Proof of Basic Properties of Conditional Expectation for Square-Integrable Random Variables

Throughout: $\mathbf{1}_{A}$ is $1$ on $A$ and $0$ off $A$; the defining conditions (i)-(iii) of a conditional expectation are those of Conditional Expectation of a Square-Integrable Random Variable; expectations are handled with Linearity and Monotonicity of the Lebesgue Integral; closure and integrability facts (sums, scalar multiples, products; square-integrable implies integrable) are from Square-Integrable Random Variables and the Mean-Square Inner Product, whose measurability preliminaries apply verbatim with $\mathcal{G}$ in place of $\mathcal{F}$; and the uniqueness and equivalence assertions of Existence and Uniqueness of Conditional Expectation for Square-Integrable Random Variables are used freely. We also use the null-support principle recorded in the proof of Mean-Square Completeness of Square-Integrable Random Variables (Riesz-Fischer): a nonnegative random variable vanishing off an event of probability $0$ has expectation $0$.

Part 1 (linearity). The variable $aY+bY'$ is $\mathcal{G}$-measurable and square-integrable. For $A\in\mathcal{G}$, pointwise $(aX+bX')\mathbf{1}_{A}=a\,X\mathbf{1}_{A}+b\,X'\mathbf{1}_{A}$ and likewise for $aY+bY'$; all products are integrable, so linearity of expectation and condition (iii) for $Y$ and $Y'$ give

$$\mathbb{E}\bigl[(aX+bX')\mathbf{1}_{A}\bigr]=a\,\mathbb{E}[X\mathbf{1}_{A}]+b\,\mathbb{E}[X'\mathbf{1}_{A}]=a\,\mathbb{E}[Y\mathbf{1}_{A}]+b\,\mathbb{E}[Y'\mathbf{1}_{A}]=\mathbb{E}\bigl[(aY+bY')\mathbf{1}_{A}\bigr].$$

Hence $aY+bY'$ satisfies (i)-(iii) for $aX+bX'$.

Part 2 (monotonicity). Fix $n\in\mathbb{N}$ and let $A=\{Y-Y'\ge\tfrac1n\}\in\mathcal{G}$ (preimage of a Borel ray under the $\mathcal{G}$-measurable variable $Y-Y'$). By condition (iii) for $Y$ and $Y'$ and linearity,

$$\mathbb{E}\bigl[(Y-Y')\mathbf{1}_{A}\bigr]=\mathbb{E}[X\mathbf{1}_{A}]-\mathbb{E}[X'\mathbf{1}_{A}]=\mathbb{E}\bigl[(X-X')\mathbf{1}_{A}\bigr].$$

Write $E=\{X\le X'\}$, so $P(\Omega\setminus E)=0$. Pointwise $(X-X')\mathbf{1}_{A}=(X-X')\mathbf{1}_{A\cap E}+(X-X')\mathbf{1}_{A\setminus E}$. The first term is $\le0$ everywhere, so its expectation is $\le0$ by monotonicity; the second term has expectation $0$, since its positive and negative parts are nonnegative variables vanishing off the probability-zero event $A\setminus E$ (null-support principle). Hence $\mathbb{E}[(Y-Y')\mathbf{1}_{A}]\le0$. On the other hand, pointwise $(Y-Y')\mathbf{1}_{A}\ge\tfrac1n\mathbf{1}_{A}$, so $0\ge\mathbb{E}[(Y-Y')\mathbf{1}_{A}]\ge\tfrac1nP(A)$ and $P(A)=0$. The event $\{Y>Y'\}$ is the countable union over $n$ of these events, so $P(Y>Y')=0$ as in the uniqueness argument of Existence and Uniqueness of Conditional Expectation for Square-Integrable Random Variables, i.e., $P(Y\le Y')=1$.

Part 3 (tower property). Let $V$ be a conditional expectation of $Y$ given $\mathcal{H}$. Then $V$ is $\mathcal{H}$-measurable and square-integrable. For $A\in\mathcal{H}$ we have $A\in\mathcal{G}$ as well, so condition (iii) for $Y$ (with respect to $X$, $\mathcal{G}$) and for $V$ (with respect to $Y$, $\mathcal{H}$) give

$$\mathbb{E}[X\mathbf{1}_{A}]=\mathbb{E}[Y\mathbf{1}_{A}]=\mathbb{E}[V\mathbf{1}_{A}].$$

Hence $V$ satisfies (i)-(iii) for $X$ and $\mathcal{H}$.

