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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 $M$ 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.

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 $\infty -\infty$.

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.

The law of total expectation makes averaging in stages reliable, provided the outer stage uses the actual probabilities of the groups. For an integrable random variable $X$ and a sub-sigma-algebra $\mathcal{G}$,Replacing each outcome by its group's conditional mean preserves the overall expected value. In the definition of conditional expectation, take the event $A=\Omega$ to obtain this identity.^[ref]

For a complete finite example, choose a group $Y=A$ with probability $1/4$ and $Y=B$ with probability $3/4$. Within group $A$, $X$ is $0$ or $4$ with equal conditional probabilities; within group $B$, it is $6$ or $10$ with equal conditional probabilities. Thus the joint probability distribution puts probabilities $1/8,1/8,3/8,3/8$ on $(A,0),(A,4),(B,6),(B,10)$. The conditional means are two and eight. Consequently,Direct averaging of the four values gives the same $0/8+4/8+18/8+30/8=13/2$. Taking the unweighted average of the two group means instead gives five, which describes a different choice of group probabilities. This is why the outer weighting matters.

The formula is useful when conditioning on a first choice makes the remaining experiment simple. It can replace a large outcome enumeration by smaller calculations within groups. In a mixture distribution, it combines component means using mixture weights; in a random sum of independent claims, it separates the number of summands from their individual expected values under the stated independence assumptions. The same averaging step appears in the law of total variance.

For integrable random variables $X_1,\dots ,X_n$ on the same probability space and fixed real numbers $a_1,\dots ,a_n$, linearity of expectation statesThe random variables can be dependent random variables. The formula follows by the linearity of integration; a finite linear combination of integrable random variables remains integrable. For example, if $X$ takes $-1,2,8$ with probabilities $1/2,3/10,1/5$, then $X$ and $2X$ are dependent random variables, yet $\mathbb{E}[X+2X]=3\mathbb{E}[X]=5.1$.

An especially useful application counts how many events occur. Put $I_i=1_{A_i}$, an indicator random variable, and $N={\sum }_i{I}_i$. Since $\mathbb{E}[I_i]=\mathbb{P}(A_i)$, we get $\mathbb{E}[N]={\sum }_i\mathbb{P}(A_i)$. There is no need to compute the whole probability distribution of $N$, or to assume independence. For a uniform random permutation of $n\ge 1$ objects, each object stays fixed with probability $1/n$, so the fixed point count of a uniform random permutation has expected value one for every $n$. ^[ref].

Linearity of expectation concerns addition and multiplication by fixed scalars. Products instead satisfy $\mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[Y]$ under the hypotheses of expectation of a product of independent random variables. For the example above, $\mathbb{E}[X^2]=14.5$, whereas $(\mathbb{E}[X]{)}^2=2.89$. Applying a nonlinear function before averaging can therefore change the answer substantially; use the law of the unconscious statistician to compute that average.

Suppose $\varphi (\lambda )=\mathbb{E}{e}^{\lambda X}<\infty$ for all $\lambda \ge 0$. Under the exponential tilting $d{\mu }^{\lambda }=e^{\lambda x}d\mu /\varphi (\lambda )$, $X$ is an integrable random variable whenever $\lambda >0$: its negative exponential tail is bounded, and its positive contribution is dominated by a larger positive exponential moment. Differentiation under the integral sign gives the displayed mean. At the endpoint $\lambda =0$, the positive part of $X$ is integrable, but the negative part need not be. The right derivative ${\varphi }_{+}^{\prime }(0)=\mathbb{E}X$ is valid as an extended value, possibly $-\infty$; a finite mean at zero requires an extra negative-tail integrability assumption.

For integrable random variable $X$ and nested sub-sigma-algebras $\mathcal{H}\subseteq \mathcal{G}$, the tower property saysThe finer-information conditional mean can be averaged within each coarser group, recovering the coarser-information conditional mean. Conversely, conditioning an already $\mathcal{H}$-measurable conditional mean on $\mathcal{G}$ leaves it unchanged. Both orders therefore recover $\mathbb{E}[X\mid \mathcal{H}]$ when the information is nested.^[ref]

For the finite example in law of total expectation, exact knowledge of $X$ gives the finer mean $X$ itself. Averaging within the groups $A,B$ recovers two and eight. Forgetting the group then gives $13/2$. The nesting condition is essential; iterated conditional expectation over nonnested sigma-algebras describes why arbitrary successive conditioning cannot simply be replaced by conditioning on an intersection.

In discrete time, the bounded-time optional stopping theorem says that an integrable supermartingale $X$ and bounded stopping times $S\le T$ satisfy $\mathbb{E}[X_T\mid {\mathcal{F}}_S]\le X_S$. For a martingale there is equality. The assertion at a possibly unbounded, almost-surely finite stopping time $T$ also holds with $S=0$ if $\{X_{T\wedge n}:n\ge 0\}$ is uniformly integrable: then $X_T$ is integrable and $\mathbb{E}{X}_T\le \mathbb{E}{X}_0$, with equality for a martingale. Boundedness of $T$ or the stated integrability condition cannot in general be omitted.

For the increment hypothesis here, $\mathbb{E}T<\infty$ implies $T<\infty$ almost surely, and telescoping givesThe right-hand side is an integrable random variable. Applying bounded-time optional stopping to $T\wedge n$ and then the dominated convergence theorem provesThis is bounded-increment optional stopping with integrable time; the same reasoning gives equality for a martingale.

