The Itô formula for $B_t^2$ yields ${\int }_0^t{B}_{u}d{B}_u=(B_t^2-t)/2$. Substituting into the expression for $Z_t$ gives the pathwise identityConsequentlyTaking expected values provesThe random variable on the preceding line is bounded by one for $\alpha \ge 0$, so its integrability is immediate.

Let $Y_t=-\beta {\int }_0^t{B}_{u}d{B}_u-{\int }_0^t{h}_{u}du$. Its quadratic variation is $[Y{]}_t={\beta }^2{\int }_0^t{B}_u^{2}du$. The Itô formula for $Z=e^Y$ givesChooseThe chosen process is continuous and adapted, hence progressively measurable. Its time integral is finite on compact intervals because Brownian paths are continuous. The resulting $Z$ is the stochastic exponential of $-\beta {\int }_0^{\cdot }{B}_{u}d{B}_u$, and is a positive local martingale.

For $t>0$, $B_t$ has the normal distribution $N(0,t)$. Evaluate its Gaussian integral:At $t=0$ the defining random variable is one, consistent with the same formula. Thus

A sufficient condition is that $x\mapsto \sqrt{2\theta (x)}$ is globally Lipschitz. A more familiar, stronger condition implying this is that $\theta$ is globally Lipschitz and $\theta \ge \epsilon >0$. Both conditions repair the existence issue in part (i).

To connect this to the suggested stochastic differential equation, apply the coordinate and coordinate-product cutoff tests to any solution of the martingale problem. As before they giveDefine, using stochastic integrals against the coordinates of $X$,These integrals exist even without a global positive lower bound: their quadratic variations are exactly $t$. Indeed,The Lévy characterization of multidimensional Brownian motion identifies $W$ as standard $d$-dimensional Brownian motion. By associativity of stochastic integration,The diffusion matrix $\sqrt{2\theta (x)}I$ is globally Lipschitz under the stated condition, and boundedness of $\theta$ gives bounded growth. The global existence theorem for stochastic differential equations with Lipschitz coefficients gives a unique strong solution of a stochastic differential equation and uniqueness in law. Every solution to the martingale problem has just been represented as a weak solution of a stochastic differential equation of this form, so they all have the same law. Conversely, the Itô formula applied to any $f\in C_c^{\infty }$ gives a true martingale, since its stochastic integrand is bounded. Thus the martingale problem is well posed under this sufficient condition.

The existence assertion is false under the hypotheses printed in the source. No measurability of $\theta$ is specified. Even requiring $\theta$ to be a Borel measurable function does not suffice: a coefficient can be positive at every point while approaching zero too rapidly near the starting point. This is the obstruction described by pointwise positivity does not ensure weak existence.

Here is a counterexample in dimension one, with $\Theta =1$:Suppose that a solution to the martingale problem with domain $C_c^{\infty }(\mathbb{R})$ existed. The cutoff argument from part (a)(ii), now with $\mathcal{L}=\theta \Delta$, makes $X$ a continuous local martingale withThe criterion for a continuous local martingale with bounded quadratic variation makes it a square-integrable martingale, and the Itô isometry gives $\mathbb{E}{X}_t^2=\mathbb{E}[X{]}_t$.

Apply the occupation-times formula with the indicator of the singleton $\{0\}$. Since that singleton has zero Lebesgue measure,Thus $X$ spends zero ordinary time at zero. Away from zero, $\theta (x)\le x^2$, soThe Gronwall inequality forces $\mathbb{E}{X}_t^2=0$ for every $t$. Continuity then forces $X\equiv 0$ almost surely, contradicting either the zero occupation time just proved or $[X{]}_t=2t$ for that constant path. Consequently

For completeness, the intended construction works if $\theta$ is a Borel measurable function and bounded away from zero on every compact set. In particular, it works for continuous strictly positive $\theta$. Let $B$ be standard $d$-dimensional Brownian motion, and define a clock and its inverse byThe Brownian path on each finite interval has compact range, so the added hypothesis makes $A_u$ finite. The clock is continuous and strictly increasing, and $A_u\ge u/(2\Theta )$, so $A_u\to \infty$. Its inverse is continuous and satisfies ${\tau }_t\le 2\Theta t$. Each ${\tau }_t$ is a stopping time for the Brownian filtration.

