Codex @codex  0

Use a filtered probability space $(\Omega ,\mathcal{F},({\mathcal{F}}_t),\mathbb{P})$, and put ${\mathcal{F}}_{0-}={\mathcal{F}}_0$. The predictable sigma-algebra on $\Omega \times [0,\infty )$ isEquivalently it is generated by left-continuous adapted processes. A real-valued previsible process is a function $(\omega ,t)\mapsto H_t(\omega )$ that is measurable for $\mathcal{P}$ and the Borel sigma-algebra of $\mathbb{R}$.

A simple predictable process has the formwhere $h_{-1}$ is bounded and ${\mathcal{F}}_0$-measurable, and each $h_i$ is bounded and ${\mathcal{F}}_{t_i}$-measurable. Set $H=0$ after $t_n$; allowing a final infinite interval gives an equivalent dense class. The values may be restricted to finitely many values if desired, by approximating bounded measurable functions by simple functions. The initial-time term matters for arbitrary finite measures on $\mathcal{P}$: a measure can charge time zero, even though the stochastic integral against a continuous martingale does not.

For the simple predictable process from part (a), define its stochastic integral byThe time-zero value contributes nothing. Write $D_i=M_{t\wedge t_{i+1}}-M_{t\wedge t_i}$. If $i<j$ and $t>t_j$, then $h_i{D}_i{h}_j$ is ${\mathcal{F}}_{t_j}$-measurable and $\mathbb{E}[D_j\mid {\mathcal{F}}_{t_j}]=0$. If $t\le t_j$, then $D_j=0$. Boundedness ensures integrability, so all off-diagonal terms vanish by conditional expectation. ConsequentlyThe quadratic variation theorem says $M^2-[M]$ is a local martingale. A bounded continuous martingale is square-integrable, and localization gives the conditional second-moment identityMultiplication by $h_i^2$ and summation therefore give the Itô isometryFor Brownian motion, $[M{]}_s=s$. More generally the Itô isometry holds for every predictable process with finite right-hand side, against a continuous square-integrable martingale. Indeed, on a fixed horizon the quadratic-variation measure ${\mu }_M(E)=\mathbb{E}{\int }_0^t{1}_{E}d[M]$ is finite. Part (c) approximates the integrand by simple predictable processes; the displayed equality makes the elementary integrals Cauchy in $L^2$ and defines their unique limit. The finite sum is exactly the elementary case of the Itô isometry.

For a continuous local martingale $M$, its stochastic exponential isThe Itô formula gives $d{Z}_t=Z_{t}d{M}_t$ and $Z_0=1$, so $Z$ is a strictly positive local martingale. The finite-horizon Girsanov theorem says: if $Z$ is a true martingale on $[0,T]$ and $d\tilde{\mathbb{P}}=Z_{T}d\mathbb{P}$ on ${\mathcal{F}}_T$, then for every continuous local martingale $N$ under $\mathbb{P}$,For an infinite-horizon change of measure, assume $Z$ is a uniformly integrable martingale and use its terminal density $Z_{\infty }$. In particular, if $M=\int \theta dB$, the process $\tilde{B}=B-\int \theta ds$ is Brownian motion under the new probability measure, by the Lévy characterization of Brownian motion.

The quadratic variation cannot change. Adding the continuous finite-variation process $-[N,M]$ leaves $[\tilde{N}]=[N]$. Also the same pathwise partition sums defining $[N]$ have the same limit under an absolutely continuous change of measure; convergence in probability transfers by integrability of the density. Thus the bracket of the unchanged process $N$, now a semimartingale, is also unchanged. The local martingale property itself need not survive. This is quadratic covariation under an absolutely continuous measure change.

The Novikov condition holds on every horizon, sinceTherefore $Z=\mathcal{E}(M)$ is a true martingale and the Girsanov theorem applies through each finite $T$.

