Discrete Time Martingales

For $n=m$, adaptedness and integrability give $E[X_m\mid {\mathcal{F}}_m]=X_m$. For the submartingale case fix $m$ and induct on $n-m$. If $n>m$, nesting and the tower identity give $E[X_n\mid {\mathcal{F}}_m]=E[E[X_n\mid {\mathcal{F}}_{n-1}]\mid {\mathcal{F}}_m]\ge E[X_{n-1}\mid {\mathcal{F}}_m]\ge X_m$. The first inequality uses the one-step hypothesis and conditional order; the second is the induction hypothesis. All conditioned variables are integrable.

For the supermartingale case the identical tower identity has both inequalities reversed, since conditional order preserves the relation $\le$. For the martingale case the inner conditional expectation equals $X_{n-1}$, so induction gives equality at every pair. Each induction uses only finitely many almost-sure identities. AC here is inherited from the existence of the conditional classes; no representatives at all pairs are selected.

Conversely, each all-pairs condition evaluated at $(m,n)=(k,k+1)$ is its defining one-step condition. This includes $k=0$, and proves each equivalence.

By conditional existence the set of real measurable integrable versions at every time is nonempty. AC selects one member for each $n\in {\mathbb{N}}_0$. Thus each $M_n$ is ${\mathcal{F}}_n$-measurable and integrable, so $M$ is an integrable adapted process. AC also covers the RN existence assumption.

Since ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$, the tower identity gives $E[M_{n+1}\mid {\mathcal{F}}_n]=E[E[X\mid {\mathcal{F}}_{n+1}]\mid {\mathcal{F}}_n]=E[X\mid {\mathcal{F}}_n]=M_n$ a.s. for each $n\ge 0$. This is the martingale condition Martingale submartingale and supermartingale. If representatives of these identities are specified, their measurable failure sets have probability zero; their countable union is measurable and null. This does not complete any ${\mathcal{F}}_n$ or alter its representatives.

First reconstruct the finite-integral interface used below. Augment every finite disjoint display of a nonnegative simple function by the complement of its displayed sets with coefficient $0$. Intersections of two augmented displays partition the whole space, and equality of the functions makes their coefficients agree on every nonempty cell. Finite additivity and $0\cdot (+\mathrm{\infty })=0$ therefore prove representation independence. Common augmented refinements give monotonicity and additivity term by term; homogeneity is direct when the scalar is zero and termwise when it is positive. Taking suprema over simple minorants gives nonnegative monotonicity, and increasing simple approximations together with the sets $\{f_j\ge cs\}$, $0<c<1$, give monotone convergence. Applying this to sums of increasing simple approximants gives nonnegative additivity. Positive/negative and real/imaginary decompositions now give the finite real and complex $L^1$ linearity used in [F3]. With this replacement for the affected foundation, the event-integral construction and uniqueness proof of the cited conditional-expectation algebra apply. For a martingale $M$, both $M_n$ and $M_{n-1}$ are ${\mathcal{F}}_n$-measurable. Their difference $D_n$ is measurable and integrable, with $E\mathrm{\mid }{D}_n\mathrm{\mid }\le E\mathrm{\mid }{M}_n\mathrm{\mid }+E\mathrm{\mid }{M}_{n-1}\mathrm{\mid }<\mathrm{\infty }$. By linearity and conditioning the known $M_{n-1}$, $E[D_n\mid {\mathcal{F}}_{n-1}]=M_{n-1}-M_{n-1}=0$. This meets the difference definition.

Conversely every summand $D_k$ for $k\le n$ is ${\mathcal{F}}_n$-measurable, as is $M_0$. The finite sum is adapted and integrable by [F3]–[F4]. Its next increment is $D_{n+1}$, so $E[M_{n+1}\mid {\mathcal{F}}_n]=E[M_n+D_{n+1}\mid {\mathcal{F}}_n]=M_n+0=M_n$. Thus it is a martingale Martingale submartingale and supermartingale. AC is inherited from the conditional classes; if versions of a class sequence are to be chosen, AC permits those countably many selections.