For $f\in C_c^{\infty }({\mathbb{R}}^d)$, the optional sampling theorem applied at these bounded random times shows thatis a true martingale in the filtration ${\mathcal{G}}_t={\mathcal{F}}_{{\tau }_t}^B$. The inverse-clock substitution $d{\tau }_t=2\theta (X_t)dt$, valid almost everywhere, rewrites it asThis proves the Brownian time change for a variable diffusion speed:The factor $2$ in the clock is necessary because standard Brownian motion has infinitesimal generator $\Delta /2$.

TakePart (i) proves that Brownian motion solves this martingale problem. For the converse, let $X$ be any continuous adapted process satisfying this martingale problem and $X_0=0$.

Set ${\tau }_n=\inf \{t\ge 0:|X_t|\ge n\}$. Choose a smooth cutoff function ${\chi }_n$ equal to one on a neighbourhood of the closed ball of radius $n$. Apply the martingale problem to ${\chi }_n(x)x_i$ and ${\chi }_n(x)x_i{x}_j$, and stop at ${\tau }_n$. Inside that ball, their Laplace operators are respectively $0$ and $2{\delta }_{ij}$. The optional sampling theorem therefore shows thatare martingales. Continuity makes ${\tau }_n\uparrow \infty$ almost surely. Hence every $X^i$ is a continuous local martingale. The Itô product rule and the fact that a continuous finite-variation local martingale is constant identify its quadratic covariations:Here ${\delta }_{ij}$ is the Kronecker delta.

The Lévy characterization of multidimensional Brownian motion now applies. To see explicitly why it identifies the entire law, for $\xi \in {\mathbb{R}}^d$ the Itô formula makesa complex local martingale. Its absolute value on $[0,T]$ is at most $e^{|\xi {|}^{2}T/2}$, so the bounded local martingale criterion makes it a true martingale there. For $0\le s\le t$,This conditional characteristic function identifies $X_t-X_s$ as having a centered multivariate normal distribution with covariance matrix $(t-s)I$, independent of ${\mathcal{F}}_s$. Together with continuity and $X_0=0$, these are exactly the defining properties of standard $d$-dimensional Brownian motion. Thus the martingale problem for Brownian motion has a unique law.

A convenient sufficient condition is that $f\in C^2({\mathbb{R}}^d)$ has a bounded gradient. The Itô formula for standard $d$-dimensional Brownian motion givesIf $|\nabla f|\le C$, the Itô isometry givesThus the right side is a square-integrable martingale, rather than merely a local martingale. In particular,There are many nonconstant examples, such as $f(x)=\sin x_1$ or $f(x)=x_1$. More generally it is enough that $\mathbb{E}{\int }_0^t|\nabla f(B_s){|}^{2}ds<\infty$ for each finite $t$. Smooth test functions with compact support satisfy the bounded-gradient condition and will provide the domain in the next part.

A second-order nondivergence-form elliptic operator on a domain $D\subseteq {\mathbb{R}}^d$ has the formOnly the symmetric part of $a$ affects the Hessian matrix term, so one may take $a(x)$ symmetric. The operator is a uniformly elliptic operator if there is a constant $\kappa >0$, independent of $x$, such thatA common bounded-coefficient convention additionally requires an upper bound $\sum a_{jk}(x){\xi }_j{\xi }_k\le \Lambda |\xi {|}^2$ with $\Lambda <\infty$. The lower bound says that the principal quadratic form cannot degenerate in any direction or at any location.

For an explicit stochastic example, standard $d$-dimensional Brownian motion has infinitesimal generatorIts principal quadratic form is $|\xi {|}^2/2$, so it is uniformly elliptic with $\kappa =\Lambda =1/2$. More generally, a sufficiently regular stochastic differential equation $d{X}_t=b(X_t)dt+\Sigma (X_t)d{B}_t$ has diffusion generator $b\cdot \nabla +\frac{1}{2}(\Sigma {\Sigma }^{\mathsf{T}}):D^2$; it is uniformly elliptic when $\Sigma {\Sigma }^{\mathsf{T}}$ has a uniformly positive lower bound. A conservative diffusion process has no zeroth-order term, since its infinitesimal generator annihilates constants.

For $\lambda \ne \mu$, apply the resolvent identity from part (i) in both orders:The two quotients are equal. For $\lambda =\mu$ the assertion is immediate. HenceThus resolvent operators from the same strongly continuous semigroup commute. This conclusion does not require commuting arbitrary bounded linear operators; it uses the fact that both arise from the same semigroup generator.

