Conditional Expectation

Let $\mu =P{\mathrm{\mid }}_{\mathcal{G}}$ and ${\nu }_{\pm }(A)={\int }_A{X}^{\pm }dP$. Restricting the measures in [F2] from $\mathcal{F}$ to $\mathcal{G}$ makes ${\nu }_{\pm }$ finite positive measures, with total mass at most $E\mathrm{\mid }X\mathrm{\mid }$. If $\mu (A)=0$, [F7] gives ${\nu }_{\pm }(A)=0$, so ${\nu }_{\pm }\ll \mu$. The constant exhaustion $\mathrm{\Omega }$ has finite $\mu$ and finite variation for both positive measures.

Apply [F3] separately to $(\mu ,{\nu }_{+})$ and $(\mu ,{\nu }_{-})$. Its AC hypothesis is [F6]; its exhaustion and absolute continuity were checked in step 1.1. It supplies real, $\mathcal{G}$-measurable integrable $f_{+},f_{-}$ with ${\int }_A{f}_{\pm }d\mu ={\nu }_{\pm }(A)$. They are nonnegative almost everywhere: on $N_{\pm }=\{f_{\pm }<0\}$ positivity of ${\nu }_{\pm }$ and nonpositivity of the integral force ${\int }_{N_{\pm }}(-f_{\pm })=0$, so [F5] makes $N_{\pm }$ null. Set them to zero there. These are $\mathcal{G}$-measurable null sets, so the modification is legitimate without completing $\mathcal{G}$. The current RN interface already supplies real integrable densities, so no infinite density is subtracted.

Put $Y=f_{+}-f_{-}$. It is real, $\mathcal{G}$-measurable and integrable, and for every $A\in \mathcal{G}$, ${\int }_{A}YdP={\nu }_{+}(A)-{\nu }_{-}(A)={\int }_A(X^{+}-X^{-})dP={\int }_{A}XdP$, with only finite subtractions. Thus [F1] makes $Y$ the required version.

The difference $D=Y-Z$ is $\mathcal{G}$-measurable and integrable, and ${\int }_{A}DdP=0$ for every $A\in \mathcal{G}$. In particular the sets $A_{+}=\{D>0\}$ and $A_{-}=\{D<0\}$ belong to $\mathcal{G}$.

On $A_{+}$, $D{\mathbf{1}}_{A_{+}}=D^{+}\ge 0$ has zero integral; on $A_{-}$, $-D{\mathbf{1}}_{A_{-}}=D^{-}\ge 0$ also has zero integral. By [F4], both parts vanish almost surely. Off the union of their two null exceptional sets, $D=D^{+}-D^{-}=0$, proving $Y=Z$ almost surely.

Choose versions $U=E[X\mid \mathcal{G}]$ and $V=E[Y\mid \mathcal{G}]$. The function $aU+bV$ is $\mathcal{G}$-measurable and integrable. For every $A\in \mathcal{G}$, ${\int }_A(aU+bV)=a{\int }_{A}X+b{\int }_{A}Y={\int }_A(aX+bY)$. It is a version of the left side, so uniqueness proves linearity.

If $X\ge 0$ almost surely, on $A=\{U<0\}\in \mathcal{G}$ we have $0\le {\int }_{A}X={\int }_{A}U\le 0$. Thus ${\int }_A(-U)=0$, and [F4] gives $P(A)=0$. For $X\le Y$ apply this to $Y-X$ and use linearity; this proves order preservation.

The constant function $c$ is integrable, $\mathcal{G}$-measurable, and has its own event integrals, so $E[c\mid \mathcal{G}]=c$ by uniqueness. Testing $A=\mathrm{\Omega }$ in [F1] gives $EU=EX$. Finally $-\mathrm{\mid }X\mathrm{\mid }\le X\le \mathrm{\mid }X\mathrm{\mid }$ and steps 1.1–2.1 give $-E[\mathrm{\mid }X\mathrm{\mid }\mid \mathcal{G}]\le U\le E[\mathrm{\mid }X\mathrm{\mid }\mid \mathcal{G}]$. Hence $\mathrm{\mid }U\mathrm{\mid }\le E[\mathrm{\mid }X\mathrm{\mid }\mid \mathcal{G}]$ and $E\mathrm{\mid }U\mathrm{\mid }\le E\mathrm{\mid }X\mathrm{\mid }$.

