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For $t>0$, write $\Phi$ for the standard normal distribution function. Since the Brownian running maximum is nonnegative, $a\le 0$ gives simply $\mathbb{P}(B_t\le b)=\Phi (b/\sqrt{t})$.

For $a>0$ and $b<a$, reflect the Brownian motion after its first hit of $a$. This is a stopping time, and the Strong Markov property together with symmetry of Brownian increments shows that reflection preserves its law. On paths that hit $a$, it sends the endpoint $B_t$ to $2a-B_t$. Thus the event with endpoint at most $b$ maps to endpoints at least $2a-b$, all of which necessarily hit $a$. This provesThe same reflection gives $\mathbb{P}(M_t\ge a)=2(1-\Phi (a/\sqrt{t}))$. If $b\ge a$, every path with $B_t>b$ has hit $a$, so subtracting that endpoint tail gives the complete answer:The expressions agree at $b=a$. This is the joint distribution of Brownian motion and its running maximum. In particular $M_t$ has the distribution of $|B_t|$ and has no atoms for $t>0$. If $t=0$, the pair is $(0,0)$ deterministically, so the requested probability is $1_{\{a\le 0,b\ge 0\}}$.

Fix $0<t<1$. The event $\{\tau \le t\}$ means that the future path through time one never exceeds the maximum already attained by time $t$. Given ${\mathcal{F}}_t$, future increments form an independent Brownian motion $W$, soHere the Brownian running maximum distribution is the one derived above. On $\{B_t<0\}$, an event of probability $1/2$, the finite quantity $M_t-B_t$ is strictly positive, so this conditional probability lies strictly between zero and one.

If $\tau$ were a stopping time, $\{\tau \le t\}$ would be ${\mathcal{F}}_t$-measurable, and the displayed conditional expectation of its indicator would equal that indicator, taking only zero and one. The contradiction proves $\tau$ is not an $({\mathcal{F}}_t)$-stopping time. Thus the Strong Markov property cannot be invoked at $\tau$ to assert a Brownian shifted process; the direct argument here does not assume that property at this random time.

Let $V_t=⟨M{⟩}_t$, and first suppose $V_{\infty }=\infty$ almost surely in addition to strict increase. The Dambis-Dubins-Schwarz theorem says thatdefine a Brownian motion $B$ in the time-changed filtration and giveEach $T_u$ is a stopping time and is finite. Continuity and strict increase of $V$ make $u\mapsto T_u$ continuous, $V_{T_u}=u$ and $T_{V_t}=t$.

Here are the martingale details behind this inverse-clock proof of the Dambis-Dubins-Schwarz theorem. Stopping a continuous local martingale when its bracket reaches $n$ makes it an L2-bounded continuous martingale. This follows from the stopped Itô isometry or from the $p=2$ estimate proved in Question 1(a), applied after localization. In particular it is uniformly integrable, and optional sampling is valid even at an unbounded stopping time by taking limits. Applying this to $M$ stopped at $T_n$ shows that $(M_{T_{u\wedge n}}{)}_{u\ge 0}$ is a martingale. Thus $B$ is a continuous local martingale. Time-changing $M^2-V$ in the same way shows $B_u^2-u$ is a local martingale, so $⟨B{⟩}_u=u$.

For completeness, the Lévy characterization of Brownian motion follows directly from the Itô formula. If a continuous local martingale $N$, starting at zero, has bracket $u$, thenis a complex local martingale. Its modulus is bounded on each deterministic finite horizon, so it is a true martingale there. ConsequentlyConditional characteristic functions give Gaussian increments independent of the past. Iterating this identity gives independent increments, and continuity completes the Brownian characterization. The identical vector argument proves the Lévy characterization of multidimensional Brownian motion when the bracket matrix is $uI$.

The printed strict-increase hypothesis does not imply $V_{\infty }=\infty$. For example, $M_t={\int }_0^t{e}^{-s}d{W}_s$ has strictly increasing bracket $(1-e^{-2t})/2$. To state the theorem under exactly the printed hypothesis, allow an independent enlargement of the probability space if the terminal clock $L=V_{\infty }$ can be finite.