Let $E$ be the underlying Banach space. By the Laplace-transform formula for a semigroup resolvent,These are Bochner integrals, defined for every $f\in E$ and $\lambda >0$. The contraction semigroup bound and semigroup property giveInterchanging and regrouping these Bochner integrals is justified by the integrable bound $e^{-\lambda s-\mu t}\Vert f\Vert$, whose integral is $\Vert f\Vert /(\lambda \mu )$. For $\lambda \ne \mu$, set $v=s+t$ and evaluate the inner scalar integral:Since $f$ was arbitrary, the resolvent identity follows:For $\lambda =\mu$, both sides vanish, so that case holds as well.

For $a<0$, set $v_{\infty }=-{\sigma }^2/(2a)>0$. The candidate invariant probability law of a Markov process isTo prove invariance rather than merely guess a limiting probability distribution, initialize $X_t\sim \pi$ independently of future Brownian increments. Integrating the conditional characteristic function from part (i) against this centered normal distribution givesbecause $v_{\Delta }=v_{\infty }(1-e^{2a\Delta })$. The uniqueness theorem for characteristic functions implies $X_{t+\Delta }\sim \pi$ for every $\Delta \ge 0$. This proves $\pi P_{\Delta }=\pi$, as required. The time-homogeneous Markov property then also makes the process with this Ornstein-Uhlenbeck stationary initialization a strictly stationary process.

Here the diffusion generator is $\mathcal{L}=ax{\partial }_x+\frac{1}{2}{\sigma }^2{\partial }_{xx}$. The Feynman-Kac formula givesTry $F(t,x)=\exp [iu\alpha (t)x+\beta (t)]$, where $\alpha (T)=1$ and $\beta (T)=0$. Substitution and collection of the coefficients of $x$ and the constant term giveThe solutions of these ordinary differential equations, with the terminal conditions imposed, areFor $\Delta =T-t$, define the conditional varianceThen the required conditional characteristic function isIn particular, for $a\ne 0$ this exponent is $iu{e}^{a\Delta }x-\frac{{\sigma }^2{u}^2}{4a}(e^{2a\Delta }-1)$. The variance $v_{\Delta }$ is nonnegative for either sign of $a$, and its limit at $a=0$ is ${\sigma }^2\Delta$. Hence the solution is bounded in absolute value by one and the Feynman-Kac formula identifies it with the required expectation. The value $u=0$ gives one directly, including that case without dividing by $u$.

For an independent check, the integrating factor $e^{-as}$ and the Itô product rule giveThe stochastic integral has a normal distribution because its integrand is deterministic; the Itô isometry gives its variance as $v_{\Delta }$. For $a<0$ this is the Ornstein-Uhlenbeck transition distribution. The formula also covers the unstable case $a>0$, and degenerates to deterministic motion when $\sigma =0$.

Use the corrected piecewise-deterministic Markov process from part (b). For a smooth test function $f$, its time-dependent infinitesimal generator isIndeed, over an interval of length $h$, deterministic motion contributes $ha(t,x)f^{\prime }(x)$, while a jump has probability $h\alpha (x)+o(h)$ and changes the conditional mean of $f$ by $K_{t}f(x)-f(x)$. Terms involving both drift and a jump contribute only to the remainder. Thus $P_{t,t+h}f=f+h{L}_{t}f+o(h)$.

The Dynkin formula, or this infinitesimal expansion integrated against the current probability distribution, gives the weak evolution identityTake $f\in C_c^{\infty }(\mathbb{R})$, so integration by parts has no boundary contribution. Assuming the displayed jump integrals are finite, Fubini's theorem givesConsequently the forward equation of a piecewise-deterministic Markov process isHere $A_t=L_t^{\ast }$ is the formal adjoint operator acting on probability density functions. Its terms are transport, loss from $x$, and incoming mass from every $y$. In particular, the incoming rate is $\alpha (y)$, evaluated at the departure state. The operator denoted $A$ in the source is generally time-dependent. Because $\int m_t(y,x)dx=1$, the gain and loss integrals cancel; with vanishing transport flux at infinity, $\frac{d}{dt}\int p_t=0$.

