Questions about what is $E[X \mid c]$

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Let $Y$ be a constant RV that equals $7$ with probability one.

Q1: Is it true that $E[X \mid Y] = E[X]$?

Q2: Is $E[X \mid Y]$ a random variable?

Q3: What is $E[Y \mid X]$?

My attempt:

For the first question I believe the answer is yes. Knowing $Y=7$ does not tell you anything about $X$ therefore $E[X|Y] = E[X]$ so $E[X \mid c] = E[X]$.

For the second question I believe the answer is yes, $E[X \mid c]$ is a constant RV.

The third question is $E[c \mid X] = c$ again due to independence.

Am I on the right track for all three questions? Thanks.

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Q1 Correct.

Prescribing $\mathbb E[X\mid Y]$ by $\omega\mapsto\mathbb EX$ it is guaranteed that it is a random variable that is measurable wrt $\sigma(Y)$.

This because the $\sigma$-algebra generated by a constant random variable is $\{\varnothing,\Omega\}$ hence is subset of any $\sigma$-algebra on $\Omega$.

Next to that for every Borel set $B$ we must check that $$\int_{\{Y\in B\}}\mathbb EX\;\mathsf P(d\omega)=\int_{\{Y\in B\}}X(\omega)\mathsf P(d\omega)$$

For this we discern two situations: $7\in B$ and $7\notin B$.

If $7\in B$ then $P(\{Y\in B\})=1$ and both sides take value $\mathbb EX$ and if $7\notin B$ then $P(\{Y\in B\})=0$ and both sides take value $0$.

Q2 Correct.

Q3 Correct

Prescribing $\mathbb E[Y\mid X]$ by $\omega\mapsto7$ it is again guaranteed that it is a random variable that is measurable wrt $\sigma(X)$

Now it must be checked that for every Borel set $B$ we have: $$\int_{\{X\in B\}}7\;\mathsf P(d\omega)=\int_{\{X\in B\}}Y(\omega)\mathsf P(d\omega)$$which is a direct consequence of $Y=7$ a.s.

More generally: if $X,Y$ are independent random variables then $\mathbb E[Y\mid X]=\mathbb EY$.

This can be applied here (as you did) because a random variable that is constant a.s. is independent wrt to any random variable defined on the same probability space.


P.S. In general there are more than one choices for $\mathbb E[X\mid Y]$, but it is enough to find a random variable that satisfies specific conditions.