The finite identities ${\sum }_{k=1}^n(M_k-M_{k-1})=M_n-M_0$ and $(M_0+{\sum }_{k=1}^n{D}_k)-(M_0+{\sum }_{k=1}^{n-1}{D}_k)=D_n$ prove inverse reconstruction. For $n=0$ the sum is empty and equals zero, so the initial value is exactly the prescribed $M_0$. For class representatives these finite equalities hold almost surely.

To justify finite integral linearity independently of the affected published proof, augment every finite disjoint simple display by its zero-coefficient complement. Pairwise intersections of two augmented displays partition the whole space and carry equal coefficients wherever nonempty, so finite additivity and $0\cdot (+\mathrm{\infty })=0$ prove representation independence. Common refinements give simple monotonicity and additivity; scalar zero is handled directly and positive scalars termwise. Supremum over simple minorants, followed by increasing simple approximation and the standard sets $\{f_j\ge cs\}$ for $0<c<1$, gives monotone convergence and nonnegative additivity. Positive/negative and real/imaginary decompositions therefore give finite $L^1$ linearity. This repairs the exact foundation used by [F4] and by the cited conditional-expectation identities. For $i<j$, Cauchy–Schwarz gives $E\mathrm{\mid }{D}_i{D}_j\mathrm{\mid }\le (E{D}_i^2{)}^{1\mathrm{/}2}(E{D}_j^2{)}^{1\mathrm{/}2}<\mathrm{\infty }$. Iterating [F6] gives ${\mathcal{F}}_i\subseteq {\mathcal{F}}_{j-1}$, so [F7] makes $D_i$ measurable for the latter sigma-algebra. Both $D_j$ and $D_i{D}_j$ are integrable, so the unbounded-factor clause applies: $E[D_i{D}_j\mid {\mathcal{F}}_{j-1}]=D_{i}E[D_j\mid {\mathcal{F}}_{j-1}]=0$. Expectation preservation gives $E[D_i{D}_j]=0$. The AC assumption [F5] meets the existence hypotheses of [F2, F3, F7]. All conditional identities are identities of almost-sure classes; this argument selects no sequence of representatives.

For a fixed positive $n$, expand the finite square as ${\sum }_{k=1}^n{D}_k^2+2{\sum }_{1\le i<j\le n}{D}_i{D}_j$. Every term is integrable by the assumptions and step 1.1. Finite integral linearity makes its expectation ${\sum }_{k=1}^{n}E{D}_k^2$, because every off-diagonal term vanishes. For $n=0$ both sides are zero by the empty-sum convention, and for $n=1$ there are no off-diagonal terms.

The function $\varphi$ is Borel, so each $\varphi (M_n)$ is ${\mathcal{F}}_n$-measurable. Integrability of the image is an explicit assumption. At time $n+1$ the input $M_{n+1}$ is integrable by the martingale definition and its image is integrable by hypothesis. Conditional Jensen therefore gives $E[\varphi (M_{n+1})\mid {\mathcal{F}}_n]\ge \varphi (E[M_{n+1}\mid {\mathcal{F}}_n])=\varphi (M_n)$ a.s. These are exactly the submartingale inequalities Martingale submartingale and supermartingale.

For a submartingale $X$, measurability and integrability of $\varphi (X_n)$ hold by the same argument. Jensen and the nondecreasing hypothesis give $E[\varphi (X_{n+1})\mid {\mathcal{F}}_n]\ge \varphi (E[X_{n+1}\mid {\mathcal{F}}_n])\ge \varphi (X_n)$ a.s. Monotonicity is applied to the submartingale inequality at that fixed time. AC is inherited from conditional Jensen and the CE existence used in both computations. No convex image of an arbitrary integrable variable is presumed integrable.