On $\{L<\infty \}$ the martingale $M$ has a finite terminal limit. Indeed, stopping at each bracket level $n$ gives an L2-bounded continuous martingale which converges; on $\{V_{\infty }<n\}$ the stopped process is the original one. This proves the finite-bracket convergence lemma. Set $T_u=\infty$ when $u\ge L$ and continue $N_u=M_{T_u}$ by that terminal limit. The optional-sampling argument just given makes $N$ a continuous local martingale with bracket $u\wedge L$. Moreover $L$ is a stopping time in ${\mathcal{G}}_u={\mathcal{F}}_{T_u}$.

On a product extension add an independent Brownian motion $\beta$ in clock time, and putThe two summands have zero quadratic covariation, and their brackets are $u\wedge L$ and $(u-L{)}^{+}$. Thus $⟨\tilde{B}{⟩}_u=u$; the proved characterization makes $\tilde{B}$ Brownian. Since $V_t<L$ at every finite $t$ when $L$ is finite, $M_t={\tilde{B}}_{V_t}$ still holds. This is the finite-lifetime extension of the Dambis-Dubins-Schwarz theorem. An infinite clock gives Brownian motion on the original space; a finite clock may require the independent extension.

Since the density martingale is uniformly integrable, $Z_t={\mathbb{E}}_{\mathbb{P}}[Z_{\infty }\mid {\mathcal{F}}_t]$. Equivalence of measures, as stipulated, means $Z_{\infty }>0$ almost surely. Write $L=\int Z^{-1}dZ$ for the stochastic logarithm. The Itô formula gives $\log Z=L-⟨L⟩/2$, and hence the quadratic covariation in the drift correction isIt involves the local-martingale part of $\log Z$; its continuous finite-variation part has zero covariation.

With $Y=X-C$, the Itô product rule now gives an exact cancellation:Thus $ZY$ is a P-local martingale. To transfer this conclusion rigorously, set ${\tau }_n=\inf \{t:|Y_t|\ge n\}$. The same computation for the stopped $Y$ givesThis is again a P-local martingale. Moreover its absolute value is at most $n{Z}_t$. Uniform integrability of $Z$ makes the family over bounded stopping times uniformly integrable, so the product is a true martingale. This is the bounded-process density-product criterion.

The Bayes formula for conditional expectation therefore gives, for $s\le t$,Continuity gives ${\tau }_n\uparrow \infty$, proving $X-⟨\log Z,X⟩$ is a Q-local martingale. This proves the needed Girsanov theorem rather than invoking it.

For the Brownian-filtration conclusion, use this precise Brownian martingale representation theorem: every continuous local martingale in the usual augmentation of the natural Brownian filtration is an Itô integral with a predictable integrand locally square integrable in time. In particularDefineOn each finite horizon a strictly positive continuous $Z$ has a positive pathwise minimum. Thus ${\int }_0^t{\alpha }_s^{2}ds<\infty$ almost surely, andThe already proved measure-change result makes $\hat{W}=W-\int \alpha ds$ a continuous Q-local martingale. Its quadratic variation is $t$, unchanged by its finite-variation correction. The Lévy characterization of Brownian motion provesNo additional exponential-integrability condition is needed, since the equivalent uniformly integrable density process is already given.

Choose a localizing sequence of discrete-time stopping times ${\tau }_j\uparrow \infty$ for $X$. For fixed integer $t$, every stopped value is bounded in absolute value by the finite sumThat sum is integrable under the hypothesis. Since $X^{{\tau }_j}$ is a martingale,The dominated convergence theorem, including its conditional version, now removes the stopping. Thus $\mathbb{E}[X_t\mid {\mathcal{F}}_{t-1}]=X_{t-1}$. The process is adapted and integrable by hypothesis, so $X$ is a true discrete-time martingale. This is the integrable discrete-time local martingale is a martingale criterion. The finite sum dominating stopped values is the crucial discrete-time feature.

Stop just before a large predictable coefficient would be used. Setwith the infimum of the empty set equal to infinity. Because $K_{t+1}$ is ${\mathcal{F}}_t$-measurable, ${\sigma }_j$ is a stopping time. The increment of the stopped martingale transform isIts coefficient is predictable and bounded by $j$: the first coefficient exceeding $j$ occurs at the step after stopping, and is never included. Part (c) makes $Y^{{\sigma }_j}$ a martingale. Since the finitely many $K_s$ on any fixed finite horizon are finite almost surely, ${\sigma }_j\uparrow \infty$ almost surely. ThusThis is predictable-coefficient localization of a martingale transform. Stopping after taking the large increment would not give the required bound.

