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A submartingale with ${\sup }_n\mathbb{E}{Y}_n^{+}<\infty$ converges almost surely to a finite integrable random variable. The positive-part bound also bounds negative parts because $\mathbb{E}{Y}_n\ge \mathbb{E}{Y}_0$. The result specializes to an $L^1$-bounded martingale, and hence to every nonnegative martingale. It does not by itself give convergence in L1; uniform integrability supplies that stronger conclusion.

For an integrable supermartingale with the displayed increment bound, $|X_{T\wedge n}|\le |X_0|+KT$. Thus the stopped values are dominated by one integrable random variable, including $X_T$. Apply optional stopping at the bounded times $T\wedge n$ and then the dominated convergence theorem. For a martingale the same proof gives equality.

A bounded stopping time is a stopping time bounded almost surely by a deterministic finite constant. In discrete time, a bound $T\le N$ makes $M_T$ integrable whenever $M_0,\dots ,M_N$ are integrable random variables, and the optional stopping theorem applies to a martingale without additional limiting hypotheses. Almost-sure finiteness alone is weaker than boundedness.

The conditional expectation is the average of a random variable that can be predicted using specified information. It is itself a random variable: the average appropriate to one observed group may differ from that appropriate to another group. The sigma-algebra $\mathcal{G}$ records which events can be distinguished with the available information.

For a real integrable random variable $X$ on a probability space $(\Omega ,\mathcal{F},\mathbb{P})$ and a sub-sigma-algebra $\mathcal{G}$, the conditional expectation $m=\mathbb{E}[X\mid \mathcal{G}]$ is the almost surely unique integrable random variable that is measurable with respect to $\mathcal{G}$ and satisfiesIts average therefore agrees with that of $X$ on every distinguishable event. The integral characterization handles discrete observations, continuous observations, and information that is not given by a single numerical value.^[ref]

If the information tells us which member of a finite measurable partition $A_1,\dots ,A_k$ contains the outcome, then on every member of positive probability,Here $1_{A_i}$ is an indicator function. Zero-probability members do not determine a unique value, which is why uniqueness is only almost surely. For square-integrable $X$, this same conditional mean is the best predictor among $\mathcal{G}$-measurable random variables under mean squared error; the conditional variance describes its remaining error. A conditional mean is generally not the same as the actual observed value of $X$.^[ref]

For independent and identically distributed random variables $X_j$ that are integrable random variables, with $S_n={\sum }_{j=1}^n{X}_j$, symmetry gives $\mathbb{E}[X_1\mid S_n,S_{n+1},\dots ]=S_n/n$. The decreasing sigma-algebras generated by these future sums permit application of the reverse martingale convergence theorem.

For a real integrable random variable $X$, a conditional expectation $Y=\mathbb{E}[X\mid \mathcal{G}]$ with the same probability distribution as $X$ must equal $X$ almost surely. No second moment is needed. For any real $c$, the nonnegative random variable $D_c=|X-c|-\mathrm{sign}(Y-c)(X-c)$ has expected value zero, by conditioning and equality of the two absolute moments. Thus $Y>c$ implies $X\ge c$, $Y<c$ implies $X\le c$, and $Y=c$ implies $X=c$, outside a null set. Using a countable family of rational $c$ separates any unequal real values and proves the claim. Equality of the probabilities of $X=c$ and $Y=c$ also gives equality of their zero sets and hence equality of the two sign functions for every fixed $c$.

Let $X$ be a real integrable random variable, let $\mathcal{G}$ be a sub-sigma-algebra, and let $\varphi :\mathbb{R}\to \mathbb{R}$ be a finite convex function with $\varphi (X)$ integrable. Thenalmost surely. These explicit integrability assumptions make both conditional expectations well defined.

A finite convex function on the real line is the supremum of a countable family of supporting affine functions, for example choosing supports at rational points and using continuity. Write them as $a_{j}x+b_j$. For each $j$, monotonicity and linearity of conditional expectation give $a_j\mathbb{E}[X\mid \mathcal{G}]+b_j\le \mathbb{E}[\varphi (X)\mid \mathcal{G}]$. Countability permits one exceptional null set for every $j$. Taking the supremum proves the inequality.

With $\varphi (x)=x^2$ and square-integrable $X$, the inequality says that retaining only the information in $\mathcal{G}$ cannot increase its second moment. More generally it is the convex-loss step behind the Rao-Blackwell theorem.

If $X$ is an integrable random variable and $Z$ is a finite, strictly positive $\mathcal{G}$-measurable random variable, thenIndeed, $0\le Z{1}_{\{|X|\ge Z\}}\le |X|$. Monotonicity of conditional expectation and the measurable-factor rule for conditional expectation giveDivision by the strictly positive $Z$ proves the claim. Only the displayed product needs to be an integrable random variable; neither $Z$ nor $|X|/Z$ needs to have finite expected value. The measurable threshold represents information already available before asking how likely the remaining exceedance is.