The deterministic initial state in part (b) does not actually yield an everywhere smooth full density. For finite times the no-jump contribution is the positive measurewhere ${\delta }_z$ is the Dirac measure at $z$. Smooth jump destination probability density functions cannot remove this positive atom. The density equation above is therefore conditional on the extra density hypothesis, for example with a suitable random initial probability density function. For the specified fixed initial state, write ${\nu }_t$ for the whole probability measure and use insteadThis weak measure evolution includes both the atom and any continuous component. It becomes the displayed partial differential equation whenever ${\nu }_t(dx)=p_t(x)dx$ with the assumed regularity. It does not describe the literal hidden-reset construction without the correction in part (b).

The identity for the killed deterministic flow follows directly from the chain rule and the fundamental theorem of calculus, independently of the construction issue in part (b). For a continuously differentiable $f$, put $z(t)=\phi (s,x,t)$ and $H(t)={\int }_s^t\alpha (\phi (s,x,u))du$. ThenHence, on such test functions,The time subscript on $L_t^{\alpha }$ is necessary because the vector field may depend on time; it is suppressed on the right-hand side of the source's displayed operator identity. The killed deterministic flow transports an observable along the flow of a vector field and multiplies it by the no-jump survival probability.

The posterior mean is a convex combination of the prior mean and the usual Poisson maximum-likelihood estimate:Both weights are positive and sum to one, so the mean lies between $r/\lambda$ and $\bar{x}$, strictly between them unless they coincide. If all counts are zero, $\bar{x}=0$ is the boundary estimate in the usual nonnegative Poisson parameter space; with a strictly positive parameter restriction it represents the likelihood supremum at that boundary. The displayed posterior-mean relation is valid in either interpretation.

Writing $s={\sum }_i{x}_i$, the Poisson distribution likelihood is proportional to ${\theta }^s{e}^{-n\theta }$. Multiplying by the gamma distribution prior givesThis is Poisson-gamma conjugacy, with the rate convention made explicit:Its normalized density is $(\lambda +n{)}^{r+s}{\theta }^{r+s-1}{e}^{-(\lambda +n)\theta }/\Gamma (r+s)$ for $\theta >0$.

An equal-tailed credible interval for a continuous posterior distribution has probability $\alpha /2$ below its lower endpoint and $\alpha /2$ above its upper endpoint. If $F$ is the posterior cumulative distribution function, the endpoints are $F^{-1}(\alpha /2)$ and $F^{-1}(1-\alpha /2)$.

A highest posterior density interval is a credible interval consisting of points of highest posterior density: every included point has density at least as high as every excluded point, apart from null-set and boundary choices. It is a posterior highest density region when that region is an interval. For a continuous unimodal density with two interior endpoints, their densities agree. A mode at a support boundary can instead force one endpoint to that boundary; a multimodal posterior can have a disconnected highest-density region rather than one interval. Equal tails and highest density need not give the same interval.

Given the observed data, a $100(1-\alpha )\%$ credible interval $[a,b]$ for the scalar parameter satisfiesThe probability refers to the posterior distribution, proportional to $\pi (\theta ){\prod }_{i}f(x_i\mid \theta )$ on the parameter space. It conditions on the observations; it is not the repeated-sampling coverage probability definition of a confidence interval.

The Neyman-Pearson lemma states that a test which rejects where the joint likelihood ratio $L_1/L_0$ is largest is a most powerful test of its size: its power is at least that of any test with no larger size. More precisely, choose a threshold $k$ and reject if $L_1/L_0>k$, accept if it is smaller, and if necessary randomize on equality to attain the specified size.

For an independent and identically distributed statistical sample, takeThe $Y_i$ are independent and identically distributed random variables under either simple hypothesis, and the log joint ratio is their sum. Thus the nonrandomized threshold form can be writtenwith $c=-\log k$, using either strict or non-strict inequality when the boundary has probability zero. The usual finite-log expression assumes positive common support; zero densities can instead be handled by the extended likelihood ratio on the relevant supports. Continuous observations alone do not ensure a continuous likelihood-ratio statistic: ties may require the boundary randomization in the lemma. This qualification preserves the exact-size formulation.

For a critical region $C$, the size of a statistical test of the simple null hypothesis $P_0$ is $\alpha =P_0(\mathbf{X}\in C)$, the Type I error probability. Its statistical power against the simple alternative hypothesis $P_1$ is $P_1(\mathbf{X}\in C)$, the probability of correctly rejecting that null. For a randomized test function $\phi (\mathbf{x})\in [0,1]$, replace these by ${\mathbb{E}}_0\phi$ and ${\mathbb{E}}_1\phi$.