If $W=Y-X>0$ almost surely, let $T=E[W\mid \mathcal{G}]\ge 0$ almost surely by step 2.1. The event $A=\{T=0\}$ has ${\int }_{A}W={\int }_{A}T=0$, so $W{\mathbf{1}}_A=0$ almost surely by [F4]. Since $W>0$ off a null set, this forces $P(A)=0$. Together with $P(T<0)=0$ this gives $T>0$ almost surely; linearity identifies $T=V-U$.

Write $U=E[X\mid \mathcal{G}]$. If $Z={\mathbf{1}}_B$ with $B\in \mathcal{G}$, then for every $A\in \mathcal{G}$, ${\int }_{A}ZU={\int }_{A\cap B}U={\int }_{A\cap B}X={\int }_{A}ZX$. The products are integrable and $ZU$ is $\mathcal{G}$-measurable, so uniqueness gives the identity. Finite sums of such indicators give the identity for bounded simple $Z$ by linearity.

If $\mathrm{\mid }Z\mathrm{\mid }\le M$, apply [F4] to $Z^{+}$ and $Z^{-}$ and subtract their approximations to obtain simple $Z_n\to Z$ with $\mathrm{\mid }{Z}_n\mathrm{\mid }\le M$. Then $\mathrm{\mid }{Z}_{n}X\mathrm{\mid }\le M\mathrm{\mid }X\mathrm{\mid }$ and $\mathrm{\mid }{Z}_{n}U\mathrm{\mid }\le M\mathrm{\mid }U\mathrm{\mid }$, both integrable majorants. By [F5] in each event identity of step 1.1, ${\int }_{A}ZU={\int }_{A}ZX$. The limit $ZU$ is measurable and integrable, so [F3] proves the bounded case.

For finite measurable $Z$ with $ZX\in L^1$, set $V=E[\mathrm{\mid }X\mathrm{\mid }\mid \mathcal{G}]\ge 0$ and $R_n=\mathrm{\mid }Z\mathrm{\mid }\wedge n$. The bounded case and expectation preservation imply $E[R_{n}V]=E[R_n\mathrm{\mid }X\mathrm{\mid }]$. MCT gives $E[\mathrm{\mid }Z\mathrm{\mid }V]=E\mathrm{\mid }ZX\mathrm{\mid }<\mathrm{\infty }$. Since $\mathrm{\mid }U\mathrm{\mid }\le V$ almost surely, $E\mathrm{\mid }ZU\mathrm{\mid }\le E\mathrm{\mid }ZX\mathrm{\mid }<\mathrm{\infty }$.

Let $Z_n=\max (-n,\min (Z,n))$. For each event $A$ the bounded case gives ${\int }_A{Z}_{n}U={\int }_A{Z}_{n}X$. Now $Z_{n}U\to ZU$ and $Z_{n}X\to ZX$, dominated by the integrable $\mathrm{\mid }ZU\mathrm{\mid }$ and $\mathrm{\mid }ZX\mathrm{\mid }$ respectively. DCT and uniqueness yield $E[ZX\mid \mathcal{G}]=ZU$. This proves both the stated two-product extension and its stronger integrability observation.

The same indicator identity gives locality: if $X=Y$ almost surely on $B\in \mathcal{G}$, then ${\mathbf{1}}_{B}E[X\mid \mathcal{G}]=E[{\mathbf{1}}_{B}X\mid \mathcal{G}]=E[{\mathbf{1}}_{B}Y\mid \mathcal{G}]={\mathbf{1}}_{B}E[Y\mid \mathcal{G}]$ as classes. No assertion is made about arbitrary values on exceptional points.

Set $U=E[X\mid \mathcal{G}]$ and take a version $V=E[U\mid \mathcal{H}]$. For $A\in \mathcal{H}\subseteq \mathcal{G}$, ${\int }_{A}V={\int }_{A}U={\int }_{A}X$. Since $V$ is $\mathcal{H}$-measurable and integrable, [F2] identifies it with $E[X\mid \mathcal{H}]$.

