Stopping Times and Optional Stopping

Conditions (1) and (3) are equivalent because $\{\tau >n\}=\mathrm{\Omega }\setminus \{\tau \le n\}$ and ${\mathcal{F}}_n$ is closed under complements.

From (1), $\{\tau =0\}=\{\tau \le 0\}\in {\mathcal{F}}_0$ and, for $n\ge 1$, $\{\tau =n\}=\{\tau \le n\}\setminus \{\tau \le n-1\}\in {\mathcal{F}}_n,$ using ${\mathcal{F}}_{n-1}\subseteq {\mathcal{F}}_n$. Thus (1) implies (2).

From (2), $\{\tau \le n\}={\bigcup }_{k=0}^n\{\tau =k\}\in {\mathcal{F}}_n,$ since each summand lies in ${\mathcal{F}}_k\subseteq {\mathcal{F}}_n$. Thus (2) implies (1), completing the equivalence.

Finally, for $n\ge 1$, $\{\tau \ge n\}=\mathrm{\Omega }\setminus \{\tau \le n-1\}\in {\mathcal{F}}_{n-1}\mathrm{.}$ The value $\mathrm{\infty }$ is automatically included in this event.

For each $n$, $\{\tau \le n\}={\bigcup }_{k=0}^n\{X_k\in B\}\mathrm{.}$ Every term is in ${\mathcal{F}}_k\subseteq {\mathcal{F}}_n$ by F1, and the union is finite.

Hence the event belongs to ${\mathcal{F}}_n$, so F2 proves that $\tau$ is a stopping time. If the path never enters $B$, it belongs to none of these events and the convention gives exactly $\tau =\mathrm{\infty }$.

The identities $\{\sigma \wedge \tau \le n\}=\{\sigma \le n\}\cup \{\tau \le n\},\{\sigma \vee \tau \le n\}=\{\sigma \le n\}\cap \{\tau \le n\}$ prove the first two claims by F1.

If $n<c$, $\{\tau +c\le n\}=\mathrm{\varnothing }$; if $n\ge c$, it is $\{\tau \le n-c\}\in {\mathcal{F}}_{n-c}\subseteq {\mathcal{F}}_n$. If $\rho$ is stopping for $\mathcal{G}$, the same event is in ${\mathcal{G}}_{n-c}={\mathcal{F}}_n$.

For the earlier shift, $\{(\tau -c{)}^{+}\le n\}=\{\tau \le n+c\}\in {\mathcal{F}}_{n+c},$ so it is stopping for the shifted filtration. The right side need not lie in ${\mathcal{F}}_n$, which is why no unshifted assertion is made.

A deterministic $N$ is a stopping time, so step 1.1 makes $\tau \wedge N$ a stopping time, and it is bounded by $N$. All empty and infinite-value cases follow from the displayed identities.

Since $\{\tau \le n\}\in {\mathcal{F}}_n$, $\mathrm{\Omega }$ passes every test. If $A$ passes, then $A^c\cap \{\tau \le n\}=\{\tau \le n\}\setminus (A\cap \{\tau \le n\})\in {\mathcal{F}}_n\mathrm{.}$ If all $A_j$ pass, distributivity gives $({\bigcup }_j{A}_j)\cap \{\tau \le n\}={\bigcup }_j(A_j\cap \{\tau \le n\})\in {\mathcal{F}}_n\mathrm{.}$ Thus ${\mathcal{F}}_{\tau }$ is a $\sigma$-algebra, and containment in $\mathcal{F}$ is part of F1.

If $\tau \equiv n$, all tests below $n$ are vacuous and the test at $n$ is $A\in {\mathcal{F}}_n$; upward nesting then supplies every later test. Hence ${\mathcal{F}}_{\tau }={\mathcal{F}}_n$.

Suppose $\sigma \le \tau$ and $A\in {\mathcal{F}}_{\sigma }$. For each $n$, $A\cap \{\tau \le n\}={\bigcup }_{k=0}^n(A\cap \{\sigma \le k\})\cap \{\tau =k\}\mathrm{.}$ On $\{\tau =k\}$ the condition $\sigma \le k$ is automatic, so the equality is exact. Each term belongs to ${\mathcal{F}}_k\subseteq {\mathcal{F}}_n$, proving $A\in {\mathcal{F}}_{\tau }$. The order hypothesis is not weakened to an untracked almost-sure order.

