Explanation for E[E[X|Y]|Y]=E[X|Y]

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.

• You have kinda already arrived at the result you wish to show. Note that $E[g(Y) | Y=y] = g(y)$. Apr 15 '21 at 4:53
• 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. Apr 15 '21 at 5:10
• @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. Apr 15 '21 at 5:35
• 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. Apr 15 '21 at 5:37
• 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. Apr 15 '21 at 8:04

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 $$E [g(Y) | Y=y] = g(y)$$

• 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:$\sum_y g(y).P(Y=y|Y=y) = g(1).P(Y=1|Y=1)+g(2).P(Y=2|Y=2)+g(3).P(Y=3|Y=3) = g(1)+g(2)+g(3) = \sum_y g(y)$ Apr 15 '21 at 5:59
• Yes to your question. You should only evaluate the part above the conditional sign. The part under the conditional sign is held constant. Apr 15 '21 at 6:16
• 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. Apr 15 '21 at 6:24
• 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 $y$. Apr 15 '21 at 6:36
• 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. Apr 15 '21 at 7:10