Fix $p\ge 1$ and set $\varphi (t)=\mathrm{\mid }t{\mathrm{\mid }}^p$ on all of $\mathbb{R}$, using $\varphi (0)=0$. The power lemma supplies finiteness, Borel measurability and convexity, including real noninteger $p$ and the endpoint $p=1$. Since $\varphi (M_n)\ge 0$, the assumed finite pth moment is exactly its $L^1$ condition. The convex-transform theorem therefore gives $E[\mathrm{\mid }{M}_{n+1}{\mathrm{\mid }}^p\mid {\mathcal{F}}_n]\ge \mathrm{\mid }{M}_n{\mathrm{\mid }}^p$ a.s. for every $n$.

At $p=1$, $E\mathrm{\mid }{M}_n\mathrm{\mid }<\mathrm{\infty }$ already follows from the martingale definition, so the first assertion requires no extra moment hypothesis. AC is inherited from the convex-transform theorem and its conditional Jensen argument. For $p>1$ the moment assumption is retained; no assertion for $p<1$ is made.

Put $D_k=M_k-M_{k-1}$. It is integrable since $E\mathrm{\mid }{D}_k\mathrm{\mid }\le E\mathrm{\mid }{M}_k\mathrm{\mid }+E\mathrm{\mid }{M}_{k-1}\mathrm{\mid }$. In the bounded case [F1] gives $E\mathrm{\mid }{H}_k{D}_k\mathrm{\mid }\le C_k(E\mathrm{\mid }{M}_k\mathrm{\mid }+E\mathrm{\mid }{M}_{k-1}\mathrm{\mid })<\mathrm{\infty }$; in the general case this is assumed. Consequently $Z=H\cdot M$ is adapted, integrable, and $Z_0=0$. This is verified before conditioning any product.

For each $n\ge 0$, $H_{n+1}$ is finite ${\mathcal{F}}_n$-measurable and $D_{n+1},H_{n+1}{D}_{n+1}$ are integrable. The unbounded clause of [F2] therefore gives $E[H_{n+1}{D}_{n+1}\mid {\mathcal{F}}_n]=H_{n+1}E[D_{n+1}\mid {\mathcal{F}}_n]=H_{n+1}(M_n-M_n)=0$. Linearity and known-variable conditioning now give $E[Z_{n+1}\mid {\mathcal{F}}_n]=Z_n$. This proves the martingale assertion Martingale submartingale and supermartingale even for signed $H$. AC is inherited from the conditional classes in this calculation.

If a uniform bound $C$ is supplied, choose $C_k=C$ in step 1.1. Another sufficient domain condition at a fixed $k$ is $H_k\in L^p$ and $D_k\in L^q$ with conjugate $p,q$ under the clauses of [F6]: then $E\mathrm{\mid }{H}_k{D}_k\mathrm{\mid }\le \mathrm{\parallel }{H}_k{\mathrm{\parallel }}_p\mathrm{\parallel }{D}_k{\mathrm{\parallel }}_q<\mathrm{\infty }$. The proof of step 2.1 only needs the resulting product integrability.

Set $Z=H\cdot X$ and $d_k=E[X_k-X_{k-1}\mid {\mathcal{F}}_{k-1}]$. By the transform domain $Z$ is adapted and integrable with $Z_0=0$. Linearity and the known-variable identity give $d_k=E[X_k\mid {\mathcal{F}}_{k-1}]-X_{k-1}\ge 0$ a.s. by the submartingale hypothesis Martingale submartingale and supermartingale.

Apply [F2] with input $X_k-X_{k-1}$ and finite known factor $H_k$. The input and its product are integrable, so [F2] also guarantees $H_k{d}_k\in L^1$ and gives $E[Z_k-Z_{k-1}\mid {\mathcal{F}}_{k-1}]=H_k{d}_k\ge 0$ a.s. Adding the known $Z_{k-1}$ yields $E[Z_k\mid {\mathcal{F}}_{k-1}]\ge Z_{k-1}$ for every $k\ge 1$. For timewise bounds, [F1] verifies the product hypothesis. AC is inherited from the conditional classes; no positivity of $X$ or boundedness of an unbounded $H$ is inferred.

