Proof of Existence and Uniqueness of Conditional Expectation for Square-Integrable Random Variables

Write $\mathcal{S}$ for the set of $\mathcal{G}$-measurable square-integrable random variables on $(\Omega,\mathcal{F},P)$; norms, inner products, distances, and their properties are those of Square-Integrable Random Variables and the Mean-Square Inner Product, expectations are handled with Linearity and Monotonicity of the Lebesgue Integral, and the triangle inequality is Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm. As noted in the proof of Mean-Square Completeness of Square-Integrable Random Variables (Riesz-Fischer), the measurability preliminaries of Square-Integrable Random Variables and the Mean-Square Inner Product apply verbatim with $\mathcal{G}$ in place of $\mathcal{F}$; consequently $\mathcal{S}$ is closed under sums, scalar multiples, and midpoints, and contains the constant $0$.

Step 1 (parallelogram identity). For square-integrable $U,V$, expanding pointwise and using linearity of expectation,

Step 2 (a minimizing sequence and its limit). The set $T=\{\lVert X-Z\rVert_{2}:Z\in\mathcal{S}\}$ is a nonempty set of nonnegative reals ($0\in\mathcal{S}$), so it has a greatest lower bound $d\ge0$. For each $n\in\mathbb{N}$: the number $\sqrt{d^{2}+1/n}$ exceeds $d$ (its square exceeds $d^{2}$, and squaring preserves the order of nonnegative reals), so it is not a lower bound of $T$, and there is $Z_n\in\mathcal{S}$ with $\lVert X-Z_n\rVert_{2}<\sqrt{d^{2}+1/n}$; squaring gives

Apply Step 1 with $U=X-Z_n$, $V=X-Z_m$: since $U+V=2\bigl(X-\tfrac{Z_n+Z_m}{2}\bigr)$ and $U-V=Z_m-Z_n$,

The midpoint $\tfrac{Z_n+Z_m}{2}$ lies in $\mathcal{S}$, so the first term is at least $4d^{2}$, whence

Given $\varepsilon>0$, choose $N>4/\varepsilon^{2}$ by the Archimedean property; for $n,m\ge N$ the right side is less than $\varepsilon^{2}$, so $\lVert Z_m-Z_n\rVert_{2}<\varepsilon$ (order-preservation of the nonnegative square root). Thus $(Z_n)$ is Cauchy in mean square, and by Mean-Square Completeness of Square-Integrable Random Variables (Riesz-Fischer) there is a $\mathcal{G}$-measurable square-integrable random variable $Y$ with $\lVert Z_n-Y\rVert_{2}\to0$.

Step 3 ($Y$ attains the infimum; property 1). For every $n$, the triangle inequality gives

where the last step uses $\sqrt{d^{2}+1/n}\le d+\sqrt{1/n}$ (both sides nonnegative, and the square of the right side is $d^{2}+1/n+2d\sqrt{1/n}\ge d^{2}+1/n$). The right side has limit $d$ (for the middle term: given $\varepsilon>0$, $\sqrt{1/n}<\varepsilon$ once $n>1/\varepsilon^{2}$, by the Archimedean property and order-preservation of squaring), so the constant $\lVert X-Y\rVert_{2}$ satisfies $d\le\lVert X-Y\rVert_{2}\le d+\varepsilon$ for every $\varepsilon>0$, hence $\lVert X-Y\rVert_{2}=d$. Since $d$ is a lower bound of $T$, property 1 holds for $Y$.

Step 4 (property 1 implies property 2). Let $Y'\in\mathcal{S}$ satisfy property 1; then $\lVert X-Y'\rVert_{2}=d$ (it is at most every element of $T$ and belongs to $T$). Fix $Z\in\mathcal{S}$ and set $b=\mathbb{E}[(X-Y')Z]$, which is defined and finite by Square-Integrable Random Variables and the Mean-Square Inner Product. For every real $t$, $Y'+tZ\in\mathcal{S}$, so expanding pointwise and using linearity,

If $\lVert Z\rVert_{2}=0$, then $0\le-2tb$ for all real $t$ forces $b=0$. Otherwise take $t=b/\lVert Z\rVert_{2}^{2}$, which yields $0\le-b^{2}/\lVert Z\rVert_{2}^{2}$, so again $b=0$. Hence property 2 holds for $Y'$.

