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I would like to ask for the proof of $E[E[X|Y]|Y]=E[X|Y]$

Per my understanding (for discrete case):

because $E[X|Y]=g(Y)$

hence, $E[E[X|Y]|Y] = E[g(Y)|Y]= \sum_y g(y)*p(y|y)=\sum_y g(y)=\sum E(X|Y=y)$

I could not arrive to the correct result $E[E[X|Y]|Y]=E[X|Y]$, it is very nice if someone can tell me what's wrong in my demonstration.

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  • $\begingroup$ You have kinda already arrived at the result you wish to show. Note that $E[g(Y) | Y=y] = g(y)$. $\endgroup$ Apr 15 '21 at 4:53
  • $\begingroup$ I think the issue is when you say $E[g(Y)|Y] = \sum_y g(y) p(y | y)$. You are summing over all the possible values that $Y$ can take. Any value other than $y$ has conditional probabiltiy of 0, so you are left with $g(y)$ in the end. $\endgroup$ Apr 15 '21 at 5:10
  • $\begingroup$ @SOULed_Outt I understand that $E[X|Y]$ is a RV which is a function of Y and E[X|Y=y] is a value evaluated at Y=y. Expected value should be evaluated at all possible value of RV, that is why the sum seems to be reasonable for me. $\endgroup$
    – Hunglk
    Apr 15 '21 at 5:35
  • $\begingroup$ The earlier comment I made was very misleading so I deleted it (apologies). Take a look at what I posted as an answer and let me know if it clears anything up. $\endgroup$ Apr 15 '21 at 5:37
  • $\begingroup$ Since $E[X|Y]$ is a (measurable) function of $Y$, it is a fixed (deterministic) and known quantity given a realisation of $Y$. Hence, in the probabilistic universe where $Y$ is observed as $y$, it is a constant, equal to its expectation. $\endgroup$
    – Xi'an
    Apr 15 '21 at 8:04
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Suppose, for example, that $Y$ could take values $1,2,$ or $3$ and we want to find $E[g(Y) | Y =3]$ for some nice function $g$. By defintion of conditional expectation we have \begin{align} E [g(Y) | Y=3] &= \sum_y g(y) \cdot P(Y=y | Y=3) \\ &= g(1) \cdot P(Y=1 | Y=3) + g(2) \cdot P(Y =2 | Y=3) + g(3) \cdot P(Y = 3 | Y = 3) \end{align}

Note that $P(Y = 1 | Y = 3) = 0$, $P(Y = 2 | Y = 3) = 0$, and that $P(Y=3 | Y=3)=1$ so $$ E [g(Y) | Y=3] = g(3) $$


I think your problem starts after you introduce the summation. $p(y|y)$ is not always equal to 1. It equals zero for any $y' \neq y$. In the summation, the $y$ in the first slot of $p(y|y)$ varies while the $y$ in the second slot is fixed (see like in the example above).

Look at what I have highlighted in red. The red $y$ serves as a dummy variable. I could replace it with any letter I want so long as I understand that it should represent all possible values that $Y$ can take. Let's say I use $a$ instead. \begin{align} E [g(Y) | Y=y] &= \sum_{\color{red}{y}} g(\color{red}{y}) \cdot P(Y = \color{red}{y} | Y = y) \\ &= \sum_{a} g(a) \cdot P(Y = {a} | Y = y). \end{align}

Notice that $P(Y = a | Y= y) = \begin{cases} 1 \text{ if } a = y \\ 0 \text{ if } a \neq y \end{cases}.$ Therefore $E [g(Y) | Y=y] = g(y)$

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  • $\begingroup$ It means that the Y under conditional sign (|) will hold constant while the Y above conditional sign (|) will be evaluated at all possible value? I thought that Y should be evaluated at both. For example, if Y takes value in $[1,2,3]$, we have:$\sum_y g(y).P(Y=y|Y=y) = g(1).P(Y=1|Y=1)+g(2).P(Y=2|Y=2)+g(3).P(Y=3|Y=3) = g(1)+g(2)+g(3) = \sum_y g(y)$ $\endgroup$
    – Hunglk
    Apr 15 '21 at 5:59
  • $\begingroup$ Yes to your question. You should only evaluate the part above the conditional sign. The part under the conditional sign is held constant. $\endgroup$ Apr 15 '21 at 6:16
  • $\begingroup$ In this case, what is the different between $E[X|Y]$ and $E[X|Y=y]$? As my knowledge, The first term is rv and the second term is the value. $\endgroup$
    – Hunglk
    Apr 15 '21 at 6:24
  • $\begingroup$ Sometimes they are used interchangeably. I believe most texts on probability say that you should view $E[X | Y]$ as a random variable and you should view $E[X|Y=y]$ as a number that depends on $y$. $\endgroup$ Apr 15 '21 at 6:36
  • $\begingroup$ That's why I got confused. my knowledge is $E[g(Y)|Y]$ is a r.v, then, when we evaluated it, we should evaluated both g(Y) and Y. $\endgroup$
    – Hunglk
    Apr 15 '21 at 7:10

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