For Borel $B$, $\{X_{\tau }\in B\}={\bigcup }_{k\ge 0}(\{\tau =k\}\cap \{X_k\in B\})\cup (\{\tau =\mathrm{\infty }\}\text{ if }{x}_{\mathrm{\infty }}\in B)\mathrm{.}$ All finite-level terms lie in $\mathcal{F}$, and $\{\tau =\mathrm{\infty }\}$ is the complement of their countable union, so the inverse image is in $\mathcal{F}$.

Intersecting that inverse image with $\{\tau \le m\}$ deletes the infinity term and all levels above $m$, leaving ${\bigcup }_{k=0}^m\{\tau =k\}\cap \{X_k\in B\}\in {\mathcal{F}}_m\mathrm{.}$ Thus every inverse image passes F2's tests and $X_{\tau }$ is ${\mathcal{F}}_{\tau }$-measurable. Apply the same proof to the bounded stopping time $\tau \wedge n$ for the final assertion.

The finite formula in F1 makes $M_{n\wedge \tau }$ ${\mathcal{F}}_n$-measurable and integrable: it is a finite sum of $M_k{1}_{\{\tau =k\}}$ for $k<n$ and $M_n{1}_{\{\tau \ge n\}}$.

Pathwise, $M_{n\wedge \tau }-M_{(n-1)\wedge \tau }=1_{\{\tau \ge n\}}(M_n-M_{n-1})\mathrm{.}$ The indicator is bounded and ${\mathcal{F}}_{n-1}$-measurable by F2, so this is the $n$th increment of the predictable transform in F3.

Taking the conditional expectation of step 1.2 and pulling out the indicator by F6 gives zero by F4. Together with adaptedness and integrability from step 1.1, this is exactly the martingale condition. The stopped process never evaluates a value at infinity; AC has only the role in F5.

Put $H_k=1_{\{\sigma <k\le \tau \}}$ for $1\le k\le N$. Both $\{\sigma <k\}=\{\sigma \le k-1\}$ and $\{\tau \ge k\}$ lie in ${\mathcal{F}}_{k-1}$, so $H$ is nonnegative, bounded, and predictable. On the probability-one event $\{\sigma \le \tau \le N\}$, the finite pathwise telescope is $X_{\tau }-X_{\sigma }={\sum }_{k=1}^N{H}_k(X_k-X_{k-1})\mathrm{.}$ The integral identities below use this almost-sure equality; no equality is asserted on a null outcome with $\tau >N$.

Fix $A\in {\mathcal{F}}_{\sigma }$. On $\{\sigma \ge k\}$, $H_k=0$; hence $1_A{H}_k=1_{A\cap \{\sigma \le k-1\}}{H}_k,$ and the first factor on the right is ${\mathcal{F}}_{k-1}$-measurable by F1. Thus $1_{A}H$ is another bounded nonnegative predictable process.

Apply F3's one-step conditional calculation and sum: for a submartingale, ${\int }_A(X_{\tau }-X_{\sigma })dP\ge 0;$ for a martingale equality holds, and for a supermartingale the inequality reverses. Almost-sure boundedness identifies every stopped value almost surely with a finite sum of integrable variables, while F2 gives ${\mathcal{F}}_{\sigma }$-measurability of $X_{\sigma }$. F4 therefore identifies the stated conditional relation. Taking $A=\mathrm{\Omega }$ gives the expectation relation. AC has exactly the role in F5.

Fix a stopping time $\rho$ equal to either $\sigma$ or $\tau$. For $r\ge n$, F2 applied to $\rho \wedge n\le r$ gives $M_{\rho \wedge n}=\mathbb{E}[M_r\mid {\mathcal{F}}_{\rho \wedge n}]\mathrm{.}$ Let $r\to \mathrm{\infty }$ in $L^1$ using F1 and conditional contraction to obtain $M_{\rho \wedge n}=\mathbb{E}[M_{\mathrm{\infty }}\mid {\mathcal{F}}_{\rho \wedge n}]\mathrm{.}$

The sigma-algebras ${\mathcal{F}}_{\rho \wedge n}$ increase with $n$. Thus the sequence in step 1.1 is a closed, hence uniformly integrable, martingale by F1. Since $\rho <\mathrm{\infty }$ almost surely, F3 gives $M_{\rho \wedge n}\to M_{\rho }$ almost surely; F1's UI convergence implication upgrades this to $L^1$, proving $M_{\rho }\in L^1$.

