Positivity and Bounds for Conditional Expectations

Why Conditional Expectations Preserve Bounds

A conditional expectation is an average taken within the information supplied by a conditioning variable. A basic consequence is that averaging cannot make a nonnegative quantity negative. More generally, if one random variable is at most another, their conditional averages retain that order. These facts are useful both as theoretical safeguards and as quick checks on calculations.

The qualifications matter: conditional distributions and conditional expectations are defined only up to sets of probability zero in the conditioning variable. Thus, a bound for a conditional expectation is generally an almost-sure statement, not a claim that holds for every possible conditioning value.

Conditional Distributions and Conditional Means

Let $X$ be a real-valued random variable and $Z$ a random variable taking values in a standard Borel space, such as the real line or a finite set. A regular conditional distribution of $X$ given $Z$ is a probability kernel $K(z,A)$ such that, for each measurable set $A$, $K(Z,A)$ is a version of $P(X\in A\mid \sigma(Z))$. In words, $K(z,\cdot)$ describes the conditional law of $X$ at conditioning value $z$, for almost every $z$ under the distribution of $Z$.

Definition: If $E[|X|]<\infty$, the conditional mean obtained from the conditional distribution is $$ g(z)=\int_{\mathbb{R}}x\,K(z,dx). $$ The random variable $g(Z)$ is a version of $E[X\mid \sigma(Z)]$, also written $E[X\mid Z]$. The integral is finite for almost every $z$ under the distribution of $Z$, since $$ \int \left(\int_{\mathbb{R}}|x|\,K(z,dx)\right)P_Z(dz)=E[|X|]<\infty. $$

Here $P_Z$ denotes the distribution of $Z$. Conditional distributions, and hence the resulting function $g$, may be changed on a set of $z$-values having $P_Z$-probability zero without changing the conditional expectation as a random variable. When $X$ is discrete, the integral is a conditional probability-weighted sum; when it has a conditional density, the integral is taken against that density.

Worked Example: Nonnegativity from a Discrete Conditional Law

Suppose $Z$ takes values $a$ and $b$, each with probability $1/2$. Given $Z=a$, let $X$ equal 0 or 4 with conditional probabilities $1/4$ and $3/4$. Given $Z=b$, let $X$ equal 2 or 5 with conditional probabilities $2/3$ and $1/3$. Find $E[X\mid Z]$ and verify its sign.

For $Z=a$, the conditional mean is $$ E[X\mid Z=a]=0\left(\frac14\right)+4\left(\frac34\right)=3. $$ For $Z=b$, it is $$ E[X\mid Z=b]=2\left(\frac23\right)+5\left(\frac13\right) =\frac43+\frac53=3. $$ Therefore a version of $E[X\mid Z]$ is 3 in either conditioning group. In particular, it is nonnegative, as expected since every possible value of $X$ in this model is nonnegative.

The calculation averages using the conditional probabilities within each group. The probabilities of $Z=a$ and $Z=b$ are not weights in these conditional means; they describe how often the groups occur overall.

Nonnegativity and Order Preservation

The order results can be proved directly by integrating under a conditional law. For random variables $U$ and $V$, use a regular conditional joint distribution $Q_z$ of $(U,V)$ given $Z=z$. Its marginal distributions give the conditional laws of $U$ and $V$. The integrability assumptions below ensure that these conditional means are finite for almost every $z$.

Theorem: Let $U$ and $V$ be integrable real-valued random variables, and let $Z$ take values in a standard Borel space.

If $U\geq 0$ almost surely, then $E[U\mid Z]\geq 0$ almost surely.
If $U\leq V$ almost surely, then $E[U\mid Z]\leq E[V\mid Z]$ almost surely.

Proof. For the first claim, take a regular conditional distribution $K(z,du)$ of $U$ given $Z=z$. The conditional distribution identity gives $$ \int K(z,(-\infty,0))\,P_Z(dz)=P(U<0)=0. $$ The integrand is nonnegative, so it must equal zero for $P_Z$-almost every $z$. For each such $z$, the conditional law assigns probability one to $[0,\infty)$. Its integral of $u$ is therefore nonnegative. This integral is finite for almost every $z$ because $E[|U|]<\infty$, so it gives $E[U\mid Z]\geq 0$ almost surely.

For the second claim, choose a regular conditional joint distribution $Q_z(du,dv)$ of $(U,V)$ given $Z=z$. Since $P(U\leq V)=1$, the conditional distribution identity yields $$ \int Q_z\bigl(\{(u,v):u\leq v\}\bigr)\,P_Z(dz)=1. $$ The integrand is at most 1, so it equals 1 for $P_Z$-almost every $z$. Thus, under $Q_z$, $u\leq v$ with probability one for almost every $z$. Both conditional means are finite for almost every $z$, by integrability and the identity for integrating absolute values under a conditional law. Integrating $u\leq v$ with respect to $Q_z$ then gives $$ \int u\,Q_z(du,dv)\leq \int v\,Q_z(du,dv). $$ The two integrals are versions of $E[U\mid Z=z]$ and $E[V\mid Z=z]$, respectively. The desired inequality follows almost surely. $\square$

The first claim is a special case of the same averaging principle: if a conditional distribution places all its probability on nonnegative values, its mean cannot be negative. The second claim also gives useful bounds immediately. If $c\leq X\leq d$ almost surely for finite constants $c,d$, then $c\leq E[X\mid Z]\leq d$ almost surely. More generally, if integrable variables $L,X,U$ satisfy $L\leq X\leq U$ almost surely, then $$ E[L\mid Z]\leq E[X\mid Z]\leq E[U\mid Z]\quad\text{almost surely}. $$ This follows by applying order preservation first to $L\leq X$ and then to $X\leq U$.

