Discrete Time Martingales — Examples

To justify the finite $L^1$ operations independently of the affected supplier proof, augment every finite disjoint simple display by the complement with coefficient $0$. Pairwise intersections of two augmented displays partition the whole space and carry equal coefficients; finite additivity and $0\cdot (+\mathrm{\infty })=0$ prove representation independence. Common refinements give simple addition and monotonicity; scalar zero is direct and positive scalars are termwise. Supremum over simple minorants, increasing simple approximation and the sets $\{f_j\ge cs\}$ for $0<c<1$ give monotone convergence and nonnegative additivity. Positive/negative and real/imaginary decompositions give finite $L^1$ linearity. This also repairs the integral base beneath the event identities defining the conditional classes used below. The generated sigma-algebras are nested because their generator families are nested. The finite sum $S_n$ is ${\mathcal{F}}_n$-measurable and $E\mathrm{\mid }{S}_n\mathrm{\mid }\le \mathrm{\mid }c\mathrm{\mid }+{\sum }_{k=1}^{n}E\mathrm{\mid }{Y}_k\mathrm{\mid }<\mathrm{\infty }$. Independence means independence of the sigma-algebras $\sigma (Y_k)$ Independent random elements. Group the first $n$ of these separately from $\sigma (Y_{n+1})$. Then $Y_{n+1}$ is independent of ${\mathcal{F}}_n$, so $E[Y_{n+1}\mid {\mathcal{F}}_n]=E{Y}_{n+1}=0$. At $n=0$ independence of the trivial sigma-algebra follows directly from its two events.

Conditioning the finite identity $S_{n+1}=S_n+Y_{n+1}$ gives $E[S_{n+1}\mid {\mathcal{F}}_n]=S_n+0=S_n$. Hence this is a martingale Martingale submartingale and supermartingale. Each $S_k$ for $k\le n$ is measurable for $\sigma (Y_1,\dots ,Y_n)$, while $Y_k=S_k-S_{k-1}$ is measurable for $\sigma (S_0,\dots ,S_n)$. Minimality in both directions proves equality of these sigma-algebras, including the trivial initial one, as required by Natural filtration of a process. AC is inherited solely from the conditional classes; the independent sequence was given.

For a concrete model let $\mathrm{\Omega }=\{-1,1{\}}^2$ with each point of mass $1\mathrm{/}4$, $Y_1(u,v)=u$, $Y_2(u,v)=2v$, and $Y_k=0$ for $k\ge 3$. The two coordinate events have product probabilities because their intersections have cardinality the product of their cardinalities; adding constant variables preserves this identity. Here $S_0=c$, $S_1=c+u$, and $S_n=c+u+2v$ for $n\ge 2$. Averaging the two values in each fixed-u fibre gives $c+u$, and the first average is $c$. This displays the calculation without an identical-distribution assumption.

The generated finite-history sigma-algebras form a filtration; finite products are adapted. Apply factorization to the Borel functions $t\mapsto \mathrm{\mid }t\mathrm{\mid }$ to get $E\mathrm{\mid }{M}_n\mathrm{\mid }={\prod }_{k=1}^{n}E\mathrm{\mid }{Y}_k\mathrm{\mid }<\mathrm{\infty }$ for $n\ge 1$; at $n=0$ it equals one. Group independence separates $\sigma (Y_{n+1})$ from the finite past, so $E[Y_{n+1}\mid {\mathcal{F}}_n]=E{Y}_{n+1}=1$ (also for the trivial past at zero).

The variable $M_n$ is finite and ${\mathcal{F}}_n$-measurable. Both $Y_{n+1}$ and $M_n{Y}_{n+1}=M_{n+1}$ are integrable by step 1.1. The unbounded taking-out clause therefore gives $E[M_{n+1}\mid {\mathcal{F}}_n]=M_{n}E[Y_{n+1}\mid {\mathcal{F}}_n]=M_n$. This proves the martingale assertion Martingale submartingale and supermartingale. AC is inherited from CE; neither positivity nor identical distribution of factors was used. If one adjoins the deterministic factor $Y_0=1$, the displayed filtration is exactly the natural filtration of the resulting zero-based factor process Natural filtration of a process, not necessarily that of the products.

