Conditional Expectation — Examples

Let $T={\sum }_i{c}_i{\mathbf{1}}_{A_i}$ with the displayed zero convention. It is G-measurable and $E\mathrm{\mid }T\mathrm{\mid }={\sum }_{i:P(A_i)>0}\mathrm{\mid }{\int }_{A_i}X\mathrm{\mid }\le {\sum }_i{\int }_{A_i}\mathrm{\mid }X\mathrm{\mid }=E\mathrm{\mid }X\mathrm{\mid }<\mathrm{\infty }$. Each positive-mass cell has ${\int }_{A_i}T=c_{i}P(A_i)={\int }_{A_i}X$; on a null cell both integrals are zero. Every G-event is a union of cells (these unions form a sigma-algebra), so adding proves every defining identity. Hence [F1]–[F2] identify T as the conditional mean.

For a numerical instance take four atoms of weight $1\mathrm{/}4$, permitted by [F3], partitioned into $A_1=\{1,2\}$ and $A_2=\{3,4\}$. For X with values $(0,2,4,6)$, the cell integrals are $1\mathrm{/}2$ and $5\mathrm{/}2$ and their masses are $1\mathrm{/}2$, so T has values $(1,1,5,5)$. Its mean is 3, matching $(0+2+4+6)\mathrm{/}4=3$. For $X={\mathbf{1}}_B$ the same positive-cell calculation is $c_i=P(B\cap A_i)\mathrm{/}P(A_i)$, the finite conditional-probability formula.

Every X is independent of the trivial sigma-algebra: for A empty both sides of the rectangle identity are zero, and for A=Omega both equal $P(X\in B)$. Thus [F1] gives the first formula. Since X is F-measurable, the known-variable clause of [F1] gives the second.

For example take two atoms a,b of masses $1\mathrm{/}4,3\mathrm{/}4$ using [F2], and X(a)=0, X(b)=4. Then $EX=0\mathrm{/}4+12\mathrm{/}4=3$. Under trivial conditioning the version has values (3,3), with integral 3 on Omega; under full conditioning it has values (0,4), with integrals 0 on {a} and 3 on {b}. These verify the two formulas numerically.

The sets $X^{-1}(B)$, B real Borel, form a sigma-algebra because preimages preserve complements and countable unions; by definition this is $\sigma (X)$. Thus the given rectangle identity is precisely independence of Y from every event of $\sigma (X)$. By [F1], $E[Y\mid \sigma (X)]=EY$ and $E[X\mid \sigma (X)]=X$. Linearity [F2] gives the stated sum formula.

Take $\mathrm{\Omega }=\{0,1{\}}^2$, each atom of mass $1\mathrm{/}4$, and $X(u,v)=u$, $Y(u,v)=2v$. This is a probability space by [F3]. Each coordinate value has mass $1\mathrm{/}2$ and each pair mass $1\mathrm{/}4=(1\mathrm{/}2)(1\mathrm{/}2)$; adding atom probabilities proves all Borel rectangle identities. Since $EY=1$, the conditional mean is $u+1$. Directly, on the u=0 fibre the values 0,2 average to 1, and on the u=1 fibre the values 1,3 average to 2.

List the at most countably many fibres $A_y=\{Y=y\}$. Every union of fibres is measurable as a countable union, and these unions are exactly $\sigma (Y)$: each fibre is a preimage of a singleton Borel set, and every preimage is a union of fibres. The proposed function T is thus $\sigma (Y)$-measurable. Its absolute integral is ${\sum }_y\mathrm{\mid }{c}_y\mathrm{\mid }P(A_y)\le {\sum }_y{\int }_{A_y}\mathrm{\mid }X\mathrm{\mid }=E\mathrm{\mid }X\mathrm{\mid }$, where [F3] applies to nonnegative finite partial sums. Null fibres have zero X integral and their countable union is null.

On each positive fibre ${\int }_{A_y}T=c_{y}P(A_y)={\int }_{A_y}X$, and on null fibres both sides are zero. For every union A of fibres, sum these identities; absolute summability follows from step 1.1 and integrability of X, with [F3] applied to positive and negative parts. Thus ${\int }_{A}T={\int }_{A}X$. By [F1]–[F2] T is the desired version.

For a concrete instance take atoms a,b,c with masses $1\mathrm{/}4,1\mathrm{/}4,1\mathrm{/}2$ by [F4], with Y values (0,0,2) and X values (2,6,10). The zero fibre has mass $1\mathrm{/}2$ and X integral $2\mathrm{/}4+6\mathrm{/}4=2$, so its conditional value is 4. The fibre at 2 has mass $1\mathrm{/}2$ and X integral 5, giving value 10. Hence T=(4,4,10), with mean $4\mathrm{/}4+4\mathrm{/}4+10\mathrm{/}2=7=EX$.

Write $X-Z=(X-U)+(U-Z)$. Since $U-Z\in L^2(\mathcal{G})$, [F1] gives $E[(X-U)(U-Z)]=0$. Expansion gives the displayed decomposition. The last term is the squared $L^2$ norm of U-Z, so it is zero exactly when Z=U almost surely. This proves both directions of the minimum-error assertion.

For an instance take four equally weighted atoms by [F2], X values (0,2,4,6), and G generated by the first pair and last pair. The G-measurable U=(1,1,5,5) has on each pair the same integral as X, namely $1\mathrm{/}2$ and $5\mathrm{/}2$, so it is the conditional mean. For the constant predictor Z=3 the total error is $(9+1+1+9)\mathrm{/}4=5$, the residual error is $(1+1+1+1)\mathrm{/}4=1$, and the prediction displacement is $(4+4+4+4)\mathrm{/}4=4$. Thus the formula reads $5=1+4$.