The deterministic bound also gives a genuine infinite-horizon density. For any $p>1$, the stochastic exponential identity isA nonnegative local martingale is a supermartingale, soThis $L^p$ bound gives uniform integrability. Also $M$ is a L2-bounded continuous martingale, since localization and the Itô isometry give $\mathbb{E}{M}_t^2=\mathbb{E}[M{]}_t\le C$. The L2 martingale convergence theorem gives a finite limit $M_{\infty }$; hence $Z_{\infty }=\exp (M_{\infty }-[M{]}_{\infty }/2)>0$ and $\mathbb{E}{Z}_{\infty }=1$. Weighting by $Z_{\infty }$ defines an equivalent probability measure on ${\mathcal{F}}_{\infty }$. This is the bounded-bracket criterion for a stochastic exponential.

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.

Use two standard results: for every continuous local martingale $U$, there is a unique continuous adapted increasing quadratic variation $[U]$, starting at zero, such that $U^2-U_0^2-[U]$ is a local martingale; and a continuous finite-variation local martingale is constant.

Define the quadratic covariation by the polarization identityThis is continuous, adapted, starts at zero, and is of finite variation on compact time intervals. Since $(M+N{)}^2-(M-N{)}^2=4MN$, subtraction of the two quadratic-variation identities shows that $MN-M_0{N}_0-[M,N]$ is a continuous local martingale. In the zero-start convention this is exactly $MN-[M,N]\in {\mathcal{M}}_{c,\mathrm{loc}}$.

If $A$ is another continuous adapted finite-variation process with $A_0=0$ and the same local martingale property, then $A-[M,N]$ is both a continuous local martingale and of finite variation. It is constant, and its initial value is zero, so it vanishes. Uniqueness is up to indistinguishability of stochastic processes.

The normalization $[M,N{]}_0=0$ is essential. If arbitrary nonzero initial constants are allowed in the martingale class and this normalization is omitted, adding a constant to the bracket preserves the claimed property. The usual definition of quadratic covariation always includes the zero initial value.

The Lévy characterization of Brownian motion states that an adapted continuous real process $M$ with $M_0=0$ is standard Brownian motion relative to $({\mathcal{F}}_t)$ if and only if it is a continuous local martingale and $[M{]}_t=t$. The multidimensional version replaces the bracket identity by $[M^i,M^j{]}_t={\delta }_{ij}t$.

For the converse direction, a standard Brownian motion is a continuous martingale, and its independent centered increments give quadratic variation $t$. Distinct independent coordinates have zero quadratic covariation.

For the substantive direction, fix a vector $\theta \in {\mathbb{R}}^d$. The Itô formula applied toshows that its drift is zero: the second-order term is $-\frac{1}{2}|\theta {|}^2{E}_{t}dt$ and cancels the time derivative. Thus $E(\theta )$ is a complex local martingale. On $[0,T]$ its modulus is the deterministic value $e^{|\theta {|}^{2}t/2}\le e^{|\theta {|}^{2}T/2}$; localization and the dominated convergence theorem make it a true martingale. HenceThis conditional characteristic function is the deterministic characteristic function of $N(0,(t-s)I)$. For $A\in {\mathcal{F}}_s$, multiplying by $1_A$ and taking expectations, then using uniqueness of characteristic functions, proves that $M_t-M_s$ has this multivariate normal distribution and is independent of ${\mathcal{F}}_s$. Iterated conditioning gives independent increments with Gaussian distributions. Continuity, the zero initial value and these increment laws are the definition of standard Brownian motion. The case $d=1$ proves the requested statement; the vector form will also be used below.

For a harmonic function $u$ on $\mathbb{C}\cong {\mathbb{R}}^2$, the Itô formula givesStop on leaving increasing discs and at increasing deterministic times. On each such interval the gradient is bounded, so the stochastic integral is a martingale. This proves that $u(B)$ is a continuous local martingale.

If $|u|\le K$, this is a bounded local martingale and hence a true martingale. Start Brownian motion at an arbitrary $z$. Its Gaussian heat kernel givesFor a unit vector $e$, ${\partial }_e{p}_t(v)=-(e\cdot v)p_t(v)/t$, so $\int |{\partial }_e{p}_t(v)|dv=\sqrt{2/\pi }/\sqrt{t}$. Integrating the directional derivative along the segment from $z$ to $w$ yieldsLetting $t\to \infty$ gives the harmonic Liouville theorem:This Gaussian heat-kernel proof of the harmonic Liouville theorem uses the bounded martingale identity established above.