For example take $\mathrm{\Omega }=\{-1,3{\}}^2$ with four equal masses, let $Y_1,Y_2$ be its coordinates and $Y_k=1$ for $k\ge 3$. Each first factor has mean $(-1+3)\mathrm{/}2=1$ and absolute mean $2$; coordinate rectangle counting proves independence. The four values of $M_2$ are $1,-3,-3,9$, so $E{M}_2=1$ and $E\mathrm{\mid }{M}_2\mathrm{\mid }=4=2\cdot 2$. Conditional on $Y_1=-1$ the product averages $(1-3)\mathrm{/}2=-1$, and conditional on $Y_1=3$ it averages $(-3+9)\mathrm{/}2=3$.

Repair first the integral foundation inherited by RN. Augment any finite disjoint display of a nonnegative simple function by the complement with coefficient $0$. Pairwise intersections of two augmented displays partition the whole space and have equal coefficients on nonempty cells; finite additivity and $0\cdot (+\mathrm{\infty })=0$ prove representation independence. Common refinements give simple addition and monotonicity; scalar zero is direct and positive scalars are termwise. Supremum over simple minorants and the sets $\{f_j\ge cs\}$, $0<c<1$, give monotone convergence; increasing simple approximations give nonnegative additivity, and positive/negative plus real/imaginary decompositions give finite $L^1$ linearity. With these facts substituted for the affected foundation, the cited RN proof applies. Apply it to $\mu =P,\nu =Q$ with the constant exhaustion $X_j=\mathrm{\Omega }$. Both masses and the total variation of the positive $Q$ are one. Thus $L$ is real measurable, integrable and ${\int }_{A}LdP=Q(A)$ for every $A\in {\mathcal{F}}_N$. It is nonnegative a.s.: for each positive integer $j$, put $B_j=\{L\le -1\mathrm{/}j\}$. Then $0\le Q(B_j)={\int }_{B_j}LdP\le -P(B_j)\mathrm{/}j$, so each $B_j$ is null, and ${\bigcup }_{j\ge 1}{B}_j=\{L<0\}$. Testing $\mathrm{\Omega }$ gives $E_{P}L=1$.

Extend the filtration constantly after $N$ to apply [F3]; up to $N$, it makes $Z$ a martingale. Conditional positivity gives $Z_n\ge 0$ a.s. For every $A\in {\mathcal{F}}_n$, the defining event identity gives ${\int }_A{Z}_{n}dP={\int }_{A}LdP=Q(A)$, exactly the restricted density assertion. At $n=N$ known-variable conditioning gives $Z_N=L$. If ${\mathcal{F}}_0$ is trivial, the constant one has the same integrals as $L$ on its two events, so it is the conditional class $Z_0$. AC covers RN and CE existence and the finite choice of versions.

For instance let $N=1$, $\mathrm{\Omega }=\{a,b\}$, $P(a)=P(b)=1\mathrm{/}2$, $Q(a)=3\mathrm{/}4$, $Q(b)=1\mathrm{/}4$, and ${\mathcal{F}}_0$ trivial. Then $L(a)=3\mathrm{/}2$, $L(b)=1\mathrm{/}2$, and $Z_0=1$, $Z_1=L$. The average $(3\mathrm{/}2+1\mathrm{/}2)\mathrm{/}2=1$ verifies the martingale equality, and $(1\mathrm{/}2)(3\mathrm{/}2)=3\mathrm{/}4$ verifies the restricted density on $\{a\}$.

Use $\mathrm{\Omega }=(0,1]$ with the trace Lebesgue sigma-algebra and length probability. Restriction is a measure by [F2] (equivalently restrict its event formula to subsets of $\mathrm{\Omega }$), and [F1] gives total mass one. Define intervals for finite color words recursively: $I_{\mathrm{\varnothing }}=(0,1]$. If $h$ has length $n$, contains $j$ red letters, and $I_h=(a,b]$, put $T_n=r+g+nc$ and $p_h=(r+cj)\mathrm{/}{T}_n$. Then $0<p_h<1$ because both $r+cj$ and $g+c(n-j)$ are positive. Set $I_{hR}=(a,a+p_h(b-a)]$ and $I_{hG}=(a+p_h(b-a),b]$. Both are positive-length half-open intervals Half-open boxes in ${\mathbb{R}}^n$ and their volume, disjoint with union $I_h$. Induction gives a finite partition at every depth, and each point has exactly one compatible word of every finite length. Thus every draw is defined on this one space by the interval containing the point; no infinite-product existence is assumed.