Step 5 (property 2 implies property 3). Let $Y'\in\mathcal{S}$ satisfy property 2 and let $A\in\mathcal{G}$. The indicator $\mathbf{1}_{A}$ is $\mathcal{G}$-measurable (its preimages are $\emptyset$, $A$, $\Omega\setminus A$, or $\Omega$) and square-integrable ($\mathbf{1}_{A}^{2}=\mathbf{1}_{A}$ has expectation $P(A)\le1$), so $\mathbf{1}_{A}\in\mathcal{S}$ and property 2 gives $\mathbb{E}[(X-Y')\mathbf{1}_{A}]=0$. The products $X\mathbf{1}_{A}$ and $Y'\mathbf{1}_{A}$ are integrable by Square-Integrable Random Variables and the Mean-Square Inner Product, and $(X-Y')\mathbf{1}_{A}=X\mathbf{1}_{A}-Y'\mathbf{1}_{A}$ pointwise, so linearity yields $\mathbb{E}[X\mathbf{1}_{A}]=\mathbb{E}[Y'\mathbf{1}_{A}]$, which is property 3.

Step 6 (property 3 implies property 1, closing the equivalence). Let $Y'\in\mathcal{S}$ satisfy property 3. We first show that any two members of $\mathcal{S}$ satisfying property 3 are almost surely equal; we then deduce property 1 for $Y'$.

Almost-sure uniqueness from property 3. Suppose $Y_1,Y_2\in\mathcal{S}$ both satisfy property 3. Fix $n\in\mathbb{N}$ and let $A=\{Y_1-Y_2\ge\tfrac1n\}$. Then $A\in\mathcal{G}$: $Y_1-Y_2$ is $\mathcal{G}$-measurable by the preliminaries of Square-Integrable Random Variables and the Mean-Square Inner Product applied with $\mathcal{G}$, and $A$ is the preimage of the Borel set $[\tfrac1n,\infty)$. Property 3 for both variables gives

Pointwise $(Y_1-Y_2)\mathbf{1}_{A}\ge\tfrac1n\mathbf{1}_{A}$ (on $A$ by definition of $A$; off $A$ both sides are $0$), so monotonicity and linearity give $0\ge\tfrac1n P(A)$, hence $P(A)=0$. The event $\{Y_1>Y_2\}$ is the union over $n\in\mathbb{N}$ of the events $\{Y_1-Y_2\ge\tfrac1n\}$, each of probability $0$, so $P(Y_1>Y_2)=0$ by countable additivity (applied to a disjointified union, or by the monotone bound of the measure of a countable union by the sum of the measures as in Borel-Cantelli Lemmas). Exchanging $Y_1$ and $Y_2$ gives $P(Y_2>Y_1)=0$, so $P(Y_1\ne Y_2)=0$, i.e., $P(Y_1=Y_2)=1$.

Property 1 for $Y'$. By Steps 3-5, the random variable $Y$ constructed in Step 2 lies in $\mathcal{S}$ and satisfies properties 1, 2, and 3. By the uniqueness just proved, $P(Y'=Y)=1$, so $\lVert Y'-Y\rVert_{2}=0$ by Square-Integrable Random Variables and the Mean-Square Inner Product. Then for every $Z\in\mathcal{S}$, the triangle inequality gives

so $Y'$ satisfies property 1. This closes the cycle: property 1 implies 2 (Step 4), 2 implies 3 (Step 5), and 3 implies 1 (this step), so the three properties are equivalent for members of $\mathcal{S}$.

Step 7 (conclusion). The random variable $Y$ of Step 2 is $\mathcal{G}$-measurable, square-integrable, and satisfies properties 1, 2, and 3 (Steps 3-5). The equivalence of the three properties for members of $\mathcal{S}$ was established in Steps 4-6, and the final uniqueness assertion is the almost-sure uniqueness proved in Step 6. $\blacksquare$