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Put $v=\mathbb{E}{X}_n^2$. For $c\ge 0$, conditional Jensen inequality applied to the convex function $x\mapsto (x+c{)}^2$ shows thatis a nonnegative submartingale. Its integrability follows from the square integrability of $X_k$. If $X_k\ge \lambda$, then $Z_k\ge (\lambda +c{)}^2$, since $c\ge 0$. The Doob maximal inequality for a nonnegative submartingale yieldswhere zero mean removes the cross term. For completeness, the maximal inequality follows by stopping at the first crossing: on the event of a crossing at $k$, the submartingale property gives $\mathbb{E}[Z_n{1}_{\{T=k\}}]\ge \mathbb{E}[Z_k{1}_{\{T=k\}}]$. Sum over $k\le n$, and use nonnegativity on the event of no crossing.

The derivative of the last ratio isFor $v>0$, the minimum over $c\ge 0$ is attained at $c=v/\lambda$. Substitution gives the one-sided maximal inequality for a centered square-integrable martingale:If $v=0$, $X_n=0$ almost surely and $X_k=\mathbb{E}(X_n\mid {\mathcal{F}}_k)=0$ almost surely for every $k\le n$, so the bound also holds. The optimization is the same one underlying the Cantelli inequality, but the submartingale argument controls the entire finite-time maximum.

Work on a filtered probability space $(\Omega ,\mathcal{F},({\mathcal{F}}_n{)}_{n\ge 0},\mathbb{P})$. A submartingale is a real-valued stochastic process satisfying the following three conditions: $Y_n$ is ${\mathcal{F}}_n$-measurable, $\mathbb{E}|Y_n|<\infty$, andThus it is adapted, has integrable random variables at every time, and its conditional expected next value is at least its current value. Repeated conditioning gives $\mathbb{E}[Y_m\mid {\mathcal{F}}_n]\ge Y_n$ for every $m\ge n$. The inequality concerns conditional expectation; individual sample paths need not increase. If no filtration is specified, the definition can be applied to the natural filtration generated by the process.

The almost sure submartingale convergence theorem says that if a submartingale $(Y_n)$ satisfiesthen there is an integrable random variable $Y_{\infty }$ such thatIndeed the positive-part bound also bounds the negative parts, since $\mathbb{E}{Y}_n\ge \mathbb{E}{Y}_0$; the theorem gives a finite limit. In particular, the martingale convergence theorem applies to any martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$.

For a nonnegative martingale, $\mathbb{E}{M}_n=\mathbb{E}{M}_0$ supplies this bound automatically. Its limit obeys $\mathbb{E}{M}_{\infty }\le \mathbb{E}{M}_0$ by the Fatou lemma. Almost sure convergence alone need not preserve the expectation or give convergence in L1. Uniform integrability is the additional condition that supplies convergence in L1 and the identity $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$.

Write $D_k=M_k-M_{k-1}$ for the martingale difference sequence. Bounded increments mean that one deterministic $C<\infty$ satisfies $|D_k|\le C$ for every $k$, almost surely. Then $\mathbb{E}{D}_k=0$ and, for $j<k$,The increments are uncorrelated random variables, uniformly bounded in $L^2$. Applying the supplied strong law for uniformly L2-bounded uncorrelated random variables gives $n^{-1}{\sum }_{k=1}^n{D}_k\to 0$ almost surely. Also $M_0/n\to 0$, because $M_0$ is finite almost surely. Therefore a martingale with bounded increments has zero linear growth:This is the strong law for martingales with bounded increments.

For the submartingale, let $\Delta Y_k=Y_k-Y_{k-1}$ and define its predictable drift and compensator byThis is the Doob decomposition in discrete time. The process $N_n$ is a martingale. Since $|\Delta Y_k|\le C$, we have $0\le a_k\le C$, so $|N_k-N_{k-1}|\le 2C$. The result just proved gives $N_n/n\to 0$ almost surely, and $A_n\ge 0$. Consequently the strong law for submartingales with bounded increments isThe compensator may grow, so the conclusion is a lower bound rather than a claim of convergence to zero.