For $A\in {\mathcal{F}}_{\rho }$, the event $A\cap \{\rho \le n\}$ belongs to ${\mathcal{F}}_{\rho \wedge n}$ (check the defining finite-level intersections). Apply the conditional identity in step 1.1 on this event, then let $n\to \mathrm{\infty }$. The left side converges by step 2.1; the right side converges by dominated convergence for $M_{\mathrm{\infty }}{1}_{A\cap \{\rho \le n\}}$. Hence ${\int }_A{M}_{\rho }dP={\int }_A{M}_{\mathrm{\infty }}dP,$ so $M_{\rho }=\mathbb{E}[M_{\mathrm{\infty }}\mid {\mathcal{F}}_{\rho }]$.

Since $\sigma \le \tau$ pointwise, F6 gives ${\mathcal{F}}_{\sigma }\subseteq {\mathcal{F}}_{\tau }$. Apply the tower property in F4 to the two representations from step 3.1: $\mathbb{E}[M_{\tau }\mid {\mathcal{F}}_{\sigma }]=\mathbb{E}[\mathbb{E}[M_{\mathrm{\infty }}\mid {\mathcal{F}}_{\tau }]\mid {\mathcal{F}}_{\sigma }]=\mathbb{E}[M_{\mathrm{\infty }}\mid {\mathcal{F}}_{\sigma }]=M_{\sigma }\mathrm{.}$ Taking expectations finishes. AC is the background hypothesis recorded in F5; the conditional-expectation identities come from F1 and F4.

For every positive integer $m$, F2 gives $\mathbb{P}(\tau =\mathrm{\infty })\le \mathbb{P}(\tau \ge m)\le \mathbb{E}\tau \mathrm{/}m$. Since $\mathbb{E}\tau <\mathrm{\infty }$, letting $m\to \mathrm{\infty }$ shows $\tau <\mathrm{\infty }$ almost surely. On that event, $\mathrm{\mid }{M}_{\tau \wedge n}-M_{\tau }\mathrm{\mid }\le C(\tau -n{)}^{+}\le C\tau ,$ because the difference telescopes over at most $(\tau -n{)}^{+}$ increments. It tends pointwise to zero and has the integrable dominator $C\tau$.

F3 gives $M_{\tau \wedge n}\to M_{\tau }$ in $L^1$; in particular $M_{\tau }$ is integrable. F1 gives $\mathbb{E}{M}_{\tau \wedge n}=\mathbb{E}{M}_0$ for every $n$, so taking the $L^1$ limit proves the equality. Bounded increments and integrability of $\tau$ are used exactly in step 1.1. AC has only the role in F4.

Since $\tau <\mathrm{\infty }$ almost surely, $M_{\tau \wedge n}$ eventually equals $M_{\tau }$ almost surely. On the null event $\{\tau =\mathrm{\infty }\}$ the stipulated cemetery value makes the stopped variable total; all limit assertions are almost-sure assertions. The assumed bound passes to $\mathrm{\mid }{M}_{\tau }\mathrm{\mid }\le Y$, proving integrability, and also gives $\mathrm{\mid }{M}_{\tau \wedge n}-M_{\tau }\mathrm{\mid }\le 2Y\mathrm{.}$ F2 therefore yields $L^1$ convergence.

F1 gives $\mathbb{E}{M}_{\tau \wedge n}=\mathbb{E}{M}_0$ for every $n$. Pass to the $L^1$ limit from step 1.1 to get $\mathbb{E}{M}_{\tau }=\mathbb{E}{M}_0$. AC has exactly the inherited role in F3.