Worked Example: Bounds from a Continuous Conditional Law

Let $Z$ be uniform on $[0,1]$, and suppose the conditional density of $X$ given $Z=z$ is $$ f_{X\mid Z}(x\mid z)= \begin{cases} \frac12, & z\leq x\leq z+2,\\ 0, & \text{otherwise}. \end{cases} $$ Find the conditional mean and check it against the bounds implied by the support.

For every $z\in[0,1]$, the conditional density integrates to one: $$ \int_z^{z+2}\frac12\,dx=\frac12\bigl((z+2)-z\bigr)=1. $$ The conditional mean is $$ E[X\mid Z=z]=\int_z^{z+2}x\frac12\,dx =\frac14\bigl((z+2)^2-z^2\bigr) =\frac14(4z+4)=z+1. $$ The conditional distribution assigns probability one to $[z,z+2]$. Therefore $z\leq X\leq z+2$ under that conditional law, and its mean must lie within the same bounds. Indeed, $$ z\leq z+1\leq z+2. $$ Thus the calculated conditional mean satisfies the support bounds for every $z\in[0,1]$, in particular for almost every value of $Z$.

Worked Example: Comparing Two Conditional Means

Let $Z$ take values $A$ and $B$, each with positive probability. Given $Z=A$, suppose $(U,V)$ equals $(1,3)$ or $(4,6)$, each with probability $1/2$. Given $Z=B$, suppose it equals $(0,2)$ with probability $1/3$ and $(3,5)$ with probability $2/3$. Verify that $E[U\mid Z]\leq E[V\mid Z]$.

In group $A$, both possible pairs satisfy $U\leq V$, and $$ E[U\mid Z=A]=\frac12(1)+\frac12(4)=\frac52, \qquad E[V\mid Z=A]=\frac12(3)+\frac12(6)=\frac92. $$ Hence $5/2\leq 9/2$. In group $B$, the two pairs also satisfy $U\leq V$, and $$ E[U\mid Z=B]=\frac13(0)+\frac23(3)=2, \qquad E[V\mid Z=B]=\frac13(2)+\frac23(5)=4. $$ Hence $2\leq4$. The conditional means preserve the pointwise order in both groups, just as the theorem guarantees almost surely.

What the Bounds Do—and Do Not—Say

If $X$ is nonnegative, its conditional mean cannot be negative. If $X$ is bounded between two constants, its conditional mean cannot leave that interval. These conclusions are useful checks on a calculation: a negative conditional mean for a nonnegative variable, for example, signals an error in the conditional law, the integration, or the arithmetic.

A common mistake is to confuse a bound on $X$ with a bound by its unconditional mean. The latter does not generally hold. For example, let $Z$ be Bernoulli with $P(Z=1)=P(Z=0)=1/2$, and set $X=Z$. Then $E[X]=1/2$, while the conditional distribution of $X$ given $Z=z$ assigns probability one to $z$. Consequently, $E[X\mid Z]=Z$, which is 1 when $Z=1$ and 0 when $Z=0$. It is not bounded above by $E[X]$ almost surely. The valid bounds come from pointwise bounds on the random variables being averaged, not from assuming that a conditional mean must lie below the unconditional mean.

Another mistake is to read an almost-sure statement as a guarantee for every $z$. Conditional distributions can be specified differently at conditioning values that have probability zero, without changing any probability statement about the model. The theorem therefore guarantees the bounds for $P_Z$-almost every $z$, which is exactly what is needed for the conditional expectation as a random variable.

Key takeaway: Integration under a conditional distribution preserves nonnegativity and pointwise order. Thus pointwise bounds on integrable random variables give corresponding almost-sure bounds on their conditional means.

Check Your Understanding

Use conditional distributions and the almost-sure qualifications in your reasoning.

If $X\geq 2$ almost surely, what lower bound must $E[X\mid Z]$ satisfy?
Suppose $U\leq V$ almost surely and both variables are integrable. What order relation holds between their conditional expectations given $Z$?
A conditional distribution assigns probability one to the interval $[-1,3]$. What bounds does this give for its conditional mean?
Why does a conditional-mean bound generally hold for almost every conditioning value rather than necessarily every value?
Give a reason that $E[X\mid Z]\leq E[X]$ need not hold even when $X$ is nonnegative.

Positivity and Bounds for Conditional Expectations

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