First repair the integral foundation inherited by RN and conditional expectation. Augment any finite disjoint display of a nonnegative simple function by the complement with coefficient $0$. Intersections of two augmented displays partition the whole space and have equal coefficients wherever nonempty, so finite additivity and $0\cdot (+\mathrm{\infty })=0$ prove representation independence. Common refinements give simple monotonicity and additivity; handle scalar $0$ directly and positive scalars termwise. Supremum over simple minorants and the sets $\{f_j\ge cs\}$, $0<c<1$, give monotone convergence; increasing simple approximations then give nonnegative additivity. Positive/negative and real/imaginary decompositions give finite $L^1$ linearity. With these facts substituted at the affected foundation, the cited RN proof gives its density, and its event-integral existence and uniqueness argument gives the conditional-expectation class and algebra in [F1], [F4] and [F5]. Each $V_k=X_k-X_{k-1}$ is therefore real measurable and integrable, since $E\mathrm{\mid }{V}_k\mathrm{\mid }\le E\mathrm{\mid }{X}_k\mathrm{\mid }+E\mathrm{\mid }{X}_{k-1}\mathrm{\mid }$. AC permits choosing a finite real integrable ${\mathcal{F}}_{k-1}$-measurable version $a_k$ of its conditional expectation for every $k\ge 1$. Set $A_0=0$, $A_n={\sum }_{k=1}^n{a}_k$, and $M_n=X_n-A_n$. For $k\le n$ one has ${\mathcal{F}}_{k-1}\subseteq {\mathcal{F}}_{n-1}$; therefore $A_n$ is predictable for $n\ge 1$. Finite sums and differences show that $A$ is integrable and $M$ is adapted and integrable.

The increment $M_n-M_{n-1}=V_n-a_n$ has conditional expectation $a_n-a_n=0$ given ${\mathcal{F}}_{n-1}$, by linearity and known-variable conditioning. Since $M_0=X_0$ is integrable and ${\mathcal{F}}_0$-measurable, [F6] makes $M$ a martingale. The displayed decomposition holds pointwise for the chosen representatives.

If $X=\tilde{M}+\tilde{A}$ is another normalized decomposition, then ${\tilde{A}}_n-{\tilde{A}}_{n-1}$ is integrable and ${\mathcal{F}}_{n-1}$-measurable: for $n=1$ use ${\tilde{A}}_0=0$, and for $n>1$ use predictability and nesting. Conditioning the decomposition increment and using the zero martingale drift gives $E[V_n\mid {\mathcal{F}}_{n-1}]={\tilde{A}}_n-{\tilde{A}}_{n-1}$ a.s. Thus these increments equal $a_n$ a.s. Induction from zero gives ${\tilde{A}}_n=A_n$ and then ${\tilde{M}}_n=M_n$ a.s. for each $n$. The sets where either equality fails are ambient measurable null sets; their countable union is null by [F7]. Off that one set both entire sequences agree. No completeness of the filtration is used.

The Doob formula and conditional linearity give $A_n-A_{n-1}=E[X_n-X_{n-1}\mid {\mathcal{F}}_{n-1}]=E[X_n\mid {\mathcal{F}}_{n-1}]-X_{n-1}$ a.s., since $X_{n-1}$ is known and integrable. If $X$ is a submartingale the right side is nonnegative for every $n\ge 1$, hence so is the compensator increment. Conversely nonnegative compensator increments imply $E[X_n\mid {\mathcal{F}}_{n-1}]\ge X_{n-1}$ for every $n\ge 1$, which is the submartingale definition Martingale submartingale and supermartingale. AC is inherited from the Doob construction and its conditional classes.

If every increment is nonnegative a.s., the measurable sets $N_n=\{A_n<A_{n-1}\}$ are null. Their union $N={\bigcup }_{n\ge 1}{N}_n$ is measurable and null. For $\omega \mathrm{\notin }N$ every successive inequality holds, so finite chaining gives $A_m(\omega )\le A_n(\omega )$ whenever $m\le n$. Conversely, if all paths off a measurable null $N$ are nondecreasing, each $N_n$ is contained in $N$, and hence has probability zero. This proves the path formulation for any chosen measurable versions.