A version $W=E[X\mid \mathcal{H}]$ is already $\mathcal{G}$-measurable and integrable. It has its own event integrals ${\int }_{A}W={\int }_{A}W$ on every $A\in \mathcal{G}$, so it is a version of $E[W\mid \mathcal{G}]$; [F2] gives the second identity. Finally if $U$ itself is $\mathcal{H}$-measurable, its $\mathcal{G}$ event identities restrict to $\mathcal{H}$, so [F2] gives the additional assertion.

If $X$ is $\mathcal{G}$-measurable it itself meets every condition for a version: integrability is assumed and every event equality is ${\int }_{A}X={\int }_{A}X$. Hence uniqueness gives the first identity.

Fix $A\in \mathcal{G}$ under the independence hypothesis. For $h={\mathbf{1}}_B$, $E[h(X){\mathbf{1}}_A]=E[h(X)]P(A)$ is exactly that hypothesis. For nonnegative Borel simple $h={\sum }_{j=1}^m{c}_j{\mathbf{1}}_{B_j}$, multiplication by $c_j$ and addition give the same equality. For any nonnegative Borel $h$, compose the increasing Borel simple approximations from [F3] with $X$ and use [F4] on both sides to obtain the equality, allowing infinite values.

Apply step 1.2 to $h(t)=t^{+}$ and $h(t)=t^{-}$. Their expectations are finite because $X\in L^1$, so subtracting yields ${\int }_{A}X=EXP(A)$. The constant $EX$ is finite, $\mathcal{G}$-measurable and integrable, and its integral on $A$ is $EXP(A)$. Since $A$ was arbitrary, [F1]–[F2] identify it with $E[X\mid \mathcal{G}]$.

For the ordered nonnegative versions $U_n$ of [F1], ordinary MCT on each event gives ${\int }_A{\lim }_n{U}_n={\lim }_n{\int }_A{U}_n={\lim }_n{\int }_A(X\wedge n)={\int }_{A}X$. Changes on the common measurable null set have zero event integral by [F5]. Thus the limit has the stated event characterization. If inputs are changed almost surely, their nonnegative integrals also agree by splitting each event into its part in and outside the measurable exceptional null set.

For uniqueness, let $Y,Z$ be two characterized versions and set $A_{k,m}=\{Y\ge Z+1\mathrm{/}k,Z\le m\}$ for positive integers $k,m$. This is in $\mathcal{G}$: the difference is formed only on the finite-Z set. On $A_{k,m}$, $\int Z\le m$, and the event identities give $\int Y=\int Z<\mathrm{\infty }$. Integration of $Y\ge Z+1\mathrm{/}k$ there yields $P(A_{k,m})\mathrm{/}k\le 0$. Their countable union is $\{Y>Z\}$, since strict extended inequality forces the smaller value to be finite. Thus $P(Y>Z)=0$; interchanging the two variables gives equality almost surely. No infinite integrals are subtracted.

More generally, if $X\le X^{\ast }$ almost surely and $Y,Z$ are their characterized versions, then ${\int }_{A}Y\le {\int }_{A}Z$ on all $\mathcal{G}$ events. On the same $A_{k,m}$ as in step 2.1, the right integral is finite and the inequality forces $P(A_{k,m})=0$. Thus $Y\le Z$ almost surely. This extends order to the nonnegative classes, including infinite values.

Choose versions $Y_n$ for the given $X_n$ using [F6]. By step 3.1 remove one $\mathcal{G}$-null union of consecutive order-exception sets and set all $Y_n$ to zero there. Their limit $Y$ is measurable by [F4]. MCT and the event identities give ${\int }_{A}Y={\lim }_n{\int }_A{X}_n={\int }_{A}X$. For almost-sure input monotonicity the common ambient measurable null set can be removed from the inputs using [F5]; this does not require that set to belong to $\mathcal{G}$. Step 2.1 now identifies $Y$ with $E[X\mid \mathcal{G}]$. The same argument works for any increasing integrable nonnegative approximations.