Pointwise, ${\sum }_{k=1}^{\tau }{X}_k={\sum }_{k\ge 1}{X}_k{1}_{\{\tau \ge k\}},\tau ={\sum }_{k\ge 1}{1}_{\{\tau \ge k\}}\mathrm{.}$ The first identity is understood through finite partial sums; the integrability calculation below shows that $\tau =\mathrm{\infty }$ is null.

By F1, F2, F3, $\mathbb{E}[\mathrm{\mid }{X}_k\mathrm{\mid }{1}_{\{\tau \ge k\}}]=\mathbb{E}\mathrm{\mid }{X}_1\mathrm{\mid }\mathbb{P}(\tau \ge k)\mathrm{.}$ MCT applied to the nonnegative partial sums and to the second identity in step 1.1 gives $\mathbb{E}{\sum }_{k\ge 1}\mathrm{\mid }{X}_k\mathrm{\mid }{1}_{\{\tau \ge k\}}=\mathbb{E}\mathrm{\mid }{X}_1\mathrm{\mid }{\sum }_{k\ge 1}\mathbb{P}(\tau \ge k)=\mathbb{E}\mathrm{\mid }{X}_1\mathrm{\mid }\mathbb{E}\tau <\mathrm{\infty }\mathrm{.}$ Thus the stopped series is absolutely integrable and $\tau <\mathrm{\infty }$ almost surely.

The signed partial sums are dominated by the integrable absolute series in step 2.1. DCT and finite linearity therefore give $\mathbb{E}{\sum }_{k=1}^{\tau }{X}_k={\sum }_{k\ge 1}\mathbb{E}[X_k{1}_{\{\tau \ge k\}}]={\sum }_{k\ge 1}\mu \mathbb{P}(\tau \ge k)=\mu \mathbb{E}\tau ,$ where F3 gives the middle factorization. This argument is choice-free: the given iid sequence and stopping time supply all indexed objects, and no conditional-expectation version is chosen.

Adaptedness and F1 make $\tau$ a stopping time. At the start of any block of $N$ fresh increments, conditional on not yet having exited, the event that all $N$ increments are $+1$ has probability $2^{-N}$ and forces an upper exit within that block. Independence of successive increments therefore gives inductively $\mathbb{P}(\tau >mN)\le (1-2^{-N}{)}^m⟶0.$ Thus $\tau <\mathrm{\infty }$ almost surely.

Since $\mathbb{E}{\xi }_k=0$ and ${\xi }_k$ is independent of the past, F2 gives $\mathbb{E}[S_k\mid {\mathcal{F}}_{k-1}]=S_{k-1}$. Before and at exit the nearest-neighbour path stays in $[0,N]$, so $\mathrm{\mid }{S}_{\tau \wedge n}\mathrm{\mid }\le N$. F3 with dominator $N$ yields $\mathbb{E}{S}_{\tau }=\mathbb{E}{S}_0=i$.

At the finite exit time, $S_{\tau }\in \{0,N\}$; the chosen value zero on the null event $\{\tau =\mathrm{\infty }\}$ preserves this assertion everywhere. Therefore $i=\mathbb{E}{S}_{\tau }=N\mathbb{P}(S_{\tau }=N),$ which gives the result. AC is used exactly through F4.

The predictable quadratic variation of the symmetric walk is $n$, since every increment has conditional square one. Thus F2 and bounded optional sampling give $\mathbb{E}[S_{\tau \wedge n}^2]-\mathbb{E}(\tau \wedge n)=i^2\mathrm{.}$

The stopped position lies in $[0,N]$ and converges almost surely to $S_{\tau }$ by F1. DCT gives $\mathbb{E}[S_{\tau \wedge n}^2]\to \mathbb{E}{S}_{\tau }^2=N^2\mathbb{P}(S_{\tau }=N)=Ni\mathrm{.}$ Meanwhile $\tau \wedge n\uparrow \tau$, so MCT gives $\mathbb{E}(\tau \wedge n)\uparrow \mathbb{E}\tau$, initially allowing infinity.

Taking limits in step 1.1 forces the latter value to be finite and gives $Ni-\mathbb{E}\tau =i^2,\mathbb{E}\tau =i(N-i)\mathrm{.}$ The argument does not assume integrability of $\tau$ before proving it. AC has exactly the role in F5.