Let ${\mathcal{F}}_n$ consist of all unions of depth-n intervals. Complements and countable unions just select subsets of this finite partition, so it is a sigma-algebra. Refinement makes it a filtration. A word interval is exactly the intersection of the corresponding first n color events; conversely a color event up to n is a union of word intervals. Hence ${\mathcal{F}}_n$ is the draw-history sigma-algebra. On $I_h$ set $R_n=r+cj$ and $U_n=R_n\mathrm{/}{T}_n$. These are finite-valued ${\mathcal{F}}_n$-measurable variables and $0<U_n<1$. If $J_{n+1}$ indicates the next red draw, then ${\int }_{I_h}{J}_{n+1}dP=P(I_{hR})=p_{h}P(I_h)={\int }_{I_h}{U}_{n}dP$. Finite addition proves this identity for every event in ${\mathcal{F}}_n$. Both variables are bounded, so [F3] identifies $E[J_{n+1}\mid {\mathcal{F}}_n]=U_n$.

The pathwise update is $R_{n+1}=R_n+c{J}_{n+1}$. The bounded ${\mathcal{F}}_n$-measurable $R_n$ is its own conditional version, since its event identities are tautologies. Since $T_n+c$ is deterministic and positive, conditional linearity gives $E[U_{n+1}\mid {\mathcal{F}}_n]=(R_n+c{U}_n)\mathrm{/}(T_n+c)=R_n(1+c\mathrm{/}{T}_n)\mathrm{/}(T_n+c)=R_n\mathrm{/}{T}_n=U_n$. Thus the bounded adapted process is a martingale Martingale submartingale and supermartingale. For $r=g=c=1$, $U_0=1\mathrm{/}2$ and the first two possible new proportions are $2\mathrm{/}3$ and $1\mathrm{/}3$, each with probability $1\mathrm{/}2$; their mean is $1\mathrm{/}2$. After the first red draw the next proportions are $3\mathrm{/}4$ and $1\mathrm{/}2$ with probabilities $2\mathrm{/}3$ and $1\mathrm{/}3$, whose mean is $1\mathrm{/}2+1\mathrm{/}6=2\mathrm{/}3$. AC supplies the countable choice assumed in Lebesgue construction and CE existence; the interval recursion itself makes no selections.

First justify the finite integral operations locally. Augment every finite disjoint simple display by the complement with coefficient $0$. Intersections of two augmented displays partition the whole space and have equal coefficients on every nonempty cell, so finite additivity and $0\cdot (+\mathrm{\infty })=0$ prove representation independence. Common refinements give simple addition and monotonicity; scalar zero is direct and positive scalars are termwise. Supremum over simple minorants, increasing simple approximation and the sets $\{f_j\ge cs\}$, $0<c<1$, give monotone convergence and nonnegative additivity. Positive/negative and real/imaginary decompositions give the finite $L^1$ linearity and triangle estimate used below. The trace restriction of Lebesgue measure to $(0,1]$ has total mass one. The displayed half-open cells Half-open boxes in ${\mathbb{R}}^n$ and their volume are disjoint, exhaust the space and have probability $2^{-n}>0$. Their unions form a sigma-algebra because unions and complements select cells from a finite partition. Each cell at n is exactly the union of its children at n+1, so the sigma-algebras increase. Every cell integral of $f$ is finite. The displayed finite-valued $M_n$ is ${\mathcal{F}}_n$-measurable, and $E\mathrm{\mid }{M}_n\mathrm{\mid }={\sum }_j\mathrm{\mid }{\int }_{I_{n,j}}fdP\mathrm{\mid }\le {\sum }_j{\int }_{I_{n,j}}\mathrm{\mid }f\mathrm{\mid }dP=E\mathrm{\mid }f\mathrm{\mid }<\mathrm{\infty }$.