Finally $0\le V_0-V_n\uparrow V_0-V$ almost surely, and all these variables are integrable because $\mathrm{\mid }{V}_n\mathrm{\mid }\le \mathrm{\mid }{V}_0\mathrm{\mid }+\mathrm{\mid }V\mathrm{\mid }$. Apply step 4.1 and linearity [F2] to obtain $E[V_0\mid \mathcal{G}]-E[V_n\mid \mathcal{G}]\uparrow E[V_0\mid \mathcal{G}]-E[V\mid \mathcal{G}]$. The fixed first term is finite almost surely, so subtraction gives the claimed decreasing convergence.

Put $Z_n={\inf }_{k\ge n}{X}_k$. These are measurable by [F5], nonnegative, and increase to ${\mathrm{lim\hspace{0.17em}inf}}_k{X}_k$. For each $k\ge n$, [F1] gives $E[Z_n\mid \mathcal{G}]\le E[X_k\mid \mathcal{G}]$ almost surely. There are only countably many pairs $(n,k)$; after removing their null union, take the infimum over $k\ge n$ and then the increasing limit in $n$. Conditional MCT gives exactly the claimed Fatou inequality.

In the dominated case, $\mathrm{\mid }X\mathrm{\mid }\le W$ almost surely, so [F3] gives $X\in L^1$. Set $T=E[W\mid \mathcal{G}]$, $U_n=E[X_n\mid \mathcal{G}]$ and $U=E[X\mid \mathcal{G}]$. These are finite almost surely. Apply step 1.1 to $W+X_n$ and $W-X_n$, which are nonnegative. By linearity this gives $T+U\le T+\mathrm{lim\hspace{0.17em}inf}{U}_n$ and $T-U\le T-\mathrm{lim\hspace{0.17em}sup}{U}_n$. The modulus bound gives $\mathrm{\mid }{U}_n\mathrm{\mid },\mathrm{\mid }U\mathrm{\mid }\le T$ outside one common null set. Subtracting the finite $T$ yields $U\le \mathrm{lim\hspace{0.17em}inf}{U}_n\le \mathrm{lim\hspace{0.17em}sup}{U}_n\le U$, hence almost-sure convergence.

The differences $\mathrm{\mid }{U}_n-U\mathrm{\mid }$ tend to zero almost surely and are bounded by $2T$, with $ET=EW<\mathrm{\infty }$. Ordinary DCT therefore gives $E\mathrm{\mid }{U}_n-U\mathrm{\mid }\to 0$, the claimed $L^1$ convergence.

By [F1], $m_q={\varphi }_{-}^{\mathrm{\prime }}(q)$ is finite and lies between ${\varphi }_{-}^{\mathrm{\prime }}(q)$ and ${\varphi }_{+}^{\mathrm{\prime }}(q)$. Therefore [F2] gives ${\mathrm{\ell }}_q(t)\le \varphi (t)$ for every $t$, with equality at $t=q$. The family is countable by [F3], and no slope choice is made.

Fix real $a<b$. For $a\le u<v\le b$, the inequalities in [F1], also applied between $a-1,u$ and $v,b+1$, bound the secant slope between the finite numbers ${\varphi }_{+}^{\mathrm{\prime }}(a-1)$ and ${\varphi }_{-}^{\mathrm{\prime }}(b+1)$. The same bounds hold for ${\varphi }_{-}^{\mathrm{\prime }}(q)$ for $q\in [a,b]$. Let $M$ be the maximum of their absolute values. Then $\mathrm{\mid }\varphi (v)-\varphi (u)\mathrm{\mid }\le M\mathrm{\mid }v-u\mathrm{\mid }$, proving Lipschitz continuity on $[a,b]$ and thus continuity everywhere; [F5] gives Borel measurability.

For fixed $x$ use step 1.2 on $[x-1,x+1]$. Given $\epsilon >0$, [F4] supplies rational $q$ in this interval with $\mathrm{\mid }q-x\mathrm{\mid }<\epsilon \mathrm{/}(2M+1)$. Then $0\le \varphi (x)-{\mathrm{\ell }}_q(x)\le \mathrm{\mid }\varphi (x)-\varphi (q)\mathrm{\mid }+\mathrm{\mid }{m}_q\mathrm{\mid }\mathrm{\mid }x-q\mathrm{\mid }\le 2M\mathrm{\mid }x-q\mathrm{\mid }<\epsilon$. Thus the supremum of the supporting lines is at least $\varphi (x)-\epsilon$ for every positive $\epsilon$, and at most $\varphi (x)$ by step 1.1, proving equality. This also handles $M=0$ and affine functions with irrational slopes.

