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My intuition told me E(X|X) is X. However,I get stuck when when I try to evaluate it from the definition. How do I define f(X|X) ? would it be constant? Thanks for your help.

zqzwxec11
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1 Answers1

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Your intuition is correct: if $X$ is any random variable with finite mean (not necessarily discrete or continuous!), then $E[X \mid X] = X$.

More generally, if $\mathcal{G}$ is a $\sigma$-field, then $E[X\mid\mathcal{G}] = X$ (almost surely) whenever $X$ is a $\mathcal{G}$-measurable random variable with finite mean. This can be proved from the definition of $E[X \mid \mathcal{G}]$ as the unique (up to almost everywhere equivalence) $\mathcal{G}$-measurable random variable $Y$ such that $E[X \mathbf{1}_A] = E[Y \mathbf{1}_A]$ for every $A \in \mathcal{G}$. To see why, note that if $X$ itself is $\mathcal{G}$-measurable, then certainly $E[X \mathbf{1}_A] = E[X \mathbf{1}_A]$ for all $A \in \mathcal{G}$.

The special case $E[X \mid X]$ is just $E[X \mid \mathcal{G}]$, where $\mathcal{G}$ is the $\sigma$-field generated by $X$, in which case $X$ is necessarily $\mathcal{G}$-measurable and hence $E[X \mid X] = X$ (almost surely).

Artem Mavrin
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    I think the OP is trying to fit $f_{X\mid Y}(x\mid y) = \displaystyle\frac{f_{X,Y}(x,y)}{f_Y(y)}$ to the case when $Y$ is the same as $X$, and then wanting to determine $E[X\mid X]$ from the conditional density $f_{X\mid X}$. – Dilip Sarwate Jan 20 '17 at 22:11
  • @DilipSarwate you are right. I still confused about how to define fX∣X – zqzwxec11 Jan 21 '17 at 13:53
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    @zqzwxec11 The point of my answer is that you don't need to consider densities or probability mass functions to conclude that $E[X \mid X] = X$. The formula holds for all random variables with finite mean, even random variables that are neither discrete nor continuous. – Artem Mavrin Jan 22 '17 at 02:03
  • The expression $E[X\mid\mathcal{G}] = X$ is a bit weird since the term $X$ that occurs twice has different meanings. The first occurance, $X$ refers to a variable. The second occurance, $X$ refers to a value. The same problem occurs in, and originates from, the OP's expression $E[X\mid X] = X$. – Sextus Empiricus Jan 28 '21 at 08:28
  • Say we consider a fair 6-sided dice roll and $X$ is a variable that is 0 or 1 depending on whether the rolled number is odd/even and $\mathcal{G}$ are the results $\lbrace 1,2,3,4,5,6 \rbrace$. Then $X$ a $\mathcal{G}$-measurable variable. But I would not know what to make of $E[X\mid\mathcal{G}] = X$. I think that the right-hand side should be, informally speaking, some function that maps the elements in the set $\mathcal{G}$ to the values of the variable $X$ – Sextus Empiricus Jan 28 '21 at 08:38
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    @SextusEmpiricus In the expression $E[X\mid\mathcal{G}]=X$, both $X$'s are random variables. That's because conditioning on a $\sigma$-algebra (and in particular, on a random variable) gives a random variable. This is different from conditioning on the value that a random variable takes (e.g., $E[X\mid Y=y]$). Please see [my answer here](https://stats.stackexchange.com/a/395328/97872). In your die rolling example, $\mathcal{G}$ doesn't appear to be a $\sigma$-algebra. – Artem Mavrin Jan 28 '21 at 17:24