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Let $\varphi :\mathbb R\to \mathbb R$ defined by $$\varphi (x)=\mathbb E[Y\mid X=x].$$ Then, $\mathbb E[Y\mid X]$ is defined by $\varphi (X)$. This definition looks a bit confusion, rigorously speaking, shouldn't it be $$\varphi (X)=\mathbb E[Y\mid X=X],$$ and thus $$\mathbb E[Y\mid X=X](\omega ):=\varphi (X)(\omega )=\varphi (X(\omega ))=\mathbb E[Y\mid X=X(\omega )] \ \ ?$$

user659895
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    The above $\varphi (x)=\mathbb E[Y\mid X=x]$ is just a notation, we invent $\varphi$. (This is not doing anything about $X,Y$ or the conditional expectation.) The real question is: What is "$\mathbb E[Y\mid X=x]$"? Because this is already a big abuse of notation, assuming $X$ has "density and jumps" repartition. So what is your definition of $\mathbb E[Y\mid X=x]$? (Then we can clarify what happens if we blindly plug in $X$ in place of $x$, and how to write this plug without confusion. Because note that $X=X$ is the full space $\Omega$ for some people that have no knowledge of our plug in.) – dan_fulea May 01 '19 at 12:41
  • @dan_fulea: I'm not sure that $\mathbb E[Y\mid X=y]$ is an abuse of notation... If $f_{Y\mid X=x}$ is the density function of $Y\mid X=x$, then $$\mathbb E[Y\mid X=x]=\int_{\mathbb R}yf_{Y\mid X=x}(y)dy.$$ – user659895 May 01 '19 at 12:45
  • @user659895 your notations implies that $Y | X = x$ is a random variable, but what exactly is it? – mihaild May 01 '19 at 12:48
  • @mihaild : Yes it is. Take for example $Y_n=\sum_{k=1}^n X_i$ and $N$ a random variable in $\mathbb N$. Then $(Y_N\mid N=n)=\sum_{k=1}^NX_i.$ – user659895 May 01 '19 at 12:51
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    This is just misunderstanding of notation. $E(X|Y)$ is a random variable $E(X|Y)(\omega)=E(X|Y=y)$ –  May 01 '19 at 12:53
  • Your notation is just making it simpler by renaming your random variable to $\phi$ –  May 01 '19 at 12:53
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    @user659895 it works if variable you condition on is discrete (so you essentially take a subspace of probability space). It doesn't work with non-discrete variables. – mihaild May 01 '19 at 12:56
  • @mihaild : I'm not sure about that. In my course I have (for at least a stoping time) that $(Y\mid \tau=t)=Y\int_0^\tau B_tdt$. – user659895 May 01 '19 at 13:01
  • Related question/answer https://math.stackexchange.com/a/3023233/291100 – Nap D. Lover May 01 '19 at 13:07
  • @LoveTooNap29: Thanks it's very helpful. In fact $\varphi (X)=\mathbb E[Y\mid X]$ a.s. right ? But do you agree with the nottion $\varphi (X)(\omega )=\varphi (X(\omega ))=\mathbb E[Y\mid X=X(\omega )]$ ? – user659895 May 01 '19 at 13:09
  • @user659895 I don’t really see what you’re trying to clarify with its use. I prefer the notation in Probability with Martingales by D. Williams which is standard and clear, imo. See the chapter on conditional expectations if you can get a copy. – Nap D. Lover May 01 '19 at 14:13

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