Set $U=E[X\mid \mathcal{G}]$ and $V=E[\varphi (X)\mid \mathcal{G}]$. For each supporting line ${\mathrm{\ell }}_q(t)=m_{q}t+b_q$ of [F3], the variable $m_{q}X+b_q$ is integrable and bounded above by $\varphi (X)$. Linearity and order give $m_{q}U+b_q\le V$ almost surely. There are countably many $q$, so remove one measurable null union to make all inequalities hold together.

On the resulting conull set take the supremum over $q$. By [F3], $\varphi (U)={\sup }_q(m_{q}U+b_q)\le V$. Measurability follows either from that countable supremum or composition with the Borel function $\varphi$. The fixed supporting line at $q=0$ also gives $m_{0}U+b_0\le \varphi (U)$. Hence $(\varphi (U){)}^{+}\le V^{+}$ and $(\varphi (U){)}^{-}\le (m_{0}U+b_0{)}^{-}$ almost surely. Both upper bounds are integrable, proving integrability as well as the inequality.

For $p>1$ and $u>0$, differentiation gives $h^{\mathrm{\prime }}(u)=p{u}^{p-1}$ and $h^{\mathrm{\prime }\mathrm{\prime }}(u)=p(p-1)u^{p-2}\ge 0$ for $h(u)=u^p$. Thus $h$ is nondecreasing and convex on $(0,\mathrm{\infty })$. The derivative and second-derivative hypotheses hold at every positive $u$.

For $0<u\le 1$, $\log u\le 0$, hence $p\log u\le \log u$ and $0<u^p=\exp (p\log u)\le \exp (\log u)=u$. The last equality is the inverse identity [F16]. The squeeze theorem gives $u^p\to 0$ as $u\downarrow 0$. With $h(0)=0$, this extends $h$ continuously to $[0,\mathrm{\infty })$.

For $a,b\ge 0$ and $0\le \lambda \le 1$, apply positive-half-line convexity to $a+\epsilon ,b+\epsilon$ and let $\epsilon \downarrow 0$ using step 1.2: $h(\lambda a+(1-\lambda )b)\le \lambda h(a)+(1-\lambda )h(b)$. The inequality remains valid at weights zero and one, where it is equality. Monotonicity extends to zero because $h\ge 0=h(0)$.

For real $x,y$, [F10]–[F11] give $\mathrm{\mid }\lambda x+(1-\lambda )y\mathrm{\mid }\le \lambda \mathrm{\mid }x\mathrm{\mid }+(1-\lambda )\mathrm{\mid }y\mathrm{\mid }$. Apply monotonicity and then step 2.1 to get $\mathrm{\mid }\lambda x+(1-\lambda )y{\mathrm{\mid }}^p\le \lambda \mathrm{\mid }x{\mathrm{\mid }}^p+(1-\lambda )\mathrm{\mid }y{\mathrm{\mid }}^p$. For $p=1$ the same inequality is already precisely the triangle inequality with the scalar absolute values evaluated. Thus [F12] proves convexity for every $p\ge 1$.

The function is finite by [F1]. Continuity of absolute value [F13] and continuity of $h$ (steps 1.1–1.2, or the identity for $p=1$) imply continuity of $h(\mathrm{\mid }t\mathrm{\mid })$: choose an output tolerance for $h$ at $\mathrm{\mid }t\mathrm{\mid }$ and then the corresponding input tolerance for absolute value. Consequently all Borel preimages are Borel by [F14], which is exactly [F15].

For $p>1$, [F1] with $P(\mathrm{\Omega })=1$ makes $X$ integrable; for $p=1$ it is integrable by assumption. Put $U=E[X\mid \mathcal{G}]$. For finite $p$, [F2] supplies the convex Borel function and $E\mathrm{\mid }X{\mathrm{\mid }}^p<\mathrm{\infty }$ supplies its integrability. Jensen yields $\mathrm{\mid }U{\mathrm{\mid }}^p\le E[\mathrm{\mid }X{\mathrm{\mid }}^p\mid \mathcal{G}]$. Taking expectations using [F4] gives $E\mathrm{\mid }U{\mathrm{\mid }}^p\le E\mathrm{\mid }X{\mathrm{\mid }}^p$ and hence the norm inequality by taking the increasing positive pth root. At $p=1$ this is also the modulus estimate of [F4].

