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First work with the filtration ${\mathcal{G}}_t={\mathcal{F}}_t^W\vee \sigma (1_{\{u\le \tau \}}:0\le u\le t)$, which records the Brownian motion and whether default has occurred. On $\{s>\tau \}$ all subsequent defaultable stock prices vanish. On $\{s\le \tau \}$, the memoryless property of the exponential distribution gives survival from $s$ to $t$ with conditional probability $e^{-\lambda (t-s)}$. The residual survival event is independent of the future independent increments of the Brownian motion. Thus, for deterministic $s<t$,Also $\mathbb{E}{\hat{S}}_t=e^{\lambda t}{e}^{-\lambda t}{S}_0=S_0$, so this is a true martingale, with no localization needed. Its natural filtration is contained in $\mathcal{G}$; applying the tower property of conditional expectation to the displayed identity yields the requested natural-filtration martingale property:The printed convention keeps the stock alive at $t=\tau$. It is a left-continuous convention at that one random time. Replacing it by $1_{\{t<\tau \}}$ gives the usual càdlàg zero-recovery default model; since $\mathbb{P}(\tau =t)=0$ at each fixed $t$, the fixed-time martingale calculations above are unchanged.

A martingale with the specified terminal value must equal its conditional expectation. If ${\hat{S}}_t=0$, default has already occurred and the terminal European call option payoff is zero. If ${\hat{S}}_t>0$, put $h=T-t$. Conditional on survival to $T$,where $Y$ has the standard normal distribution; the conditional probability of this survival is $e^{-\lambda h}$. The conditional expectation of the payoff in the enlarged filtration used in part (c) is consequentlyThis expression is already measurable with respect to the natural filtration of $\hat{S}$, so the tower property of conditional expectation gives the same value there. The survival factor cancels the compensating growth factor:The separate zero branch avoids division by zero. At $t=T$ it gives the required payoff, using $F(0,m)=(1-m{)}^{+}$; integrability follows from $({\hat{S}}_T-K{)}^{+}\le {\hat{S}}_T$.

For fixed maturity $T$, define $b_t^T={\int }_t^T\sigma (t,u)du$ and $I_t^T={\int }_t^{T}f(t,u)du$. The stochastic Fubini theorem and the moving lower endpoint giveThe factor $1/2$ is the integral over one of the two triangles in the square $[t,T{]}^2$. Applying the Itô formula to $P(t,T)=e^{-I_t^T}$, its quadratic-variation correction cancels that factor:Set $D_t=e^{-{\int }_0^t{r}_{s}ds}$, the reciprocal of the continuous-time bank account. The Itô product rule then givesIf $|\sigma |\le M$, then $|b_t^T|\le MT$ on this finite horizon. The Novikov condition holds, so this stochastic exponential is a true martingale, not merely a local martingale. The discounted price is thereforeThe authoritative PDF discounts to $t$ in this part. The TeX transcription's upper endpoint $T$ would include future short rates and is incorrect here.

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.

Apply the Itô formula to ${\xi }_t=V(t,S_t)$. The backward equation cancels its drift, leavingBounded $a$ and $V_S$ make this Itô integral square-integrable on the finite horizon; in addition $V$ itself is bounded. Hence $\xi$ is a true martingale, not just a local martingale, and ${\xi }_T=g(S_T)$. Taking conditional expectations givesThe stock is understood to be adapted to the stated Brownian filtration, with its initial value fixed there. This is the Feynman-Kac formula in the zero-potential, zero-drift case.

The positive density process is the stochastic exponentialSince $a^{\prime }$ is bounded, the Novikov condition holds; $Z$ is a true martingale with $\mathbb{E}{Z}_T=1$ and $Z_T>0$. Thus it defines the stated equivalent probability measure. The positive sign in the exponent means that the Girsanov theorem givesUnder this change of measure, applying the Itô formula to ${\pi }_t=U(t,S_t)$ produces precisely the drift in its backward equation, which vanishes:Bounded $a$ and $U_S$ make $\pi$ a true martingale under $\hat{\mathbb{P}}$. Its terminal value is $g^{\prime }(S_T)$. ConsequentlyThe sign can also be checked before changing measure: under the original measure $d\pi =-a{a}^{\prime }{U}_{S}dt+a{U}_{S}dW$, while the quadratic covariation term in $d(Z\pi )$ is $Za{a}^{\prime }{U}_{S}dt$, exactly cancelling its drift.

Relative to a probability space $(\Omega ,\mathcal{F},\mathbb{P})$ and a filtration $({\mathcal{F}}_n)$, the real stochastic process $(M_n)$ is a martingale if it is adapted, each $M_n$ is an integrable random variable, andHere adapted means $M_n$ is ${\mathcal{F}}_n$-measurable, and integrability means $\mathbb{E}|M_n|<\infty$. By the tower property of conditional expectation, the equivalent multi-step condition is $\mathbb{E}[M_j\mid {\mathcal{F}}_i]=M_i$ for all $i\le j$. The filtration is part of the definition; specifying only the marginal means is insufficient.

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 $(M_k-M_0)$.

Use the natural filtration ${\mathcal{F}}_k=\sigma (X_1,\dots ,X_k)$ and define $M_k=S_k-mk$. The increments $X_k-m$ are centered, independent, and square integrable, so their conditional expectations given the past are zero. Thus $(M_k)$ is a square-integrable martingale with $M_0=0$. Independence givesApplying the Doob L2 maximal inequality yieldsThis includes ${\sigma }^2=0$, when every centered increment vanishes almost surely.

