What prevents the Expectation conditional on a sub sigma-algebra to be the the variable being conditioned?

Question

I have come to a very simple issue which I have not figured out yet. In fact there are two questions. Could you please help me out on this?

Let $(\Omega,\mathcal{F},P)$ be a probability space, $X:\Omega\rightarrow R$ be a random variable, and $\mathcal{A}$ be a sub $\sigma$-algebra in $\mathcal{F}$. The conditional expectation of $X$ given $\mathcal{A}$ is the a.s. unique random variable, denoted as $E[X/\mathcal{A}]:\Omega\rightarrow R$ that satisifies: (i) $E[X/\mathcal{A}]$ is $\mathcal{A}/\mathcal{B}$ measurable and (ii) $\int_BE[X/\mathcal{A}]dP=\int_B XdP \hspace{5mm } \forall B\in\mathcal{A}$.

What prevents $X$ to be this random variable regardless of $\mathcal{A}$?

$X$ seems to satisfy both conditions, (i) $\mathcal{A}/\mathcal{B}$ measurability and (ii) $\int_BE[X/\mathcal{A}]dP=\int_B XdP$.

The second question is:

In the same setting as above, it is defined $P[A/\mathcal{A}]=E[I_A/\mathcal{A}] \hspace{5mm}\forall A\in \mathcal{F}$, and if $\mathcal{A}=\sigma(Y)$ for a random element $Y:\Omega\rightarrow\Lambda$.

How does $P[A/Y]=P[A/\mathcal{A}]$ relate with $P[A/Y=y]$ for $A\in\mathcal{F}$ and $y\in\Lambda$.

Best regards,

Juan Manuel

Answer 1

I'm not sure what is meant by $\mathcal A/\mathcal B$-measurability. The definition of conditional expectation that I learned requires it to be $\mathcal A$-measurable, which $X$ usually isn't.

Answer 2

What prevents $X$ from being its own conditional expectation given $\mathcal{A}$ is indeed the measurability part. It can't really fail to satisfy the other condition, can it? In your setting $X$ is a random variable meaning that $X$ is $(\mathcal{F},\mathcal{B}(\mathbb{R}))$-measurable. But $\mathcal{A}$ is a sub sigma-algebra of $\mathcal{F}$, and thus it can be smaller than $\mathcal{F}$ and hence $X$ need not be $(\mathcal{A},\mathcal{B}(\mathbb{R}))$-measurable. However, if $X$ is $(\mathcal{A},\mathcal{B}(\mathbb{R}))$-measurable then $$ {\rm E}[X\mid\mathcal{A}]=X. $$

The relationship between $P(A\mid Y)$ and $P(A\mid Y=y)$ is the following: If $\varphi:y\mapsto P(A\mid Y=y)$ for $y\in\Lambda$, then $P(A\mid Y)=\varphi(Y)$.