For $A={\bigcup }_{j\in J}{I}_{n,j}\in {\mathcal{F}}_n$, finite addition and the cell mass give ${\int }_A{M}_{n}dP={\sum }_{j\in J}{2}^n({\int }_{I_{n,j}}fdP)2^{-n}={\int }_{A}fdP$. Thus $M_n$ meets every defining conditional-expectation requirement. Apply [F6] to obtain the martingale. For $f={\mathbf{1}}_{(0,1\mathrm{/}2]}$, the first average is $M_0=1\mathrm{/}2$ and every finer cell lies entirely in $(0,1\mathrm{/}2]$ or its complement, so $M_n=f$ for $n\ge 1$. In particular $E[M_1\mid {\mathcal{F}}_0]=1\mathrm{/}2$. AC supplies the Lebesgue construction and CE existence; the explicit finite averages require no version selection.

The generated sigma-algebras are nested, and finite sums are measurable by Arithmetic and lattice operations preserve measurability whenever they are defined. Cauchy–Schwarz with the constant one gives $E\mathrm{\mid }{Y}_k\mathrm{\mid }\le (E{Y}_k^2{)}^{1\mathrm{/}2}<\mathrm{\infty }$. Repeated use of $(a+b{)}^2\le 2a^2+2b^2$ shows that every finite sum $S_n$ has finite second moment, and hence finite first moment. Group independence and [F2] give $E[Y_{n+1}\mid {\mathcal{F}}_n]=0$, while the known $S_n$ conditions to itself. Thus $E[S_{n+1}\mid {\mathcal{F}}_n]=S_n$, proving the martingale property Martingale submartingale and supermartingale.

The measurable variable $Y_k^2$ is integrable and its Borel events belong to $\sigma (Y_k)$, independent of ${\mathcal{F}}_{k-1}$. Thus $E[Y_k^2\mid {\mathcal{F}}_{k-1}]=E{Y}_k^2={\sigma }_k^2$. Since $S_k-S_{k-1}=Y_k$, the bracket formula gives $⟨S{⟩}_n={\sum }_{k=1}^n{\sigma }_k^2$. The square-minus-bracket theorem now gives the asserted martingale. AC is inherited from these conditional classes and the bracket construction.

To see why the optional sum differs, take $\mathrm{\Omega }=\{-1,0,1\}$ and $P=\frac{1}{4}{\delta }_{-1}+\frac{1}{2}{\delta }_0+\frac{1}{4}{\delta }_1$. This is a measure by [F7]–[F8], and its total mass is one. Set $Y_1(\omega )=\omega$ and $Y_k=0$ for $k\ge 2$. The family $(Y_k{)}_{k\ge 1}$ is independent because all but one member have only probability-zero or probability-one events. Here $E{Y}_1=0$ and $E{Y}_1^2=1\mathrm{/}2$. Thus $[S{]}_1=Y_1^2$ takes values $0,1$ with probabilities $1\mathrm{/}2,1\mathrm{/}2$, whereas $⟨S{⟩}_1=1\mathrm{/}2$ everywhere. The compensated square takes values $-1\mathrm{/}2,1\mathrm{/}2$ of equal mass and thereafter stays fixed.

Let $\mathrm{\Omega }=\{\ast \}$ with $P={\delta }_{\ast }$ and ${\mathcal{F}}_n=\{\mathrm{\varnothing },\mathrm{\Omega }\}$ for all $n$. This is a probability space and a filtration. Define $X_n(\ast )=n$. Each $X_n$ is measurable for ${\mathcal{F}}_n$ and $E\mathrm{\mid }{X}_n\mathrm{\mid }=n<\mathrm{\infty }$, so the process is adapted and integrable at every time.

Conditional expectation fixes constants, so $E[X_{n+1}\mid {\mathcal{F}}_n]=n+1$, while $X_n=n$. The two values differ on $\mathrm{\Omega }$, which has probability one, at every $n$, including $0$. Hence the martingale equality Martingale submartingale and supermartingale fails. AC is inherited from the CE class convention; the singleton calculation itself is explicit.

