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I am given that $E[X | Y] = 0$ where $X, Y$ are both random variables. So this expectation is a random variable. By the tower rule we have $E[E[X | Y]] = E[X] = E[0] = 0$.

Does this also imply that $E[x| Y] = 0$? Where $x$ is any realization of $X$?

This question was inspired by https://web.stanford.edu/~mrosenfe/soc_meth_proj3/matrix_OLS_NYU_notes.pdf, pg 5, #3, where it shows that $E[\epsilon | X] = 0$ apparently means $E[\epsilon_i | X] = 0 \ \ \forall i$. On second thought, I may have misunderstood these notes. The $\epsilon_i$ here do not appear to be constants. They appear to be random variables as well.

student010101
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    How are $E[x \mid Y]$ or $E[X=x \mid Y]$ defined? – angryavian May 05 '21 at 15:37
  • @angryavian I removed the $X = x$ syntax. I think I was trying to be unnecessarily verbose there and end up being confusing. – student010101 May 05 '21 at 15:38
  • What happens when $X$ only takes finitely many values, each with positive probability? ( ... so has a PMF with finite support ... or is a mixed PMF/PDF ...) – Eric Towers May 05 '21 at 15:39
  • It isn't clear to me what your notation means. When we refer to $E[X]$ we need a probability distribution (possibly discrete, possibly continuous) for the outcomes of $X$ taking values. You can ask about conditional expectations, but this is defined in terms of a modified probability distribution. Perhaps you have in mind a larger setting where $X,Y$ have a joint probability distribution. In any case we need more words to sort out what you mean. – hardmath May 05 '21 at 15:40
  • In response to your update: indeed $\epsilon = (\epsilon_1, \ldots, \epsilon_n)$ is a random vector whose components are random variables $\epsilon_i$. – angryavian May 05 '21 at 15:47
  • @angryavian I think you're right. This wasn't clear from the notes initially. Now my question has a false premise. – student010101 May 05 '21 at 15:50

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Since $x$ is a known constant and not a random variable we have $$\mathbb{E}[x|Y]=x\mathbb{E}[1|Y]=x$$ Because $1$ is constant too.

As for the notes here the notation $\mathbb{E}[\varepsilon_j|X]$ indicates the conditional expectation of the $j$-th component of the random vector $\varepsilon $, which as a consequence of the model assumptions is zero. So it is an expectation of a random variable, not of a constant.

Snoop
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  • Why the downvote? Should I go more into detail? – Snoop May 05 '21 at 15:40
  • Downvote for an unnecessary answer. It's not clear what the OP is missing (or even asking). The answer presupposes an interpretation or else assumes the OP is capable of answering their own question. –  May 05 '21 at 15:42
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    I'm not a downvoter, but you jumped to answer a rather trivial part of the Question. Please note that poorly presented Questions should be clarified before answering. See Enforcement of Quality Standards on meta Math.SE. – hardmath May 05 '21 at 15:45
  • Added details about OP's notes (which have been added) – Snoop May 05 '21 at 15:50