For the converse in the characterization of a martingale by stopped expectations, fix $n\ge 0$ and $A\in {\mathcal{F}}_n$. Define the bounded stopping timeIndeed, its only nontrivial sublevel event is $\{T_A\le n\}=A^c\in {\mathcal{F}}_n$. Applying the assumed stopped-expectation equality to $T_A$ and to the deterministic stopping time $n$ givesThis holds for every $A\in {\mathcal{F}}_n$. Since $M_n$ is adapted and both variables are integrable, the defining test-event property of conditional expectation yieldsTogether with the given adaptation and integrability, this is exactly the martingale property.

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.

The mean increment is $\mathbb{E}{X}_1=-6/7+2/7=-4/7$, so $M_n=S_n+(4/7)n$ is a martingale. Applying the bounded optional stopping theorem at $n\wedge T$ givesThe terminal sums are bounded by the exit-state bounds from (b), while $n\wedge T$ increases to the integrable stopping time $T$. The dominated convergence theorem on the left and the monotone convergence theorem on the right therefore proveThis is also Wald's equation here. In fact, the stopped identity already implies $\mathbb{E}(n\wedge T)\le 7m/4$ from $S_{n\wedge T}\ge -m$, giving another direct finite-expectation justification.

The Martingale convergence theorem in its $L^1$-bounded form says that a martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$ has an integrable limit $M_{\infty }$ andThe norm bound on the limit follows from Fatou lemma. Boundedness in $L^1$ by itself does not imply convergence in L1. The uniformly integrable martingale convergence theorem gives the stronger conclusion: if $(M_n)$ is uniformly integrable, then $M_n\to M_{\infty }$ both almost surely and in L1 norm, and $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$. Conversely, convergence in L1 implies uniform integrability.

For the distinction, the fair-coin doubling martingale $M_n=2^n{1}_{\{\text{first }n\text{ tosses are heads}\}}$ has expectation one for every $n$ but converges almost surely to zero. Its L1 norm remains one, so its convergence is not in L1 norm. These two formulations specify exactly which hypothesis is needed in part (d).

For $1<p<\infty$, the Lp martingale convergence theorem states that a martingale with ${\sup }_n\mathbb{E}|M_n{|}^p<\infty$ has a limit $M_{\infty }\in L^p$ such thatMoreover, $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$ and $\mathbb{E}|M_{\infty }{|}^p\le {\sup }_n\mathbb{E}|M_n{|}^p$. To see why $p>1$ matters, the Doob Lp maximal inequality gives an integrable dominating variable $({\sup }_n|M_n|{)}^p$. The Martingale convergence theorem first supplies the almost-sure limit, then the dominated convergence theorem supplies convergence in the Lp norm. This maximal estimate is unavailable at $p=1$ in the required form.

Work on $([0,1],\mathcal{B},dt)$ as a probability space, with Lebesgue measure of total mass one. Let ${\mathcal{F}}_n$ be the filtration generated by the level-$n$ half-open dyadic cells together with the separate null cell $\{1\}$. Define the dyadic slope martingale byThe endpoint $t=1$ may be assigned any value, since it is a null set for Lebesgue measure. The two child slopes average to their parent slope, by telescoping the two increments. Therefore $\mathbb{E}[G_{n+1}\mid {\mathcal{F}}_n]=G_n$ almost everywhere, so this is a martingale. The Lipschitz condition gives $|G_n|\le K$ everywhere except possibly at the freely chosen endpoint, where we take zero.

Apply the Lp martingale convergence theorem with $p=2$. Its limit $g$ has $|g|\le K$ almost everywhere and $G_n\to g$ in L1 norm as well, by Cauchy-Schwarz inequality. Choose a measurable representative of $g$ and set it to zero on any exceptional null set; it is then a bounded measurable function on the entire interval.

Set $f_n(x)=f(0)+{\int }_0^x{G}_n(t)dt$. Telescoping at the grid points shows that $f_n$ is the linear interpolation of $f$ on the dyadic grid. The Lipschitz condition gives $\Vert f_n-f{\Vert }_{\infty }\le K{2}^{-n}\to 0$. AlsoThe two uniform limits coincide, giving the absolutely continuous function representationThe chosen bound holds for every $t$ after the null-set modification; the integral identity holds for every $x$ simultaneously.

Let ${\tau }_r=\inf \{t\ge 0:|B_t|=r\}$, ${\tau }_R=\inf \{t\ge 0:|B_t|=R\}$, and $\tau ={\tau }_r\wedge {\tau }_R$. The function $u(y)=|y{|}^{-1}$ is harmonic away from the origin: its radial Laplacian in dimension three isThe Itô formula therefore makes $u(B_{t\wedge \tau })$ a local martingale, bounded between $1/R$ and $1/r$, hence a true martingale. The annulus exit time is finite almost surely. For instance, exit from the containing open ball is finite because from every point in the open ball a sufficiently large Gaussian vector increment has a uniform positive probability of leaving it in unit time; iterate the geometric tail bound from a uniform escape probability.

The optional stopping theorem at $t\wedge \tau$, followed by the dominated convergence theorem, gives $\mathbb{E}u(B_{\tau })=u(x)$. By continuity, there is no radial overshoot and the two exit spheres cannot be hit simultaneously. Writing $p={\mathbb{P}}_x({\tau }_r<{\tau }_R)$ givesAll harmonicity and boundedness statements apply only in the annulus, which stays away from the singularity at zero.