Let $\mathrm{\Omega }=\{(j,e):j\ge 1,e\in \{-1,1\}\}$ with its full power-set sigma-algebra and masses $P\{(j,e)\}=2^{-j-1}$. The weighted Dirac sum is a measure by [F1]–[F2]. Its first m two-point blocks have total mass ${\sum }_{j=1}^m{2}^{-j}=1-2^{-m}$, by multiplying the finite sum by two and subtracting; since $2^m\ge m+1$, its limit is one. Hence $P(\mathrm{\Omega })=1$. Let ${\mathcal{F}}_0$ be all unions of the blocks $B_j=\{(j,-1),(j,1)\}$ and ${\mathcal{F}}_n$ the full power set for $n\ge 1$. Complements and countable unions preserve block unions, so this is a filtration.

Set $M_0=0$, $M_n(j,e)=e$ for $n\ge 1$. These variables are adapted and bounded by one. For any $A\in {\mathcal{F}}_0$, its positive-sign and negative-sign portions have equal probability: each equals ${\sum }_{j:B_j\subseteq A}{2}^{-j-1}$. Thus ${\int }_{A}edP=0$, a finite subtraction, so the zero function meets all CE event identities. Consequently $E[M_1\mid {\mathcal{F}}_0]=0=M_0$. At later times $M_{n+1}=M_n=e$ is already known and integrable, so it conditions to itself. Thus $M$ is a bounded martingale Martingale submartingale and supermartingale.

Define $H_1(j,e)=2^j$ and $H_k=0$ for $k\ge 2$. Every value is finite, $H_1$ is constant on each block and hence ${\mathcal{F}}_0$-measurable, and later H values are known constants. Thus $H$ is predictable Predictable discrete time process and unbounded. The algebraic gain at every positive time is $G_n(j,e)=2^{j}e$. For $f_m=2^j{\mathbf{1}}_{\{j\le m\}}$, the finite simple integral is $E{f}_m={\sum }_{j=1}^m{\sum }_{e=\pm 1}{2}^j{2}^{-j-1}=m$. To justify the limiting step locally, augment every finite disjoint simple display by its complement with coefficient $0$; intersections of two augmented displays partition $\mathrm{\Omega }$ and carry equal coefficients, so finite additivity and $0\cdot (+\mathrm{\infty })=0$ prove representation independence. Common refinements give monotonicity. For any $0\le u_r\uparrow u$, a simple $s\le u$, and $0<c<1$, the sets $A_r=\{u_r\ge cs\}$ increase to $\mathrm{\Omega }$, including on the zero level of $s$, and continuity from below for the finite-sum measure $A\mapsto {\int }_{A}s$ gives $c\int s\le {\sup }_r\int u_r$. Let $c\uparrow 1$ and take the supremum over $s$ to obtain MCT. Applying this to $f_m\uparrow \mathrm{\mid }{G}_1\mathrm{\mid }$ gives $E\mathrm{\mid }{G}_1\mathrm{\mid }=\mathrm{\infty }$. Hence the product at time one is not integrable and violates [F7], and the gains cannot be a martingale. No signed conditional expectation of $G_1$ is formed. AC is inherited only from the CE notation used for the bounded M; all masses and factors are explicit.

The measure $P=\frac{1}{2}{\delta }_{-1}+\frac{1}{2}{\delta }_1$ is a measure of total mass one. The displayed filtration is increasing. All $X_n$ are adapted and bounded by one, hence integrable. On the only two events of ${\mathcal{F}}_0$, the zero function has the same integrals as $X_1$, since $E{X}_1=(-1+1)\mathrm{/}2=0$. It is therefore its conditional version. At later times $X_{n+1}=X_n$ is known, so $E[X_{n+1}\mid {\mathcal{F}}_n]=X_n$. Thus $X$ is a martingale and hence a submartingale Martingale submartingale and supermartingale.

On the measurable atom $\{-1\}$, whose probability is $1\mathrm{/}2$, one has $X_1=-1<0=X_0$. Any exceptional set outside which paths are nondecreasing must contain this atom and therefore cannot have probability zero. This refutes even almost-sure nondecreasing paths. AC is inherited from the CE class convention; the finite averages supply the displayed versions explicitly.