Stopping Times and Optional Stopping — Examples

The exit target $(-\mathrm{\infty },a]\cup [b,\mathrm{\infty })$ is Borel, so F1 applies. Explicitly, $\{\tau \le n\}={\bigcup }_{k=0}^n(\{X_k\le a\}\cup \{X_k\ge b\})\in {\mathcal{F}}_n\mathrm{.}$

If a path remains in $(a,b)$ forever, the defining set of indices is empty and $\tau =\mathrm{\infty }$. Thus the result asserts neither almost-sure finiteness nor integrability and leaves the cemetery convention relevant.

Put $r=q\mathrm{/}p$, so $r>0$ and $r\mathrm{\ne }1$. Independence of the next increment and $pr+q\mathrm{/}r=q+p=1$ give $\mathbb{E}[r^{S_{n+1}}\mid {\mathcal{F}}_n]=r^{S_n}(pr+q\mathrm{/}r)=r^{S_n}\mathrm{.}$ Thus $r^{S_n}$ is a martingale.

The same block argument as for symmetric ruin works because an all-up block of $N$ increments has positive probability $p^N$: conditional on survival, it forces exit. Hence $\mathbb{P}(\tau >mN)\le (1-p^N{)}^m\to 0$. F1 gives stopping and this bound gives almost-sure finiteness.

Before exit, $S_{\tau \wedge n}\in [0,N]$, so $r^{S_{\tau \wedge n}}$ is bounded by $\max (1,r^N)$. F3 gives $r^i=\mathbb{E}{r}^{S_{\tau }}=1\cdot \mathbb{P}(S_{\tau }=0)+r^N\mathbb{P}(S_{\tau }=N)\mathrm{.}$ Writing the first probability as one minus the second and solving yields $(r^i-1)\mathrm{/}(r^N-1)$, equal to the displayed formula. AC has exactly the role in F4.

Apply F1 to obtain $\mathbb{E}\tau =i(N-i)$.

If $N$ is even and $i=N\mathrm{/}2$, direct substitution gives $\mathbb{E}\tau =(N\mathrm{/}2)(N-N\mathrm{/}2)=N^2\mathrm{/}4.$ This calculation illustrates that almost-sure exit can have a quadratic mean duration. AC has exactly the inherited role in F2.

For every $n$, the event $\{\tau \le n\}$ is determined by $X_1,\dots ,X_n$, so $\tau$ is a stopping time for the stated filtration; it is bounded by $N$. The event $\{\tau \ge k\}$ for $1\le k\le N$ says the first $k-1$ trials failed, so it has probability $(1-p{)}^{k-1}$. The tail sum therefore gives $\mathbb{E}\tau ={\sum }_{k=1}^N(1-p{)}^{k-1}=\frac{1-(1-p{)}^N}{p},$ including $p=1$, when the geometric sum is $1$.

The stopped sum is exactly the indicator that at least one of the first $N$ trials succeeds: after the first success the sum stops, while if all fail it is zero. Its expectation is $1-(1-p{)}^N$. Since $\mathbb{E}{X}_1=p$, F1 also gives it as $p\mathbb{E}\tau$, agreeing with step 1.1. The argument is choice-free.

F1 and conditional positivity make every $L_n$ nonnegative; F2 makes the process a martingale. F4 makes $L_{\tau }$ ${\mathcal{F}}_{\tau }$-measurable. Applying F3 between $\tau$ and deterministic $N$ gives $L_{\tau }={\mathbb{E}}_P[L_N\mid {\mathcal{F}}_{\tau }],{\mathbb{E}}_P{L}_{\tau }={\mathbb{E}}_P{L}_N=Q(\mathrm{\Omega })=1.$

For $A\in {\mathcal{F}}_{\tau }$, the defining conditional-expectation identity in step 1.1 gives ${\mathbb{E}}_P[1_A{L}_{\tau }]={\mathbb{E}}_P[1_A{L}_N]=Q(A),$ where the final equality is the Radon–Nikodym identity and $A\in {\mathcal{F}}_{\tau }\subseteq {\mathcal{F}}_N$. AC has precisely the role in F5.

The last success occurs no later than time $1$ exactly when the second toss fails. Hence $\{\rho \le 1\}=\{X_2=0\}\mathrm{.}$ This event has probability $1\mathrm{/}2$.

If it belonged to ${\mathcal{F}}_1$, F2 would make it independent of itself, because it is also an event determined by $X_2$. That would give $1\mathrm{/}2=P(A)=P(A{)}^2=1\mathrm{/}4$, impossible. Thus $\{\rho \le 1\}\mathrm{\notin }{\mathcal{F}}_1$, violating F1. The counterexample is finite and choice-free.

For $a\ge 1$, let $E_a$ be the event that the walk hits $1$ before $-a$. Translating the symmetric ruin interval to $\{0,a+1\}$ with starting point $a$, F1 gives $P(E_a)=\frac{a}{a+1}\mathrm{.}$ The $E_a$ increase, and their union is $\{\tau <\mathrm{\infty }\}$: a path that reaches $1$ has a finite minimum before that time and therefore belongs to some $E_a$. Continuity from below gives $P(\tau <\mathrm{\infty })=1$.

By definition, $S_{\tau }=1$ on this probability-one event, whereas $S_0=0$. Thus their expectations differ. This explicitly shows that almost-sure finiteness alone cannot justify passing from $\tau \wedge n$ to $\tau$ in bounded optional sampling, as F2 warns. AC has exactly the inherited role in F3.

F1 proves that $\tau <\mathrm{\infty }$ almost surely. Suppose for contradiction that $\mathbb{E}\tau <\mathrm{\infty }$.

The walk is a martingale and $\mathrm{\mid }{S}_n-S_{n-1}\mathrm{\mid }=1$, so F2 would imply $\mathbb{E}{S}_{\tau }=\mathbb{E}{S}_0\mathrm{.}$ But F1 computes the two sides as $1$ and $0$. This contradiction proves $\mathbb{E}\tau =\mathrm{\infty }$. AC has exactly the inherited role in F3.

On $([0,1],\mathcal{B},\lambda )$ put $A_n=(0,2^{-n}]$, ${\mathcal{F}}_0$ trivial, ${\mathcal{F}}_n=\sigma (A_1,\dots ,A_n)$, $M_0=1$, and $M_n=2^n{1}_{A_n}$ for $n\ge 1$. On the atom $A_n$, $M_{n+1}$ equals $2^{n+1}$ on a half-measure subatom and zero on the other half, so its conditional average is $2^n$; off $A_n$ both variables vanish. Thus F1 proves directly that $M$ is a nonnegative martingale with $\mathbb{E}{M}_n=1$.

Let $\tau =\inf \{n\ge 1:M_n=0\}$. For $n\ge 1$, $\{\tau >n\}=A_n\in {\mathcal{F}}_n$, so F2 makes $\tau$ a stopping time. Its tail sum is $\mathbb{E}\tau ={\sum }_{n\ge 0}P(\tau >n)=1+{\sum }_{n\ge 1}{2}^{-n}=2.$

The intersection of the $A_n$ is empty, so every path eventually leaves and $M_{\tau }=0$. Hence $M_{\tau }$ is integrable but $\mathbb{E}{M}_{\tau }=0\mathrm{\ne }1=\mathbb{E}{M}_0\mathrm{.}$ On $A_n\setminus A_{n+1}$ the next increment has magnitude $2^n$, so no deterministic increment bound exists; this is exactly the missing hypothesis flagged by F4. The proof reconstructs its martingale locally and does not depend on a B-page supplier. AC has only the role in F5.