Put $D_n=M_n-M_{n-1}$ for $n\ge 1$. It is in $L^2$ by $D_n^2\le 2M_n^2+2M_{n-1}^2$, and its conditional mean given ${\mathcal{F}}_{n-1}$ is zero. Cauchy–Schwarz gives $E\mathrm{\mid }{M}_{n-1}{D}_n\mathrm{\mid }\le (E{M}_{n-1}^2{)}^{1\mathrm{/}2}(E{D}_n^2{)}^{1\mathrm{/}2}<\mathrm{\infty }$. The factor $M_{n-1}$ is finite and known at time $n-1$, so taking-out is legitimate and gives $E[M_{n-1}{D}_n\mid {\mathcal{F}}_{n-1}]=M_{n-1}E[D_n\mid {\mathcal{F}}_{n-1}]=0$.

For the finite integral and conditional linearity used here, augment every finite disjoint nonnegative-simple display by its zero-coefficient complement. Intersections of two augmented displays partition the space and carry equal coefficients on nonempty cells, 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 $\{f_j\ge cs\}$ for $0<c<1$ give monotone convergence and nonnegative additivity. Positive/negative and real/imaginary decompositions give finite $L^1$ linearity; substituting these facts at the base validates the event-integral construction and algebra of the cited conditional expectations. The square and bracket are adapted and integrable, hence so is $Z$. Expand $M_n^2-M_{n-1}^2=2M_{n-1}{D}_n+D_n^2$. Every term is integrable. Conditional linearity and step 1.1 give $E[M_n^2-M_{n-1}^2\mid {\mathcal{F}}_{n-1}]=E[D_n^2\mid {\mathcal{F}}_{n-1}]=⟨M{⟩}_n-⟨M{⟩}_{n-1}$. The last difference is known at time $n-1$. Subtracting it and conditioning the known $Z_{n-1}$ proves $E[Z_n\mid {\mathcal{F}}_{n-1}]=Z_{n-1}$. The initial bracket is zero, so $Z_0=M_0^2$.

Set $N_n=M_n-M_0$. Since $M_0$ is ${\mathcal{F}}_0$-measurable, it is known for every ${\mathcal{F}}_n$. Linearity gives $E[N_{n+1}\mid {\mathcal{F}}_n]=M_n-M_0=N_n$. Also $N_n^2\le 2M_n^2+2M_0^2$ is integrable, so $N$ is a square-integrable martingale with $N_0=0$. Its increments equal $D_n$, hence $⟨N⟩=⟨M⟩$ as classes. Apply the already proved step 2.1 to $N$ to conclude that $W=N^2-⟨M⟩$ is a martingale with $W_0=0$. AC is inherited from CE and bracket version construction; no unproved assertion about the product $M_0{M}_n$ is used.

By [F1], $Z_n=M_n^2-⟨M{⟩}_n$ is integrable and $E[Z_k\mid {\mathcal{F}}_{k-1}]=Z_{k-1}$ for each $k\ge 1$. Expectation preservation gives $E{Z}_k=E{Z}_{k-1}$. Induction over the finitely many times up to $n$ yields $E{Z}_n=E{Z}_0=E{M}_0^2$, also at $n=0$. The invocations of F1 and F2 are made under the AC assumption F4.

For the finite integral linearity used here, augment each finite disjoint display of a nonnegative simple function by its zero-coefficient complement. Pairwise intersections of two augmented displays partition the whole space and have equal coefficients on nonempty cells, so finite additivity and $0\cdot (+\mathrm{\infty })=0$ prove representation independence. Common refinements give simple addition and monotonicity, while 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 then give finite $L^1$ linearity. The square is integrable by the square-integrability hypothesis, and the bracket is integrable by [F5], so this local linearity gives $E{Z}_n=E{M}_n^2-E⟨M{⟩}_n$. Rearranging the finite equality from step 1.1 proves the formula. If $M_0=0$ its second moment is zero. At $n=0$ the bracket is zero and the equation reads $E{M}_0^2=E{M}_0^2$.