Put $q=1-p$ and first assume $0<p<1$. Let $\tau$ be the first occurrence time of the specified seven-letter word. Its probability in a prescribed block is $r=p^4{q}^3>0$. Disjoint seven-flip blocks are independent, and a match in one forces $\tau$ to have occurred by its end. Consequently $\mathbb{P}(\tau >7m)\le (1-r{)}^m$ and $\mathbb{E}\tau \le 7/r<\infty$; this establishes the integrability needed for stopping before computing its value.

Start a new gambler with unit capital immediately before each flip. A gambler bets all current capital on the next required letter of the word, receives capital divided by $p$ for a correct head or by $q$ for a correct tail, and receives zero on failure. After completing all seven bets, the gambler retains the winnings without further betting. Each individual bet preserves its conditional expected capital. If $C_n$ is the total capital of all gamblers who have entered through flip $n$, then $C_n-n$ is a martingale.

At any time at most seven gamblers are still betting, and each one's capital is at most $L=1/(p^4{q}^3)$, since probabilities of shorter prefixes are at least the full-word probability. Completed winnings do not change. Thus the increments of $C_n-n$ have a deterministic bound, for example $14L+1$. The preceding optional-stopping argument therefore gives $\mathbb{E}(C_{\tau }-\tau )=0$.

At the first full match, no earlier gambler has completed the word. A surviving gambler who has placed $k$ successful bets corresponds to a suffix of the completed word which is also its length-$k$ prefix, namely a border of a word. The only such lengths are $3$ and $7$: the shorter common string is HHT. Their capitals are $1/(p^{2}q)$ and $1/(p^4{q}^3)$, respectively. Hence $C_{\tau }$ is this deterministic sum andFor a fair coin this is 136 flips. This is a waiting time for a word in independent nonuniform symbols; the overlap contribution is essential. If $p=0$ or $p=1$, the word cannot occur because it contains both types of flip, so the expected waiting time is infinite.

A convenient precise version is the almost sure submartingale convergence theorem: if $(S_n)$ is a real submartingale and ${\sup }_n\mathbb{E}{S}_n^{+}<\infty$, then $S_n$ converges almost surely to a finite integrable random variable. In particular the martingale convergence theorem applies to any martingale with ${\sup }_n\mathbb{E}|S_n|<\infty$. It asserts neither convergence in L1 nor preservation of the initial expected value without additional uniform integrability.

Here is a proof including the crossing estimate. Fix real $a<b$, and let $U_n[a,b]$ count completed upcrossings by time $n$. Use a predictable indicator $H_k$ which holds one unit during each crossing attempt: enter when $S_k\le a$, exit when $S_k\ge b$. Witheach completed trade gains at least $b-a$, while the possible unfinished trade loses no more than $(S_n-a{)}^{-}$. Hence $G_n\ge (b-a)U_n[a,b]-(S_n-a{)}^{-}$. The submartingale property and $0\le H_k\le 1$ give $\mathbb{E}{G}_n\le \mathbb{E}(S_n-S_0)$: the complementary predictable gains have nonnegative expected value. ThereforeThe right side is uniformly bounded, since $(S_n-a{)}^{+}\le S_n^{+}+|a|$. By the monotone convergence theorem, every rational interval has finitely many upcrossings almost surely. A real sequence whose lower and upper limits differ crosses some rational interval infinitely often, so the countable intersection of these events gives an extended-real limit. Furthermoreso ${\sup }_n\mathbb{E}|S_n|<\infty$. The Fatou lemma makes the limit finite and integrable, completing the proof. Applying this theorem to $-Z_n$ gives the almost sure supermartingale convergence theorem for nonnegative supermartingales.

For the given adapted multiplicative drift, putThe next denominator $P_{n+1}$ is ${\mathcal{F}}_n$-measurable, and $P_n\ge 1$ makes $Z_n$ integrable. ThusThe nonnegative supermartingale $Z_n$ has a finite almost sure convergence limit $Z_{\infty }$. On the probability-one event where the specified series is finite,Consequently $P_n\uparrow P_{\infty }\in [1,\infty )$ and the multiplicative correction for summable adapted drift givesThe argument does not require a finite expected value for the infinite product or for the total drift.

The positive state-price density alternative characterizes one-period arbitrage. Interpret $A_0$ as the vector of current asset prices and $A_1$ as their terminal payments. A fixed vector of holdings produces initial cost $x^{\top }{A}_0$ and terminal payment $x^{\top }{A}_1$. The first alternative is a nonpositive-cost investment with a nonnegative payment and strict improvement either initially or on an event of positive probability. A strictly positive state-price density prices every such payment and excludes both forms of arbitrage, as proved below.

If there is a traded bank account costing one and paying a deterministic $R>0$, the density satisfies $1=R\mathbb{E}\nu$. Hence $dQ/dP=R\nu$ defines an equivalent martingale measure andConversely this pricing relation under an equivalent martingale measure gives $\nu =R^{-1}dQ/dP$. Without such an account, $\nu$ is a pricing density, not necessarily a probability density or even an integrable random variable by itself; the stated requirement is integrability of $\nu A_1$. This distinction keeps the abstract alternative valid without an unstated asset paying one.