For nonnegative random variables $Y_n\uparrow Y$, the corresponding nonnegative conditional expectations increase almost surely to $\mathbb{E}[Y\mid \mathcal{G}]$, allowing the extended value $+\infty$. When $Y$ is an integrable random variable, all conditional expectations can be taken in their ordinary finite sense.

To prove this integrable case, choose versions simultaneously ordered outside a single null set, and put $V={\lim }_n\mathbb{E}[Y_n\mid \mathcal{G}]$. This limit is $\mathcal{G}$-measurable. For every $A\in \mathcal{G}$, the ordinary monotone convergence theorem twice givesIn particular $\mathbb{E}V=\mathbb{E}Y<\infty$, and this is exactly the defining property of conditional expectation. For a general nonnegative $Y$, define its extended conditional expectation by increasing bounded truncations; the same nonnegative integral argument proves the statement.

A conservative variance estimator, in the finite-sample expectation sense used here, is an estimator $\hat{V}$ of a sampling variance $V_{\theta }$ satisfyingfor every statistical parameter value $\theta$ in the stated statistical model. Assume $V_{\theta }$ and ${\mathbb{E}}_{\theta }|\hat{V}|$ are finite. In design-based inference, the same definition applies for every permitted fixed population, with the sampling design supplying the randomness. The bias of an estimator is then nonnegative for this target. Aronow and Samii use this expectation criterion for their correction to the Horvitz-Thompson variance estimator when some pairs cannot be sampled together.^[ref]

A samplewise variance upper bound would require the stronger inequality $\hat{V}\ge V_{\theta }$ almost surely. An expectation bound allows some estimates below $V_{\theta }$. It does not even require that every realized value be nonnegative; in practice a variance estimate is commonly chosen to be a nonnegative random variable.

One construction starts with an integrable random variable $W$ that is an unbiased estimator of $V_{\theta }$ and might take negative values. Its positive part $\hat{V}=W_{+}=\max (W,0)$ satisfies $W_{+}\ge W$, hence is a conservative variance estimator. The amount of upward bias of an estimator is exactlyThe identity follows from the positive part and negative part decomposition $W=W_{+}-(-W{)}_{+}$. The Horvitz-Thompson variance estimator supplies a concrete setting in which the untruncated unbiased estimator can be negative.

A deliberately constructed example proves that conservative variance estimation does not guarantee Wald interval coverage. Observe $Y$ from a normal distribution $N(\theta ,1)$ and an independent $B$ from a Bernoulli distribution with success probability $0<\epsilon <1$. Use $\hat{\theta }=Y$ and $\hat{V}=B/\epsilon$. This nonnegative random variable hasThus it is even an unbiased estimator of the sampling variance. Yet the usual Wald confidence intervalhas coverage probabilitywhere $\Phi$ is the standard normal distribution function. When $B=0$, the interval is the singleton $\{Y\}$, which contains $\theta$ with probability zero because $Y$ is a continuous random variable. When $B=1$, its conditional coverage probability is the bracketed expression. At $\epsilon =0.01$, overall coverage probability is at most 1 percent, despite the nominal 95 percent rule and the exactly normal centre.

This example concerns the logical strength of the expectation criterion. Controlling the distribution of an estimated standard error, for example through an appropriate consistent estimator in an asymptotic argument, is additional work needed to justify confidence interval coverage.

Let $M$ be a Poisson random measure of intensity $\mu$, let $\mu (A)<\infty$, and let $f\ge 0$ be measurable. Apply the Laplace functional of a Poisson random measure to $f+t{1}_A$ and differentiate from the right at $t=0$. On the random side, differentiation is dominated by the integrable random variable $M(A)$; on the intensity side, by $1_A$. This proves the displayed identity when $\int (1-e^{-f})d\mu$ is finite. When this integral is infinite, $e^{-M(f)}=0$ almost surely, and the same identity holds with both sides zero.

For a nonnegative integrable random variable $Z$ on a probability space with probability measure $P$, its entropy functional isusing $0\log 0=0$ and allowing the value $+\infty$. If $m>0$, define the probability measure $Q$ by $dQ/dP=Z/m$. Substitution into the definition of relative entropy givesThus the normalization factor $m$ is essential. Jensen inequality shows that the entropy functional is nonnegative; it vanishes precisely when $Z$ is constant almost surely. If $m=0$, nonnegativity forces $Z=0$ almost surely and the entropy functional is zero.

For an event $A$ of probability $p>0$, take $Z=1_A$. Then $Q=P(\cdot \mid A)$ and ${\mathrm{Ent}}_P(Z)=p\log (1/p)$. This example connects the entropy functional to the information cost of conditioning on a rare event.