Recall the stopping-time sigma-algebra:For the proposed random time,The first set belongs to ${\mathcal{F}}_t$. Since $S\le T$, the second can be writtenwhich also belongs to ${\mathcal{F}}_t$. Therefore $U$ is a stopping time, as in pasting ordered stopping times. Moreover $S\le U\le T$, so $U$ is a bounded stopping time.

Fix $t>0$ and partition $[0,t]$ into $m$ equal intervals, with $r_k=kt/m$. Round $t\wedge T$ upward within this interval. The corresponding sampled value isEvery time in this sum is at most $t$. The stopping time property makes each indicator function ${\mathcal{F}}_t$-measurable, while the adapted process property makes each sampled value ${\mathcal{F}}_t$-measurable. Consequently $Y_m$ is ${\mathcal{F}}_t$-measurable.

The rounded times approach $t\wedge T$ from the right, so right continuity gives $Y_m\to X_{t\wedge T}$ for the chosen pathwise càdlàg version. If path regularity is instead stated only almost surely, the usual complete filtration handles the exceptional null set; on an incomplete filtration one should formulate that case as existence of an adapted version. At $t=0$, the value is simply $X_0$. Thus $X^T$ is an adapted process, by adaptedness of a stopped right-continuous process. The argument actually needs neither the martingale property nor boundedness of $T$.

Fix deterministic $0\le s<t$ and $A\in {\mathcal{F}}_s$. The pasting ordered stopping times argument shows that $U=s{1}_A+t{1}_{A^c}$ is a bounded stopping time. The given stopped-expectation property, applied to $U$ and to the deterministic stopping time $t$, yieldsThis holds for every $A\in {\mathcal{F}}_s$. Since $X_s$ is ${\mathcal{F}}_s$-measurable and both time values are integrable, it is exactly the defining test for conditional expectation:Hence $X$ is a martingale. This proof of the characterization of a martingale by bounded continuous-time stopped expectations uses only deterministic two-time pastings; the càdlàg assumption and the usual conditions for a filtration are stronger than needed for this implication.

For $A>0$, the first time $T_A=\inf \{n\ge 0:X_n<-A\}$ is a stopping time. On $\{T_A\ge 1\}$, the bounded increments assumption controls the overshoot:Before that time, $X_n\ge -A$. Allowing the possibility $T_A=0$, the processis a nonnegative martingale. Indeed, the stopped martingale in discrete time is a martingale, and the integrable ${\mathcal{F}}_0$-measurable variable $X_0^{-}$ is a constant-in-time martingale. If $T_A=0$, nonnegativity follows from $X_0+X_0^{-}=X_0^{+}$. Its expectations are constant and finite, so the Martingale convergence theorem makes $Y_n$ converge to a finite value almost surely.

On $\{T_A=\infty \}$ this proves finite convergence of $X_n$. Taking the countable union over positive integer $A$, we conclude that $X_n$ converges finitely on the event $\{{\inf }_n{X}_n>-\infty \}$. Apply the same reasoning to the martingale $-X$ to obtain finite convergence on $\{{\sup }_n{X}_n<\infty \}$.

Outside the event of finite convergence, both the infimum and the supremum of the sequence must therefore be infinite in the respective directions. Removing any finite initial segment cannot change this, because each of its values is finite. HenceThis is the bounded-increment martingale convergence-or-oscillation dichotomy. The $X_0^{-}$ term ensures the proof also covers an unbounded integrable initial value.

Use $c_{\mathrm{prem}}$ for the premium income rate, reserving $c$ later for the smaller exponential decay rate. In the classical risk model, the surplus isHere $\rho$ is the relative safety loading. The aggregate claims form a Compound Poisson process. Define $L_t={\sum }_{j=1}^{N_t}{X}_j-c_{\mathrm{prem}}t$, so ruin occurs when $L_t>u$. By independent increments and the exponential formula for a marked Poisson sum,Thus $Z_t=e^{R{L}_t}$ is a nonnegative continuous-time martingale with $Z_0=1$, because the adjustment coefficient makes the exponent vanish.

