Conditional Expectation of a random variable with itself

Question

Suppose $Y$ is either positive or integrable on $(\Omega, \mathcal{A}, P)$.

Then, how do I show that $E(Y|Y)=Y$ a.s.?

Seems so basic, but I know there's got to be a trick to it.

Answer 1

Assuming that $Y$ generates $\mathcal{G}$. Then

$\mathbb{E}\left[ {{1_G}\mathbb{E}\left[ {Y|Y} \right]} \right] = \mathbb{E}\left[ {{1_G}\mathbb{E}\left[ {Y|\mathcal{G}} \right]} \right] = \mathbb{E}\left[ {{1_G}Y} \right],\forall G \in \mathcal{G}$

since a product of $\mathcal{G}$-measurable functions is $\mathcal{G}$-measurable and by using properties of conditional expectation, from the definition of conditional expectation it follows that $Y = \mathbb{E}\left[ {Y|Y} \right]$ $P$-a.s.

Answer 2

Let $\left( {\Omega ,\mathcal{A},P} \right) = \left( {\left[ 0 \right.,\left. 1 \right\rangle ,\mathcal{B}\left[ 0 \right.,\left. 1 \right\rangle ,{{\left. \lambda \right|}_{\left[ 0 \right.,\left. 1 \right\rangle }}} \right)$ and $Y:\left[ 0 \right.,\left. 1 \right\rangle \to \mathbb{R},Y\left( \omega \right) = \omega $.

Consider a sequence of $\sigma $-algebras ${\mathcal{G}_n} = \sigma \left\{ {\left[ {\frac{i}{n}} \right.,\left. {\frac{{i + 1}}{n}} \right\rangle :i \in \left\{ {0,1, \ldots ,n - 1} \right\}} \right\}$ (it can be shown that a sigma-algebra generated by a finite partition is equal to all possible unions of that partition).

Consider $\mathbb{E}\left[ {Y|{\mathcal{G}_2}} \right]$. ${\mathcal{G}_2} = \left\{ {\emptyset ,\left[ 0 \right.,\left. {\frac{1}{2}} \right\rangle ,\left[ {\frac{1}{2}} \right.,\left. 1 \right\rangle ,\left[ 0 \right.,\left. 1 \right\rangle } \right\}$ and for each $G \in {\mathcal{G}_2}$, the following property must hold:

$\mathbb{E}\left[ {{1_G}\mathbb{E}\left[ {Y|{\mathcal{G}_2}} \right]} \right] = \mathbb{E}\left[ {{1_G}Y} \right]$. So,

$\mathbb{E}\left[ {{1_{\left[ 0 \right.,\left. {\frac{1}{2}} \right\rangle }}\mathbb{E}\left[ {Y|{\mathcal{G}_2}} \right]} \right] = \mathbb{E}\left[ {{1_{\left[ 0 \right.,\left. {\frac{1}{2}} \right\rangle }}Y} \right] = \frac{1}{4}$

$\mathbb{E}\left[ {{1_{\left[ {\frac{1}{2}} \right.,\left. 1 \right\rangle }}\mathbb{E}\left[ {Y|{\mathcal{G}_2}} \right]} \right] = \mathbb{E}\left[ {{1_{\left[ {\frac{1}{2}} \right.,\left. 1 \right\rangle }}Y} \right] = \frac{3}{4}$

Since ${\mathbb{E}\left[ {Y|{\mathcal{G}_2}} \right]}$ is ${{\mathcal{G}_2}}$-measurable, it must be constant on ${\left[ 0 \right.,\left. {\frac{1}{2}} \right\rangle ,\left[ {\frac{1}{2}} \right.,\left. 1 \right\rangle }$ (show it as an exercise), so

$\mathbb{E}\left[ {Y|{\mathcal{G}_2}} \right]\left( \omega \right) = \left\{ \begin{gathered} \frac{1}{4} & ,\omega \in \left[ 0 \right.,\left. {\frac{1}{2}} \right\rangle \\ \frac{3}{4} & ,\omega \in \left[ {\frac{1}{2}} \right.,\left. 1 \right\rangle \\ \end{gathered} \right.$

Analogously, it can be seen that $\mathbb{E}\left[ {Y|{\mathcal{G}_n}} \right]\left( \omega \right) = \frac{{\frac{i}{n} + \frac{{i + 1}}{n}}}{2},\omega \in \left[ {\frac{i}{n}} \right.,\left. {\frac{{i + 1}}{n}} \right\rangle $.

So, as we make ${{\mathcal{G}_n}}$ finer and finer, ${\mathbb{E}\left[ {Y|{\mathcal{G}_n}} \right]}$ resembles $Y$ more and more, and is equal to restricted expecation of $Y$ on all sets where it must be constant, namely on the basic "building blocks" of ${{\mathcal{G}_n}}$.

I suggest you draw first 4 or 5 conditional expectations with respect to given sigma algebras, the image is most illustrative.