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What is $E[E[Y|X]|X]$ ?
Can I use the law of total expectation and say that it's equal to $E[Y|X]$ ?
If it's correct, could someone explain why?

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Let $\mathscr F$ be the $\sigma$-algebra generated by $X.$ Then $E[Y|X]=E[Y|\mathscr F]$ is an $\mathscr F$-measurable random variable and we know that if $Z$ is a $\mathscr F$-measurable random variable, we have $E[Z|\mathscr F]=Z.$

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  • $\begingroup$ How do you know that $E[Z|F]=Z$ ? It's a known formula in the expectation rules? $\endgroup$
    – user869856
    Commented Jan 4, 2021 at 15:04
  • $\begingroup$ @Xavi by definition of conditional probability $\mathbf{E}[X\mid\mathscr F]$ is $\mathscr F$-measurable, thus $\mathbf{E}[\mathbf{E}[X\mid\mathscr F]\mid\mathscr F]=\mathbf{E}[X\mid\mathscr F]$ since if $X$ is $\mathscr F$-measurable then $\mathbf{E}[X\mid\mathscr F]=X$. Note that $\mathbf{E}[Y\mid X]$ means $\mathbf{E}[Y\mid \sigma(X)]$. $\endgroup$
    – jdods
    Commented Jan 4, 2021 at 15:23
  • $\begingroup$ @Xavi $E[Z|\mathscr F]=Z$ follows immediately from the definition of conditional expectation. You just have to check that $Z$ satisfies the two conditions of the definition. $\endgroup$
    – UBM
    Commented Jan 4, 2021 at 16:59
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I do not think this is the low of total expectation, at least not in its standard form.

Recall instead that:

  • $E[A|B]=A$ if $A$ is $B-$measurable.
  • $E[Y|X]$ is $X-$measurable.

So now put in the first property $A=E[Y|X]$ and $B=X$, that is your result.

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  • $\begingroup$ How do you know that $E[A|B] = A$ ? $\endgroup$
    – user869856
    Commented Jan 4, 2021 at 14:52
  • $\begingroup$ @Xavi it is because $A$ is $B$−measurable. $\endgroup$ Commented Jan 4, 2021 at 20:47
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The law of total expectation would $E[Y]=E[E[Y|X]]$. In this expression $E[Y|X]$ is a random variable that is a function of $X.$ $E[Y|X]$ changes with $X,$ assuming different realizations depending on different outcomes of $X=x,$ and by the law of total expectation, its expected value, across the support of $X$ is just $E[Y].$ Conditioning on $X$ the expression $E[Y|X]|X$ is redundant $E[Y|X]|X=E[Y|X],$ and in this sense it amounts to applying the same law of total expectation $E[E[Y|X]]=E[E[Y|X]|X]=E[Y].$ $\;$

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    $\begingroup$ Would it be possible for the gentleman who downvoted my answer to explain why it is incorrect so we can all learn? $\endgroup$
    – user869837
    Commented Jan 4, 2021 at 15:19
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    $\begingroup$ @Paul We want to get the result of $E[Y|X]$ not $E[Y]$ $\endgroup$
    – user869856
    Commented Jan 4, 2021 at 17:52
  • $\begingroup$ My earlier comment that this answer was essentially correct was wrong... I hadn't read the last part closely. $E[E[Y\mid X]\mid X]=E[Y\mid X],$ not $E[Y].$ $\endgroup$ Commented Jan 4, 2021 at 18:11
  • $\begingroup$ @Paul That's not true: $E[E[Y|X]|X]] \ne E[Y]$, the correct answer is $E[E[Y|X]|X]] = E[Y|X]$ $\endgroup$
    – user840425
    Commented Jan 4, 2021 at 18:15