The overshoot hypothesis as printed cannot hold for every $R>0$. For $R\le 1$, $T_R=0$ and $M_{T_R}=1$; choosing $R<1/\tau$ contradicts $M_{T_R}\le \tau R$. The intended and sufficient assumption is the overshoot bound for $R\ge 1$, on the event $\{T_R<\infty \}$. We use those thresholds below; when $R\le 1$, the requested stopped-time bound is trivial because $T\wedge T_R=0$.

Fix $R\ge 1$ and set $S=T\wedge T_R$. By part (a), the stopped martingale satisfies $0\le M_{S\wedge n}\le \tau R$: before stopping it is below $R$, at $T$ it is zero, and at $T_R$ the assumed overshoot bound applies. Thus the square of this stopped process is integrable. For $\Delta M_n=M_{n+1}-M_n$,The event $\{S>n\}$ is ${\mathcal{F}}_n$-measurable and contained in $\{T>n\}$. Conditional expectation of the linear term is zero, while the variance assumption makes the remaining conditional expectation nonnegative. The stopped increments here are bounded by the stopping and overshoot bounds, so all conditional expectations are legitimate. ConsequentlySince $X_0=1$, we have ${\sigma }^2\mathbb{E}(S\wedge n)\le \mathbb{E}{M}_{S\wedge n}^2-1$. Bounded-time optional stopping theorem also gives $\mathbb{E}{M}_{S\wedge n}=1$, soApply the monotone convergence theorem to $S\wedge n$. This proves the stronger estimate $(\tau R-1)/{\sigma }^2$ and hence the requested absorption-time bound from conditional variance and overshoot:In particular this stopping time is finite almost surely. For $0<R\le 1$, $S=0$ and $X_n=1$ is a constant submartingale, while the stopping-time expectation is zero.

Under the usual no-arbitrage interpretation from part (a), let $Q^N$ be an equivalent martingale measure for the strictly positive numéraire $N$. Then $M_t=S_t/N_t$ is a martingale. Define the discounted European call option payoff $Y_t=(S_t-K{)}^{+}/N_t=(M_t-K/N_t{)}^{+}$. Because $N_{t+1}\ge N_t$ and $K>0$, we have $K/N_{t+1}\le K/N_t$. The conditional Jensen inequality therefore givesFor the middle step, the strike $K/N_t$ is fixed conditionally on ${\mathcal{F}}_t$; alternatively use $\mathbb{E}[X^{+}\mid {\mathcal{F}}_t]\ge \max \{\mathbb{E}[X\mid {\mathcal{F}}_t],0\}$. Thus $Y$ is a submartingale. Risk-neutral valuation in units of $N$ gives $C(T,K)=N_0{\mathbb{E}}^{Q^N}{Y}_T$, with the usual fixed initial prices. ConsequentlyThe source's word “increasing” means nondecreasing: a European call option can have zero payoff at successive maturities, so strict increase is not guaranteed. Integrability is understood in the claim class for which the stated replication prices exist.

Backward induction makes every $U_t$ adapted and integrable: the conditional expectation of an integrable random variable is integrable, and $|\max (a,b)|\le |a|+|b|$. The Snell envelope recursion directly gives $U_t\ge \mathbb{E}[U_{t+1}\mid {\mathcal{F}}_t]$, so $U$ is a supermartingale.

If $Z$ is a submartingale, iterating its defining inequality gives $\mathbb{E}[Z_T\mid {\mathcal{F}}_t]\ge Z_t$. Starting at $U_T=Z_T$ and applying the tower property of conditional expectation backwards showsThis last conditional expectation is a true martingale. The conclusions areIn the second case waiting until $T$ attains the optimal stopping value. It is not generally true that $U_t=Z_t$: that further equality holds when $Z$ itself is a martingale.

