Conditional Monotone Convergence

Why Increasing Limits Matter

Conditional expectation summarizes the average value of a random variable given specified information. A useful question is whether this summary respects limits: if nonnegative quantities increase toward a limit, do their conditional expectations also increase toward the conditional expectation of that limit?

The answer is yes. The result is a conditional version of the Monotone Convergence Theorem. Its proof uses the defining integral identity for conditional expectation: test on each event in the conditioning sigma-algebra, and then apply ordinary monotone convergence to the resulting integrals.

Nonnegative Conditional Expectations

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and let $\mathcal G\subseteq\mathcal F$ be a sub-sigma-algebra. For an integrable random variable $X$, $\mathbb E[X\mid\mathcal G]$ is a $\mathcal G$-measurable random variable whose integral over every event in $\mathcal G$ agrees with that of $X$. For a nonnegative $X$ that may not be integrable, the same integral identity is used with values in $[0,\infty]$.

Definition: If $X$ is a nonnegative, possibly extended-valued random variable, its conditional expectation given $\mathcal G$ is a $\mathcal G$-measurable random variable $Y:\Omega\to[0,\infty]$ such that, for every $A\in\mathcal G$, $$ \int_A Y\,d\mathbb P=\int_A X\,d\mathbb P. $$ It is written $\mathbb E[X\mid\mathcal G]$, and is unique up to almost-sure equality.

For completeness, such a conditional expectation can be obtained from the integrable truncations $X\wedge k=\min(X,k)$, $k\in\mathbb N$. Choose versions of $\mathbb E[X\wedge k\mid\mathcal G]$ that increase almost surely, using order preservation, and take their pointwise limit outside a common null set. For every $A\in\mathcal G$, the Monotone Convergence Theorem gives

\int_A \lim_{k\to\infty}\mathbb E[X\wedge k\mid\mathcal G]\,d\mathbb P = \lim_{k\to\infty}\int_A (X\wedge k)\,d\mathbb P = \int_A X\,d\mathbb P.

Thus the limit has the required defining property. If $X$ is integrable, this extended definition agrees with the usual integrable conditional expectation. If $X$ is not integrable, the conditional expectation may take the value $+\infty$ on a set of positive probability.

The Conditional Monotone Convergence Theorem

Theorem: Let $X_n$, $n\in\mathbb N$, be nonnegative random variables such that $X_n\uparrow X$ almost surely. For any sub-sigma-algebra $\mathcal G$, $$ \mathbb E[X_n\mid\mathcal G]\uparrow \mathbb E[X\mid\mathcal G] \quad\text{almost surely}. $$ The variables and conditional expectations may be extended-valued. In particular, if $X$ is integrable, the limit on the right is finite almost surely.

Proof. By order preservation, $\mathbb E[X_n\mid\mathcal G]\leq\mathbb E[X_{n+1}\mid\mathcal G]$ almost surely for each $n$. Since there are countably many such inequalities, there is one event of probability one on which they all hold. Define $Y=\lim_{n\to\infty}\mathbb E[X_n\mid\mathcal G]$ there, assigning any value on the exceptional null set. Then $Y$ is $\mathcal G$-measurable and nonnegative. For any $A\in\mathcal G$, apply the Monotone Convergence Theorem first to the conditional expectations and then to $X_n$: $$ \int_A Y\,d\mathbb P = \lim_{n\to\infty}\int_A \mathbb E[X_n\mid\mathcal G]\,d\mathbb P = \lim_{n\to\infty}\int_A X_n\,d\mathbb P = \int_A X\,d\mathbb P. $$ Thus $Y$ is a conditional expectation of $X$ given $\mathcal G$, so uniqueness gives $Y=\mathbb E[X\mid\mathcal G]$ almost surely. If $X$ is integrable, taking $A=\Omega$ shows that $Y$ has finite expectation, hence is finite almost surely. $\square$

The simultaneous choice of versions matters. Each order inequality may initially fail on its own null set; taking the union of countably many such sets gives a single null set outside which the whole sequence is increasing. The theorem asserts almost-sure convergence, not pointwise convergence for every choice of versions at every $\omega$.

Worked Examples

Worked Example: A Finite Conditioning Partition

Let $\Omega=\{a,b,c,d\}$ assign probability $1/4$ to each point, and let $\mathcal G$ be generated by the two atoms $A=\{a,b\}$ and $C=\{c,d\}$. For $n\in\mathbb N$, define $$ X_n(a)=1-2^{-n},\qquad X_n(b)=2-\frac1n,\qquad X_n(c)=0,\qquad X_n(d)=1-2^{-n}. $$ Each coordinate is nonnegative and increases with $n$. The limit $X$ has values $1,2,0,1$ at $a,b,c,d$, respectively.