Let $a<i<b$ and $a<j<b$ be integers. A simple symmetric random walk started at $i$ exits the integer interval at $\tau =\inf \{n\ge 0:S_n\in \{a,b\}\}$. Its expected occupation of $j$ before exit, including its initial state, isA block of $b-a$ consecutive positive steps forces exit, so the geometric tail bound from a uniform escape probability proves $\tau <\infty$ almost surely. Optional stopping at the bounded stopping time $n\wedge \tau$, followed by the bounded convergence theorem for $S_{n\wedge \tau }\in [a,b]$, gives the gambler's ruin probabilitiesTranslate the absolute value martingale of a simple symmetric random walk from zero to $j$. Bounded-time optional stopping yieldsUse monotone convergence on the increasing visit count and the bounded convergence theorem on the right. Substituting the exit probabilities giveswhich simplifies to the displayed expression by considering $i\le j$ and $i\ge j$. In particular, $G_{a,b}(0,0)=-2ab/(b-a)$ when $a<0<b$.

The proof establishes finite expected value of the visit count. Therefore each translated stopped absolute-value martingale is dominated in absolute value by a constant plus that total count, an integrable random variable; its family of values is uniformly integrable. This supplies the limit justification without assuming uniform integrability in advance.

An expected value is the probability-weighted arithmetic mean of a real-valued random variable. On a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, its general definition is the Lebesgue integralwhere ${\mu }_X$ is the probability distribution of $X$. For an integrable random variable, this is a finite real number. With a probability mass function $p_X$, use ${\sum }_{x}x{p}_X(x)$; with a probability density function $f_X$, use ${\int }_{\mathbb{R}}x{f}_X(x)dx$. These are two ways of computing the same weighted average, rather than different definitions for different experiments.

For example, suppose $X$ takes the values $-1,2,8$ with probabilities $1/2,3/10,1/5$. ThenThe most likely outcome is $-1$, and $1.7$ is not a possible outcome at all. The expected value summarizes all outcomes and their weights; it need not predict the result of one trial. Geometrically, placing masses equal to the probability masses at the outcome positions gives a center of mass at $1.7$, since ${\sum }_x{p}_X(x)(x-1.7)=0$. This balance-point interpretation also explains the connection to a weighted arithmetic mean. ^[ref].

For the dimensionless random variable $Y$ with probability density function $f_Y(y)=2y$ on $[0,1]$, the expected value is $\mathbb{E}[Y]={\int }_0^{1}y(2y)dy=2/3$. Integrating the probability density function itself gives one; to compute the expected value, multiply by the outcome first. ^[ref].

Why is this useful? Linearity of expectation lets us compute an average total by adding average contributions, even for dependent random variables. And for independent and identically distributed integrable random variables, the strong law of large numbers says their sample mean approaches their common expected value almost surely. The theorem supplies the link to repeated observations; the weighted-average definition itself does not promise that any particular finite run will be close to the expected value.

There is a useful historical connection to valuing uncertain rewards. In his 1657 De ratiociniis in ludo aleae, Christiaan Huygens assigns the value $(a+b)/2$ to equally likely rewards $a,b$, and $(pa+qb)/(p+q)$ when they have $p,q$ equally likely chances. These are the finite weighted-average formulas for expected value. His opening discussion asks how to divide the stakes when a game stops before completion, connecting probability to a concrete question about an unfinished contest. ^[ref].

A real or complex random variable $X$ is integrable when $\mathbb{E}|X|<\infty$. This is integrability with respect to its probability measure, and ensures that its expected value is finite. A bounded random variable is always an integrable random variable; so is the example taking $-1,2,8$, for which $\mathbb{E}|X|=2.7$.

For a real random variable, put $X_{+}=\max (X,0)$ and $X_{-}=\max (-X,0)$, its positive part and negative part. Each has a nonnegative Lebesgue integral, possibly infinite. The expression $\mathbb{E}[X]=\mathbb{E}[X_{+}]-\mathbb{E}[X_{-}]$ is defined when at least one of these two expected values is finite; if both are infinite, the expected value is undefined. Integrability makes both finite. A nonnegative random variable can have expected value $+\infty$, even though it is not an integrable random variable.

The standard Cauchy distribution is a useful warning. With its probability density function $f(x)=1/[\pi (1+x^2)]$, the positive contribution is ${\int }_0^{R}xf(x)dx=\log (1+R^2)/(2\pi )$, which tends to infinity; the negative contribution has the same magnitude. Integrating symmetrically over $[-R,R]$ gives zero for each $R$, but this cancellation does not define an expected value. The Cauchy sample mean has an unchanged distribution, showing why the integrability hypothesis in the strong law of large numbers matters.

For an integrable random variable, conditional expectation preserves integrability, and linearity of expectation can safely combine finite averages without forming $\infty -\infty$.