Let $\tau =\inf \{t:U_t<0\}$. Apply the optional stopping theorem at the bounded stopping time $\tau \wedge T$. On $\{\tau \le T\}$, $L_{\tau }>u$, soLetting $T$ increase proves the Lundberg inequality and the unscaled limit:

For the precise asymptotic, putThe given exponential integral identity makes $k$ a probability density. Multiplying the given defective renewal equation by $e^{Ru}$ turns it into the ordinary renewal equationFor clarity, the version of the key renewal theorem used here is: if the interarrival law is nonarithmetic, has mean $m_k\in (0,\infty )$, and $g$ is directly Riemann integrable, the locally bounded solution of this renewal equation satisfies $z(u)\to m_k^{-1}{\int }_0^{\infty }g(v)dv$. The infinite-mean version gives zero for nonnegative directly Riemann integrable $g$.

All the hypotheses can be checked here. The density $k$ gives a nonarithmetic distribution. The Tonelli theorem givesFurthermoreThus $g$ is continuous and integrable, and ${\int }_0^{\infty }|g^{\prime }(u)|du\le R\int g+1<\infty$. On a mesh of width $\delta$, the difference between its upper and lower sums is at most $\delta \int |g^{\prime }|$; its upper sum is at most $\int g+\delta \int |g^{\prime }|$. This proves direct Riemann integrability rather than assuming it. Also $0\le z(u)\le 1$ by the Lundberg inequality, so the solution is locally bounded. Its renewal representation is $z=g\ast U_k$, where $U_k={\sum }_{j\ge 0}{K}^{\ast j}$ and $K(dx)=k(x)dx$; the residual after iteration tends to zero on compact intervals because sums of positive interarrivals tend to infinity.

Writing $J={\int }_0^{\infty }x{e}^{Rx}{f}_I(x)dx$, the tilted interarrival expected value is $m_k=hJ$. The key renewal theorem gives the Cramér–Lundberg ruin asymptoticIf $J=\infty$, the same formula is interpreted as $A=0$. A positive finite asymptotic constant requires $J<\infty$; this extra integrability is not explicitly stated in the paper.

For the final two-exponential case, evaluate the defective renewal equation at zero:One can identify the adjustment coefficient without silently assuming $A>0$. For $0<r<c$, setIt is finite and positive. Integrating the nonnegative terms of the defective renewal equation, using the Tonelli theorem, first shows that $F_I(r)={\int }_0^{\infty }{e}^{rx}{f}_I(x)dx$ is finite and then givesAs $r\uparrow c$, $P(r)\to \infty$ because $a>0$. By monotone convergence theorem, $h{F}_I(c)=1$. The integrated tail distribution in the classical risk model has density $f_I(x)=\mathbb{P}(X_1>x)/\mu$, so $F_I(r)=[M(r)-1]/(\mu r)$ by the tail integral formula for moments. Hence $c$ solves the adjustment equation, and its stipulated uniqueness implies $R=c$. Finally the displayed form of $\psi$ gives the remaining constantsIn particular the decay exponent $d$ and the coefficient $b$ do not affect $R$ or $A$. The $c$ in these final answers is the printed decay rate, not $c_{\mathrm{prem}}$.

The time $T_R$ is a stopping time, because $\{T_R\le n\}={\bigcup }_{k\le n}\{M_k\ge R\}\in {\mathcal{F}}_n$. Apply the optional stopping theorem to the bounded time $T_R\wedge n$:On $\{T_R\le n\}$ the stopped value is at least $R$, and elsewhere it is nonnegative. Letting $n\to \infty$ gives the maximal bound for a nonnegative martingale:Stop at a bounded time first, then pass to the increasing event. This avoids assuming uniform integrability of the original process. For $0<R\le 1$, $T_R=0$ and the bound simply says $1\le 1/R$.

A continuous martingale is an adapted process with almost surely continuous paths, $\mathbb{E}|M_t|<\infty$ for every $t$, andA continuous local martingale is continuous and adapted and has a localizing sequence of stopping times $T_n\uparrow \infty$ almost surely such that every $M_t^{T_n}=M_{t\wedge T_n}$ is a true martingale. The initial value is normally integrable; the frequently used class ${\mathcal{M}}_{c,\mathrm{loc}}$ is normalized to start at zero.

The usual conditions for a filtration are completeness and right continuity. Completeness means that ${\mathcal{F}}_0$ contains every subset of each $\mathbb{P}$-null set in the ambient $\mathcal{F}$; right continuity means ${\mathcal{F}}_t={\bigcap }_{u>t}{\mathcal{F}}_u$. These conditions concern the filtered probability space, whereas continuity of a martingale concerns its sample paths.