The four masses are nonnegative and sum to one, so [F1] constructs the probability space. Each U and V marginal has mass one half at zero and at one, and $P(U=u,V=v)=1\mathrm{/}4=P(U=u)P(V=v)$ for all four pairs. Summing over coordinate subsets gives independence for every Borel rectangle. Both variables are bounded and integrable, with $EU=EV=1\mathrm{/}2$.

By [F2]–[F3], $E[X\mid \mathcal{G}]=U+1\mathrm{/}2$. The residual is $V-1\mathrm{/}2$, whose square equals 1/4 at every atom. Thus [F5] and the constant rule [F3] give $\mathrm{Var}(X\mid \mathcal{G})=1\mathrm{/}4$ and its expectation 1/4.

The four X values are (0,1,1,2), with mean 1, so $\mathrm{Var}(X)=(1+0+0+1)\mathrm{/}4=1\mathrm{/}2$. The conditional mean takes values one half and three halves, each with probability one half; its mean is 1 and its variance is $((1\mathrm{/}2{)}^2+(1\mathrm{/}2{)}^2)\mathrm{/}2=1\mathrm{/}4$. Therefore the three computed quantities satisfy [F4] as $1\mathrm{/}2=1\mathrm{/}4+1\mathrm{/}4$.

Let $\mathrm{\Omega }=\{a,b\}$, $\mathcal{F}=\mathcal{P}(\mathrm{\Omega })$ and $P(A)={\mathbf{1}}_{\{b\in A\}}$. This is a probability measure: P(Omega)=1, P(empty)=0, and in a disjoint sequence at most one event contains b, so countable additivity holds. Take $T={\mathbf{1}}_{\{a\}}$. It is F-measurable and $E\mathrm{\mid }T\mathrm{\mid }=1\cdot 0+0\cdot 1=0$.

For every A subset Omega, ${\int }_{A}T=T(b){\mathbf{1}}_{\{b\in A\}}=0={\int }_{A}0$. Thus [F1] makes T a version of the conditional expectation of zero. But T(a)=1, so the claimed pointwise identity fails at a. The discrepancy set {a} has probability zero, consistent with almost-sure uniqueness.

On $\mathrm{\Omega }=\{a,b\}$ take the full sigma-algebra $\mathcal{G}=\mathcal{F}$ and $P(A)={\mathbf{1}}_{\{b\in A\}}$. Countable additivity holds because at most one member of a disjoint sequence contains b, and total mass is one. Let X=0 and Y=1 everywhere. Then X<Y at both points. Set S(a)=2,S(b)=0 and T=1 everywhere; these are measurable and integrable.

For every A, ${\int }_{A}S=0={\int }_{A}X$ and ${\int }_{A}T=P(A)={\int }_{A}Y$, so [F1] makes S,T conditional versions. At a, however, S(a)=2 is larger than T(a)=1. At the mass-one point b, S(b)=0<T(b)=1. Hence the failed strict pointwise inequality is fully consistent with the strict almost-sure order theorem [F2].

Let $\mathrm{\Omega }={\mathbb{N}}_{\ge 1}\times \{-1,1\}$ with its power-set sigma-algebra. By [F5]–[F6], $S={\sum }_{n\ge 1}{2}^{-3n}=(1\mathrm{/}8)\mathrm{/}(1-1\mathrm{/}8)=1\mathrm{/}7$. Set $w_n=2^{-3n-1}\mathrm{/}S$. These weights are positive finite numbers.

Define $P={\sum }_{n\ge 1}{w}_n{\delta }_{(n,-1)}+{\sum }_{n\ge 1}{w}_n{\delta }_{(n,1)}$. The Dirac probabilities [F1]–[F2] and weighted-sum construction [F3]–[F4] make this a measure on all subsets. Its mass is $2{\sum }_n{w}_n=S\mathrm{/}S=1$, so [F7] makes it a probability measure. In particular each atom (n,s) has mass w_n.

Let $\mathcal{G}=\sigma ((n,s)\mapsto n)$, $X(n,s)=s{2}^n$ and $Z(n,s)=2^{2n}$. Z is finite G-measurable, X is real measurable, and [F10] evaluates their absolute moments as sums. Using [F6], $E\mathrm{\mid }X\mathrm{\mid }=S^{-1}{\sum }_{n\ge 1}{2}^{-2n}=7\mathrm{/}3$ and $E\mathrm{\mid }Z\mathrm{\mid }=S^{-1}{\sum }_{n\ge 1}{2}^{-n}=7$. Thus each is integrable.

Each G-event is a union of two-point fibres. On the nth fibre the X integral is $w_n(-2^n)+w_n{2}^n=0$; summing is legitimate by the finite absolute moment in step 3.1. Consequently the zero function has every defining event integral and is a version of $E[X\mid \mathcal{G}]$ by [F8]. Hence $ZE[X\mid \mathcal{G}]=0$ is integrable.

But $ZX(n,s)=s{2}^{3n}$. Its positive-part integral is ${\sum }_{n\ge 1}{w}_n{2}^{3n}={\sum }_{n\ge 1}(2S{)}^{-1}=\mathrm{\infty }$, and the negative-part integral has exactly the same value. These sums are nonnegative integrals by [F10]. Thus the signed expectation would require infinity minus infinity; ZX is not an $L^1$ input to [F8]. The left side $E[ZX\mid \mathcal{G}]$ of [F9] is undefined in that sense although the proposed right side is zero. This proves the failure when the product-integrability hypothesis is omitted.