Part 4 (taking out what is known). Pointwise $(ZX)^{2}\le C^{2}X^{2}$ and $(ZY)^{2}\le C^{2}Y^{2}$, so $ZX$ and $ZY$ are square-integrable by monotonicity; $ZY$ is $\mathcal{G}$-measurable as a product of $\mathcal{G}$-measurable variables. Fix $A\in\mathcal{G}$. The variable $Z\mathbf{1}_{A}$ is $\mathcal{G}$-measurable and square-integrable ($(Z\mathbf{1}_{A})^{2}\le C^{2}$ everywhere, and constants are square-integrable on a probability space). Since $Y$ is a conditional expectation of $X$ given $\mathcal{G}$, the orthogonality property of Existence and Uniqueness of Conditional Expectation for Square-Integrable Random Variables (equivalent to the averaging property) applies with the test variable $Z\mathbf{1}_{A}$:

$$\mathbb{E}\bigl[(X-Y)\,Z\mathbf{1}_{A}\bigr]=0,\qquad\text{i.e.}\qquad \mathbb{E}\bigl[(ZX)\mathbf{1}_{A}\bigr]=\mathbb{E}\bigl[(ZY)\mathbf{1}_{A}\bigr],$$

using the pointwise identity $(X-Y)Z\mathbf{1}_{A}=(ZX)\mathbf{1}_{A}-(ZY)\mathbf{1}_{A}$ and linearity (all products integrable). Hence $ZY$ satisfies (i)-(iii) for $ZX$.

Part 5 (independence). Let $c=\mathbb{E}[X]$, which is defined and finite since square-integrable variables are integrable. The constant function with value $c$ is $\mathcal{G}$-measurable (its preimages are $\emptyset$ or $\Omega$) and square-integrable. Fix $A\in\mathcal{G}$. The pair $X$, $\mathbf{1}_{A}$ is independent: for Borel sets $B,B'$, the event $\{X\in B\}$ lies in $\sigma(X)$ and the event $\{\mathbf{1}_{A}\in B'\}$ is one of $\emptyset$, $A$, $\Omega\setminus A$, $\Omega$, each lying in $\mathcal{G}$; the required product formulas (for the pair and for each singleton subfamily) are instances of the defining product formula of Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras for the independent pair $\sigma(X)$, $\mathcal{G}$, with $\Omega$ inserted for omitted factors. Both $X$ and $\mathbf{1}_{A}$ are integrable, so Expectation of a Product of Independent Random Variables gives

$$\mathbb{E}[X\mathbf{1}_{A}]=\mathbb{E}[X]\,\mathbb{E}[\mathbf{1}_{A}]=c\,P(A)=\mathbb{E}[c\,\mathbf{1}_{A}],$$

the last step by linearity applied to the simple function $c\,\mathbf{1}_{A}$. Hence the constant $c$ satisfies (i)-(iii) for $X$, and $P(Y=c)=1$ follows from the uniqueness assertion of Existence and Uniqueness of Conditional Expectation for Square-Integrable Random Variables.

Part 6 (mean-square contraction). $Y$ itself is $\mathcal{G}$-measurable and square-integrable, so the orthogonality property of Existence and Uniqueness of Conditional Expectation for Square-Integrable Random Variables with test variable $Y$ gives $\mathbb{E}[(X-Y)Y]=0$. Expanding $X^{2}=\bigl((X-Y)+Y\bigr)^{2}$ pointwise and using linearity,

$$\lVert X\rVert_{2}^{2}=\mathbb{E}\bigl[(X-Y)^{2}\bigr]+2\,\mathbb{E}\bigl[(X-Y)Y\bigr]+\mathbb{E}[Y^{2}]=\mathbb{E}\bigl[(X-Y)^{2}\bigr]+\lVert Y\rVert_{2}^{2}\ge\lVert Y\rVert_{2}^{2}.$$

Since the nonnegative square root preserves the order of nonnegative reals (as recalled in the proof of Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm), $\lVert Y\rVert_{2}\le\lVert X\rVert_{2}$. $\blacksquare$