Use the random walk interpretation and optimal stopping value function recursion specified in part (b). If $h$ is convex, translating and integrating its convexity inequality gives, for $0<\theta <1$,Thus the transition operator preserves convex functions. Also the pointwise maximum of convex functions is convex: each of two convex functions at an intermediate point is bounded by the same convex combination of their pointwise maximum at the endpoints. Starting with $V(T,\cdot )=f$, backward induction intherefore proves the asserted convexity in the intended model:The statewise integrability convention in part (b) gives finite convex functions on all real states. The same inequality holds for the canonical extended value function wherever the expectations are well-defined. Integrability only along one started process need not make that canonical function finite at unused states: for $T=1$, $S_0=0$, $f(s)=e^{s^2}$ and increment density proportional to $e^{-y^2-|y|}$, $\mathbb{E}f(S_1)<\infty$, but $\mathbb{E}f(s+Y_1)=\infty$ for $s>1/2$. This concerns the canonical recursion away from visited states, rather than the almost-sure identity for $U$. Arbitrary off-state versions of $V$ need not be convex; the recursive version is the one meant here. No zero-mean assumption on the increments was used, and convexity alone does not make $f(S_t)$ a submartingale for an arbitrary drift.

The discrete-time Doob L2 maximal inequality states that a square-integrable martingale satisfiesEquivalently, the Lp norm of the maximum at $p=2$ is at most $2\Vert M_n{\Vert }_2$. The same assertion holds for a nonnegative square-integrable submartingale. No assumption $M_0=0$ is required; for a bound in terms of fluctuations from the initial value, apply it to $(M_k-M_0)$.

Write $N_2=\Vert M_{\infty }{\Vert }_2$ and $N_{\ast }=\Vert {\sup }_{t\ge 0}|M_t|{\Vert }_2$. The L2 martingale convergence theorem ensures the existence of $M_{\infty }$ and yields $|M_{\infty }|\le {\sup }_t|M_t|$ almost surely. Hence $N_2\le N_{\ast }$.

For a finite horizon $T$, the Doob L2 maximal inequality for the submartingale $|M|$ givesThe inequality does not require $M_0=0$. Send $T$ to infinity: the left side increases to $N_{\ast }^2$ by the monotone convergence theorem, and the right side tends to $4N_2^2$ by convergence in the Lebesgue space $L^2$. Thus the equivalent terminal and maximal norms for L2-bounded continuous martingales satisfyBoth are genuine norms on L2-bounded continuous martingales modulo indistinguishability of stochastic processes: a zero terminal norm gives $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]=0$, first at rational times and then at all times by continuity. This proves that the two norms are equivalent norms.

For a submartingale with ${\sup }_n\mathbb{E}|X_n|=C<\infty$, its positive part is a submartingale by the conditional Jensen inequality. Thus the displayed conditional expectations increase in $p$. Their limits are finite almost surely and have expectations at most $C$, by the monotone convergence theorem. Conditional monotone convergence and the tower property of conditional expectation give $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$. Moreover $M_n\ge X_n^{+}$. Consequently $Y_n=M_n-X_n$ is a nonnegative supermartingale, ${\sup }_n\mathbb{E}{Y}_n\le 2C$, and $X=M-Y$. The resulting martingale need not have uniform integrability.

If a submartingale has uniformly bounded increments, then ${\mathrm{liminf}}_n{Y}_n/n\ge 0$ almost surely. In its Doob decomposition in discrete time, the predictable compensator is nondecreasing, while the martingale part has bounded increments. The strong law for martingales with bounded increments makes the latter negligible on the linear scale.

For a nonnegative submartingale with maximum $M_t$, the first-moment maximal estimate is $\lambda \mathbb{P}(M_t\ge \lambda )\le \mathbb{E}[X_t{1}_{\{M_t\ge \lambda \}}]$. Integrate $2\lambda$ times the tail probability only up to $R$, obtaining $\mathbb{E}(M_t\wedge R{)}^2\le 2\mathbb{E}[X_t(M_t\wedge R)]$. The Cauchy-Schwarz inequality gives $\Vert M_t\wedge R{\Vert }_2\le 2\Vert X_t{\Vert }_2$, and monotone convergence gives the result as $R\to \infty$. Truncation avoids assuming beforehand that the maximum has a finite second moment.

// Target: brownian-motion.bigb