The vertices visited by a simple symmetric random walk form an integer interval: nearest-neighbour steps cannot skip a vertex. At the stopping time $T_k$, write that interval as $[\ell ,r]$, with $r-\ell +1=k$. The walk is at one of its endpoints, because the $k$th new vertex must extend the earlier interval. This also covers $k=1$, when both endpoints coincide.

The Strong Markov property restarts the walk from this endpoint. Visiting a new vertex is exactly exiting $[\ell ,r]$, or reaching $\ell -1$ or $r+1$. Translate these absorbing boundaries to $0,k+1$. The starting point becomes either $1$ or $k$, so the expected duration of symmetric gambler's ruin is $k$ in both cases. Taking conditional expectations and then expected values gives the expected time to expand a random-walk range:

For the forward implication, let $(M_n)$ be a martingale and let the stopping time $T$ satisfy $0\le T\le N$ for a deterministic integer $N$. The stopped variable is integrable because it uses only finitely many integrable values. The pathwise identityexpresses it through martingale increments. Since $\{T\ge j\}=\{T>j-1\}\in {\mathcal{F}}_{j-1}$, each summand has expectation zero by conditional expectation. ConsequentlyThis proves the bounded-time case of the optional stopping theorem directly; no assumptions about an unbounded stopping time are being used.

Use the usual completed right-continuous Brownian filtration. The strict passage process $(T_a{)}_{a\ge 0}$ is finite at every level on one probability-one event: finiteness at all integer levels from (a) suffices by monotonicity. Also $T_0=0$ almost surely by (b).

For each fixed $a$, $T_a$ is a stopping time with $B_{T_a}=a$. The Strong Markov property shows that the passage times above this level depend on a new independent standard Brownian motion. Thus for $0\le a<b$, $T_b-T_a$ is independent of the past at $T_a$ and has the same law as $T_{b-a}$. Iteration gives independent increments and stationary increments in the level parameter. From (a) and fixed-level equality,For $\lambda ,\eta >0$, this also proves continuity in probability at zero, sinceFinally, with $M_t={\sup }_{s\le t}{B}_s$, we have $T_a=\inf \{t:M_t>a\}$. This strict generalized inverse of a nondecreasing function is right-continuous: if $a_n\downarrow a$, then for any $t>T_a$ we have $M_t>a$, and eventually $M_t>a_n$, which forces $T_{a_n}\le t$. Monotonicity gives the reverse bound. Monotonicity and local finiteness also give finite left limits. Hence $T$ is the Brownian first-passage subordinator, with càdlàg paths.

Now condition on this clock, which is independent of $W$. For a deterministic partition $0=a_0<a_1<\cdots <a_k$, write $\Delta T_j=T_{a_j}-T_{a_{j-1}}$ and $\Delta X_j=W_{T_{a_j}}-W_{T_{a_{j-1}}}$. Conditional on the clock these are independent centered Gaussian increments, with respective variances $\Delta T_j$. ConsequentlyFactorization and dependence only on interval lengths prove independent increments and stationary increments for $X$. For small $h$, $T_h\to 0$ in probability, and independence and continuity of $W$ imply $W_{T_h}\to 0$ in probability. For example, bound its deviation probability by $\mathbb{P}(T_h>\eta )+\mathbb{P}({\sup }_{t\le \eta }|W_t|>\epsilon )$ and then let $h\downarrow 0$ and $\eta \downarrow 0$. Stationary increments give stochastic continuity at every deterministic level. Composition of the continuous path of $W$ with the nondecreasing càdlàg clock gives càdlàg paths for $X$, and $X_0=0$. ThusThis is subordination of a Lévy process. Using strict passage times ensures the required right-continuous path choice, despite their fixed-level equality with the non-strict times.

The Exponential martingale for Brownian motion is a true martingale: independent increments and the normal distribution identity $\mathbb{E}{e}^{aN}=e^{a^{2}v/2}$ for $N\sim N(0,v)$ give $\mathbb{E}[Z_t\mid {\mathcal{F}}_s]=Z_s$. Its second moment isThe uniform integrability from bounded second moments criterion now applies: for $K>0$,Thus $\{Z_t:0\le t\le T\}$ has uniform integrability. In fact, all values stopped at stopping times bounded by $T$ have uniform integrability, since $Z_{\tau }=\mathbb{E}[Z_T\mid {\mathcal{F}}_{\tau }]$ and uniform integrability of conditional expectations applies. This is a finite-horizon assertion, not an assertion of uniform integrability over all $t\ge 0$.