-\infty}^\infty f(x, y)dy}$ as $E[Y|X = x]$.


Technical Notes

To reinforce the understanding of the convention $\eqref{1}$, let's verify that the statement "$E[Y|X]$ of $\omega$ is constant a.e. on each set on which $X(\omega)$ is constant." in the quotation block and try to understand why Chung said writing $\varphi(x)$ as $E[Y|X = x]$ is "an abuse of notation".

First of all, for a fixed version $\varphi(X)$ of $E[Y|X]$, on each $\omega$ in the set $\{X = x\}$, it is clear that $\varphi(X)(\omega) = \varphi(X(\omega)) = \varphi(x)$, which is a constant. The stake of this statement is that if $\psi(X)$ is another version of $E[Y|X]$, then \begin{align*} P(\psi(X) \neq \varphi(x), X = x) = 0, \end{align*} this is a consequence of \begin{align*} P(\psi(X) \neq \varphi(x), X = x) = P(\psi(X) \neq \varphi(X), X = x) \leq P(\psi(X) \neq \varphi(X)) = 0. \end{align*} The last equality holds because any two versions of the conditional expectation are equal with probability $1$. Therefore, an accurate interpretation of $\eqref{1}$ should be: given a fixed version $\varphi(X)$ of $E[Y|X]$, for each $x \in \mathbb{R}^1$ and any other version $\psi(X)$ of $E[Y|X]$, it holds that for each $\omega \in \{X = x\}$ outside a null set (depending on $\psi$), we have \begin{align*} \psi(X)(\omega) = \varphi(x). \end{align*} Note that it is possible to have an $\omega' \in \{X = x\}$ such that $\psi(X)(\omega') \neq \varphi(x)$, but such $\omega'$ is within a null set. Since technically $\psi(X)$ and $\varphi(X)$ do not need to agree everywhere on $\{X = x\}$, yet $E[Y | X = x]$ in appearance is a single value, that's probably the reason why Chung added the qualifier "By an abuse of notation". However, this is a really a minor "abuse", just like people always use the notation $E[Y|X]$ (technically it is an equivalence class of almost surely identical random variables) to stand for a fixed version of it.

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๐”ผ[๐‘Œ|๐‘‹] is a random variable of X, where ๐”ผ[๐‘Œ|๐‘‹=๐‘ฅ] is a constant when the random variable X is evaluated at x of its domain.

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    $\begingroup$ Sure--by definition a conditional expectation is a random variable. But what exactly do you mean by "random variable of $X$"? $\endgroup$
    – whuber
    Commented May 20 at 2:43
  • $\begingroup$ "a random variable of X" means the function mapping from the state space to the real line, i.e. for any state there is a unique real value for it. $\endgroup$
    – Colin Fang
    Commented May 26 at 8:12
  • $\begingroup$ People would usually just say "random variable:" the "of $X$" part is superfluous and unconventional, potentially causing confusion. $\endgroup$
    – whuber
    Commented May 29 at 17:16
  • $\begingroup$ @whuber Probably the OP meant "a function of the random variable $X$", which for clarity he/she needs to correct for. $\endgroup$
    – Zhanxiong
    Commented May 30 at 12:29

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