Let $T$ be a measure-preserving transformation of a probability space and $X$ an integrable random variable. If the displayed duality holds for every bounded random variable $Z$, test it on indicator functions to identify $\mathbb{E}[X\mid T^{-1}(\mathcal{F})]=X\circ T$. The measure-preserving transformation gives the same probability distribution to $X$ and $X\circ T$. The principle of conditional expectation preserving a distribution then forces equality almost surely. This does not require an inverse for $T$ or require $X$ to be square-integrable.

Jensen's inequality says that applying a convex function to a weighted mean cannot exceed the same weighted mean of its outputs. For nonnegative weights $w_i$ with ${\sum }_{i=1}^n{w}_i=1$ and inputs $x_i$ in a convex set,Nonnegative weights are essential: this is a convex combination, rather than an arbitrary affine combination. For two inputs it is precisely the defining inequality for a convex function; repeatedly grouping inputs gives the finite weighted version.

Take $f(x)=x^2$, weights $1/4$ and $3/4$, and inputs $1$ and $3$. Their weighted mean is $2.5$, so the two sides areThe Jensen gap is $0.75$. The inequality measures the effect of replacing distinct inputs by one representative input before applying a nonlinear function.

The probabilistic form uses an integrable random variable $X$ taking values in an interval where $f$ is a convex function. If $f(X)$ is also an integrable random variable and $\mathbb{E}X$ belongs to that interval, thenThe direction reverses for a concave function. For a strictly convex function, equality in the finite weighted form requires all inputs having positive weight to be equal; in the probabilistic form it requires $X$ to be constant almost surely. An affine function gives equality for any admissible distribution. ^[ref].

For the quadratic function $f(x)=x^2$ and a random variable with a finite second moment, the difference between the two sides is exactly its variance:Consider two equally likely values $2-d$ and $2+d$. Their expected value stays at $2$ as $d$ increases, but their variance is $d^2$. Thus the expected value of the square grows from $4$ to $4+d^2$. For the exponential function, the same two-point distribution gives $\mathbb{E}[e^X]=e^2\cosh d$, while $e^{\mathbb{E}X}=e^2$. Both are examples of Jensen's inequality, but their Jensen gaps depend differently on the spread of the inputs.

Another useful choice is the concave function $\log x$ on positive inputs. Its version of Jensen's inequality gives ${\sum }_i{w}_i\log x_i\le \log ({\sum }_i{w}_i{x}_i)$. Applying the increasing exponential function shows that the weighted geometric mean cannot exceed the weighted arithmetic mean. For the equally weighted inputs $1$ and $9$, these are $3$ and $5$. This recovers the arithmetic-geometric mean inequality and illustrates how choosing one function can generate an entire family of bounds.

A short proof idea explains the probabilistic form. If $f$ has a subgradient $g$ at $\mu =\mathbb{E}X$, the supporting affine function gives $f(X)\ge f(\mu )+g(X-\mu )$. Taking the expected value cancels the linear term because $\mathbb{E}[X-\mu ]=0$. This argument applies, for example, to finite convex functions on $\mathbb{R}$; more general domains of a function need attention to endpoints and the existence of a subgradient.

Johan Jensen's 1906 paper “Sur les fonctions convexes et les inégalités entre les valeurs moyennes” is an early source for the connection between convex functions and inequalities for means. ^[ref].

For an integrable random variable $Y$, the conditional expectation satisfies $|\mathbb{E}[Y\mid \mathcal{G}]|\le \mathbb{E}[|Y|\mid \mathcal{G}]$ almost surely. Taking expected values gives the L1 contraction of conditional expectation. Applying it to $Y_n-Y$ shows that convergence in L1 is preserved by conditional expectation with respect to any fixed sigma-algebra.

For integrable random variables $Z,Y$,Apply this with $Z=X_n$ and $Y=X$ to prove that convergence in L1 implies uniform integrability; control finitely many early terms separately.

The law of the unconscious statistician computes the expected value of a measurable function of a random variable directly from the original probability distribution. If $g(X)$ is an integrable random variable, thenUse ${\sum }_{x}g(x)p_X(x)$ for a probability mass function, or ${\int }_{\mathbb{R}}g(x)f_X(x)dx$ for a probability density function. There is no need to derive the probability distribution of $g(X)$ first. The identity is the integration rule for a pushforward measure, and also holds with an infinite value when $g$ is nonnegative. ^[ref].

For $X$ taking $-1,2,8$ with probabilities $1/2,3/10,1/5$, choosing $g(x)=x^2$ gives $\mathbb{E}[X^2]=\frac{1}{2}+4\frac{3}{10}+64\frac{1}{5}=14.5$. Its variance is therefore $14.5-1.7^2=11.61$. Squaring the expected value would instead give $2.89$, because squaring does not commute with averaging.