Conditional expectation on either atom is the average over that atom. Therefore, on $A$, $$ \mathbb E[X_n\mid\mathcal G] =\frac{(1-2^{-n})+(2-1/n)}2 =\frac{3-2^{-n}-1/n}{2} \longrightarrow \frac32. $$ On $C$, $$ \mathbb E[X_n\mid\mathcal G] =\frac{0+(1-2^{-n})}{2} \longrightarrow \frac12. $$ These limits are exactly the conditional averages of $X$: $(1+2)/2=3/2$ on $A$, and $(0+1)/2=1/2$ on $C$.

Worked Example: Increasing Events and Conditional Probabilities

Take $\Omega=(0,1)$ with Lebesgue probability, let $B=(0,1/2)$, and set $\mathcal G=\sigma(B)$. For $n\in\mathbb N$, let $$ A_n=\left(0,\frac12\frac{n}{n+1}\right). $$ The events increase to $B$. For the indicator $X_n=\mathbf 1_{A_n}$, conditional expectation is constant on each atom of $\mathcal G$. On $B$, it equals the fraction of $B$'s length occupied by $A_n$: $$ \mathbb E[\mathbf 1_{A_n}\mid\mathcal G] =\frac{\lambda(A_n)}{\lambda(B)} =\frac{\frac12\frac{n}{n+1}}{\frac12} =\frac{n}{n+1}. $$ On $B^c$, the conditional expectation is zero because $A_n\cap B^c=\varnothing$. Hence $$ \mathbb E[\mathbf 1_{A_n}\mid\mathcal G] =\frac{n}{n+1}\mathbf 1_B \uparrow \mathbf 1_B =\mathbb E[\mathbf 1_B\mid\mathcal G]. $$ The limit is the conditional probability of $B$ given $\mathcal G$, which is $1$ on $B$ and $0$ on $B^c$.

Worked Example: A Finite Limit with Infinite Conditional Expectation

Let $U(\omega)=\omega$ on $(0,1)$ with Lebesgue probability. Use $B=(0,1/2)$ and $\mathcal G=\sigma(B)$, and define $X_n=\min(n,1/U)$. Then $X_n$ increases to $X=1/U$, which is finite at every $\omega\in(0,1)$ but has infinite expectation.

For $n\geq 2$, on $B$ the conditional expectation is the average over an interval of length $1/2$: $$ \mathbb E[X_n\mid\mathcal G]\big|_B =2\left(\int_0^{1/n}n\,du+\int_{1/n}^{1/2}\frac{du}{u}\right) =2\left(1+\log\frac n2\right). $$ On $B^c=(1/2,1)$, $1/U\leq 2\leq n$, so $$ \mathbb E[X_n\mid\mathcal G]\big|_{B^c} =2\int_{1/2}^1\frac{du}{u} =2\log 2. $$ The first value tends to $+\infty$, while the second stays equal to $2\log 2$. Indeed, the conditional expectation of $1/U$ is infinite on $B$, since $\int_0^{1/2}du/u=+\infty$, and equals $2\log 2$ on $B^c$. Thus the theorem allows a finite almost-sure limit whose conditional expectation is infinite on part of the space.

How the Proof Works

The proof has two essential ingredients. First, order preservation makes the conditional expectations an increasing sequence, after versions are chosen consistently outside one null set. Second, the conditional expectation identity transfers each event integral back to the original random variable. Monotone convergence then identifies the limiting integrals exactly.

Choose an increasing version.
Use order preservation for each adjacent pair and discard the union of the resulting countably many null sets.

Take the limit.
The pointwise limit outside that null set is nonnegative and $\mathcal G$-measurable.

Test on conditioning events.
For each $A\in\mathcal G$, apply monotone convergence to the integrals over $A$.

Identify the limit.
The integral identity characterizes the conditional expectation, so uniqueness gives the asserted almost-sure equality.

Why the Result Matters

Conditional monotone convergence justifies approximating a nonnegative quantity by simpler quantities that increase to it. The approximations may be indicators, truncations, or finite-valued functions. Their conditional expectations converge to the desired conditional expectation without requiring an integrable upper bound.

A common mistake is to assume that the limit must be integrable because every term in the sequence is integrable. The third example rules this out: every $X_n$ is bounded and integrable, but the limit $1/U$ is not integrable. The correct conclusion is convergence to an extended conditional expectation; finiteness follows only when the limit itself is integrable.

Another important distinction is between almost-sure equality and equality at every point. Conditional expectations are defined only up to almost-sure equality. The theorem therefore concerns versions that can be chosen to increase outside one null set, and it identifies their limit almost surely.

Check Your Understanding

Use the defining integral identity to explain each conclusion.

Why is it important to combine the exceptional null sets for the inequalities between consecutive conditional expectations?
In the finite partition example, what are the conditional expectations on each atom, and why do their limits match the conditional averages of $X$?
For increasing events $A_n\uparrow B$, what does the theorem say about $\mathbb E[\mathbf 1_{A_n}\mid\mathcal G]$?
Can the limit $X$ be finite almost surely while $\mathbb E[X\mid\mathcal G]$ is infinite on an event of positive probability?
Which step in the proof identifies the pointwise limit of the conditional expectations as a conditional expectation of $X$?