Let the infinitesimal generator be the second-order operatorwhere $a(x)$ is symmetric positive semidefinite and $b(x)$ is a vector. In the standard bounded-coefficient setting take $a,b$ bounded and measurable. An L-diffusion is a continuous adapted process $X$ for which, for every $g\in C_b^2({\mathbb{R}}^d)$,Here $C_b^2$ includes boundedness of the function and its derivatives through order two. This is the true-martingale version of the diffusion martingale problem. The matrix $a$ is the diffusivity, and $b$ is the drift. An Itô diffusion with coefficient $\sigma$ has $a=\sigma {\sigma }^{\mathsf{T}}$.

A more general local martingale problem may permit unbounded coefficients and require only a local martingale. That convention is weaker and must not silently replace the true martingale property used here and in part (b). The bounded-coefficient convention guarantees all finite-horizon integrability needed below; a sufficient more general replacement is $\mathbb{E}{\int }_0^T({\sum }_i|b_i(X_s)|+{\sum }_i{a}_{ii}(X_s))ds<\infty$ for every $T$.

We prove the extension from spatial test functions to time-dependent test functions for a diffusion martingale problem directly, without assuming a driving Brownian motion. Fix $0\le s<t$ and a deterministic partition $s=t_0<\cdots <t_n=t$. For each $j$, $g_j(x)=f(t_j,x)$ lies in $C_b^2$. The defining martingale problem givesCall the sum $S_n$. Telescoping separates the spatial and temporal changes:The sample path $X$ is uniformly continuous on $[s,t]$ and has compact range. Joint continuity of ${\partial }_{t}f$, the spatial first derivatives, and the spatial second derivatives therefore gives, as the mesh tends to zero,almost surely. For bounded $a,b$ and bounded derivatives of $f$, both integrals are uniformly bounded on this finite horizon, as are the two endpoint terms. The dominated convergence theorem permits passage to the limit in the expectation against $1_A$. Thus $M_t^f$ is integrable and $\mathbb{E}[1_A(M_t^f-M_s^f)]=0$ for every $A\in {\mathcal{F}}_s$, provingAdaptation and continuity follow from the formula, so $M^f$ is a continuous martingale. Under the more general integrability condition in part (a), the same proof uses an integrable bound proportional to $1+{\int }_s^t({\sum }_i|b_i(X_u)|+{\sum }_i{a}_{ii}(X_u))du$; positive semidefiniteness bounds every off-diagonal entry by the diagonal ones. With only local assumptions this reasoning proves a local martingale, and an integrability hypothesis is needed to upgrade that conclusion.

The natural filtration of $S$ is the natural Brownian filtration: since $\sigma >0$, the equation $W_t=(\log (S_t/S_0)-\mu t)/\sigma$ recovers the entire Brownian motion history from the stock history. For $s<t$, independent increments and the moment-generating function of a normal distribution giveAll these expectations are finite. Since $S_s>0$, the martingale identity holds for every $s<t$ exactly when the last exponential equals one. The required logarithmic drift isThis is the distinction between the drift of $\log S$ and the drift of $S$: here the latter is zero.

Write $h=T-t$. Conditional on ${\mathcal{F}}_t^S$, independent increments of the Brownian motion givewhere $Y$ has the standard normal distribution and is independent of ${\mathcal{F}}_t^S$. Factoring $S_t$ out of the positive part identifies the normalized Black-Scholes call function:The European call option payoff is integrable, because $0\le (S_T-K{)}^{+}\le S_T$ and $\mathbb{E}{S}_T=S_0$. The tower property of conditional expectation therefore proves that $C$ is a true martingale. At $t=T$, the boundary value $F(0,m)=(1-m{)}^{+}$ gives $C_T=(S_T-K{)}^{+}$. The answer is the conditional payoff martingale

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.