If $p=\mathrm{\infty }$, let $M=\mathrm{\parallel }X{\mathrm{\parallel }}_{\mathrm{\infty }}$. The inequalities $\mathrm{\mid }X\mathrm{\mid }\le M+1\mathrm{/}n$ for all positive integers hold outside a common null set; their limit gives $\mathrm{\mid }X\mathrm{\mid }\le M$ almost surely. Conditional order and constants imply $-M\le U\le M$ almost surely. Thus $\mathrm{\parallel }U{\mathrm{\parallel }}_{\mathrm{\infty }}\le M$. Equality of input representatives preserves the conditional class, so the maps are well defined on [F5]; linearity is [F4].

The inclusion sends the class of a $\mathcal{G}$-measurable function to its ambient class. Two such functions agree almost surely for the restricted measure exactly when they do for P; their squared integrals are identical. Thus inclusion is well defined, injective, linear and isometric. If a sequence in its image converges in ambient $L^2$, its preimages are Cauchy, converge by [F2] under [F3], and their images converge to the same ambient limit by the isometry and uniqueness of metric limits. Hence the image is closed.

By [F1], $U\in L^2(\mathcal{G})$. Fix $Z\in L^2(\mathcal{G})$. Both $ZX$ and $ZU$ are integrable by [F4]. Taking-out [F5] gives $E[ZX\mid \mathcal{G}]=ZU$. Taking ordinary expectations by [F7] yields $E[ZX]=E[ZU]$, hence $E[Z(X-U)]=0$. This establishes orthogonality for every Z directly, and in particular for bounded G-measurable tests, without a density argument.

For every $Z\in L^2(\mathcal{G})$, expand $X-Z=(X-U)+(U-Z)$. All products are integrable by [F4], and step 1.2 annihilates the cross term. Therefore $E[(X-Z{)}^2]=E[(X-U{)}^2]+E[(U-Z{)}^2]$. The last term is nonnegative and is zero exactly when $U=Z$ as an $L^2$ class, since $L^2$ is a normed space. The minimizer is therefore unique. Finally [F6] fixes every member of the subspace, so the conditional map is indeed the projection onto it.

Write $U=E[X\mid \mathcal{G}]$. It is in $L^2$ by [F2]. Since $(X-U{)}^2\le 2X^2+2U^2$, the squared residual is integrable; it is nonnegative and measurable. Thus [F1] exists as an integrable nonnegative class by [F5]. If X or U is replaced by an almost-surely equal measurable representative, the square changes only on the union of those two measurable null sets. Its event integrals are unchanged by [F6], so its conditional class is unchanged.

The product XU is integrable by [F3]; U is finite and G-measurable. Hence [F4] yields $E[XU\mid \mathcal{G}]=UE[X\mid \mathcal{G}]=U^2$. Also U squared is integrable and G-measurable, so [F7] gives $E[U^2\mid \mathcal{G}]=U^2$. Expanding the residual square and using [F5] gives $E[(X-U{)}^2\mid \mathcal{G}]=E[X^2\mid \mathcal{G}]-2U^2+U^2=E[X^2\mid \mathcal{G}]-U^2$. All three conditional inputs are integrable, so the subtraction involves finite classes only.

Put $U=E[X\mid \mathcal{G}]$. By [F1], $U^2=E[X^2\mid \mathcal{G}]-\mathrm{Var}(X\mid \mathcal{G})$ is a difference of integrable functions, so U is square integrable. Taking expectations in that formula gives $E[\mathrm{Var}(X\mid \mathcal{G})]=E{X}^2-E{U}^2$. Moreover $EU=EX$ by [F2].

For any square integrable real V, expansion of the centered square in [F3] gives $\mathrm{Var}(V)=E{V}^2-(EV{)}^2$. Therefore $E[\mathrm{Var}(X\mid \mathcal{G})]+\mathrm{Var}(U)=E{X}^2-E{U}^2+E{U}^2-(EX{)}^2=\mathrm{Var}(X)$, proving the formula.