How can I show that the conditional expectation $E(X\mid X)=X$?

Question

I tried to show that $E(X\mid X=x)=x$, which would lead me to get $E(X\mid X)=X$ but I am having trouble doing so. I know that the definition of conditional expectation (continuous case) is: $$E(X\mid Y=y)=\int_{-\infty}^{\infty}x f_{X\mid Y}(x\mid y)\,dx$$ This led me to write: $$E(X\mid X=x)=\int_{-\infty}^{\infty}x f_{X\mid X}(x\mid x)\,dx$$ but I don't know what $f_{X\mid X}(x\mid x)$ is (if that's even a valid notation...)

I do note though that I don't know anything about $\sigma$-fields or Borel sets, as someone had tried to explain me once. Is it possible to prove this in an "elementary" way?

Answer 1

It is actually easier to define $E(X\mid Y)$ than $E(X\mid Y=y)$ hence let us recall how this is done. By definition, $E(X\mid Y)$ is any random variable $Z$ such that:

$\qquad$ (1.) $Z$ is a measurable function of $Y$.

$\qquad$ (2.) For every bounded random variable $v(Y)$, $E(Zv(Y))=E(Xv(Y))$.

It happens that when $X$ is integrable, such a random variable $Z$ exists and is almost surely unique, in the sense that if $Z$ and $Z'$ both satisfy conditions (1.) and (2.), then $P(Z=Z')=1$.

When $X=Y$, things are quite simple since $Z=X$ obviously satisfies conditions (1.) and (2.), thus, $E(X\mid X)=X$.

Note finally that $Z=X$ always satisfies (2.) but does not satisfy (1.) in general, and, likewise, that $Z=E(X)$ always satisfies (1.) but does not satisfy (2.) in general.

Answer 2

Maybe there is an elementary way to illustrate why $\mathbb{E}\left [g(X)\mid X \right ]=g(X),$ for any Borel-measurable function $g(x)$.

We start with a definition.

Definition. Let $X$ and $Y$ be random variables defined on a probability space $(\mathcal{Ω},\mathcal{S},\mathbb{P})$, and let $g$ be a Borel-measurable function. Then the conditional expectation of $g(X)$,given $Y$, written as $\mathbb{E}[g(X)\mid Y]$, is an RV that takes the value $\mathbb{E}[g(X)\mid Y=y]$, defined by $$\mathbb{E}[g(X)\mid Y=y]=$$$$ \begin{cases} \sum_{x}g(x)\mathbb{P}(X=x\mid Y=y)& \text{if} (X,Y)\text{ is of the discrete type and}\mathbb{P}(Y = y) > 0, \\ \int^{+\infty}_{-\infty}g(x)f_{X\mid Y}(x\mid y)dx& \text{if} f(X,Y) \text{is of the continuous type and} f_{Y}(y) > 0. \end{cases}$$ when the RV $Y$ assumes the value $y$.

Whatever the RV $X$ is discrete type or continuous type, let $X=x_{0}$ , then we have $$\mathbb{P}\left [ g(X)=k\mid X=x_{0} \right ]=\begin{cases} 1,& k=g(x_0) \\ 0,& k\ne g(x_0) \end{cases}.$$ Note that $k$ is a variable defined on real line and $x_{0}$ is a fixed real constant such that $\left \{\omega\mid X(\omega )=x_0 \right \}\ne \emptyset$ .

The conditional probability mass function of $g(X)$ given $X=x_{0}$ is given by the following table. $$\begin{array}{c|c|c} g(X)& g(X)=g(x_{0})& g(X)\ne g(x_0) \\ \hline \mathbb{P}[g(X)\mid X=x_{0}] & 1 & 0 \end{array}$$

From above definition of the discrete type, The conditional expectation of $g(X)$ given $X = x_{0}$ is $$\mathbb{E}[g(X)\mid X=x_{0}]=g(x_0)\cdot\mathbb{P}[g(X)=g(x_0)\mid X=x_{0}]=g(x_0).(\ast)$$

Finally,let's replace the fixed constant $x_{0}$ in $(\ast)$ with a variable $x$ which varies over the range of $X$,then it will get $$\mathbb{E}\left [g(X)\mid X\right ]=g(X).$$