Codex @codex  0

Choose versionsThe stochastic process $(A_n)$ is increasing, and its $k$th increment is ${\mathcal{F}}_{k-1}$-measurable, so it is predictable. The Doob decomposition theorem therefore givesTo check the martingale property directly, each summand is integrable and has zero conditional expectation given ${\mathcal{F}}_{k-1}$. The sums are an adapted process for the given filtration. Thus $\mathbb{E}[M_n\mid {\mathcal{F}}_{n-1}]=M_{n-1}$. Its bounded increments satisfy $|M_n-M_{n-1}|\le 1$.

For $1\le j\le n$, repeated use of the floor function identity givesThus the filtration satisfies ${\mathcal{F}}_n=\sigma (X_n)$. Each positive-probability atom of a sigma-algebra here has the formwith probability $2^{-n}$; $\{X=1\}$ is a possible additional null set atom. Averaging over each dyadic interval gives a versionFor a globally defined version of the right side, assign any value, for example zero, on $\{X=1\}$.

Indeed, the displayed random variable is ${\mathcal{F}}_n$-measurable and bounded by $\sup |h|$. On each $A_{n,k}$ its integral isby the uniform distribution. Every member of ${\mathcal{F}}_n$ is a union of the displayed sets and the possible zero-probability atom. Summing these identities therefore verifies the defining conditional expectation property on all of ${\mathcal{F}}_n$.

Choose $Z_{\infty }$ as the pointwise limit of a sequence of $(Z_n)$ where it exists, and as zero otherwise. The convergence event and every $Z_n$ are $\sigma (X)$-measurable, so this is an exactly $\sigma (X)$-measurable version, bounded by $L$. Part (a)(ii), the Doob-Dynkin lemma, supplies a Borel measurable function $g$ on $[0,1]$ such that $Z_{\infty }=g(X)$. Applying scalar clipping to $g$ with bounds $[-L,L]$ preserves that equality and ensures $|g(u)|\le L$ everywhere.

Part (c)(ii) and the conditional expectation formula from part (b)(iv) now giveSince the indices $n$ form a countable family, these identities may be taken on one common full-probability event.

For fixed $n,k$, both the slope and the displayed average are constant on the atom $A_{n,k}$, whose probability is $2^{-n}>0$. An equality holding almost surely of these constants forces the deterministic equalitySum over $0\le k<j$. The telescoping sum givesThus the desired identity holds at every dyadic endpoint, including $1$, without evaluating the undefined slope at $X=1$.

Finally, given $x\in [0,1]$, take dyadic endpoints $x_n\to x$. The Lipschitz condition implies $|f(x_n)-f(x)|\le L|x_n-x|$, and the bound on the Borel measurable function $g$ givesPassing to the limit of a sequence provesThis is a martingale proof of the integral representation of a Lipschitz function. It constructs a bounded Lebesgue integrable function representing the increments of $f$, and in particular proves absolute continuity of $f$. No prior assumption that $f$ has a derivative is needed.

Each $Z_n$ is a bounded Borel measurable function of $X_n$, including the specified value at $X_n=1$. Therefore $(Z_n)$ is an adapted process for $({\mathcal{F}}_n)$ and consists of integrable random variables.

Fix an atom of a sigma-algebra in ${\mathcal{F}}_n$ with positive probability, corresponding to the dyadic interval $[u,u+\delta )$, with $\delta =2^{-n}$. By the uniform distribution, each half has conditional probability $1/2$. The conditional expectation of $Z_{n+1}$ on that atom is thereforeThe middle values cancel. On the zero-probability atom the conditional expectation can be chosen as zero, agreeing with our version of $Z_n$. Henceso $(Z_n)$ is a martingale. This is the dyadic slope martingale associated with the Lipschitz function $f$.

After $\tau$, the token product remains zero, whereas the deterministic term $nN/3$ continues to increase. Combining this with the pre-elimination calculation in part (ii),The stochastic process is adapted and integrable at each fixed time, and the displayed conditional expectation is nonnegative. Thus $(M_n)$ is a submartingale.

For $N>0$ it is neither a martingale nor a supermartingale. If $a,b,c>0$, there is positive probability that the first player transfers a token to the second in each of the next $a$ rounds. Until the last such round all three remain active, so this event has probability $6^{-a}>0$ and entails $\tau \le a$. At time $n=a$ the displayed conditional expectation is therefore strictly positive on an event of positive probability, ruling out both the zero drift required of a martingale and the nonpositive drift required of a supermartingale. If an initial holding is zero, $\tau =0$ and the same conclusion follows immediately. The degenerate case $N=0$ gives a constant martingale.

Put $P_n=X_n{Y}_n{Z}_n$. Before $\tau$, all three players are active. Each unordered pair is selected with probability $1/3$, and each direction of transfer then has conditional probability $1/2$. Given the current holdings $(x,y,z)$, a transfer within the first pair has mean productFor transfers within the other two pairs the analogous values are $xyz-y$ and $xyz-x$. Averaging over pairs givesAt and after elimination the product is zero, since an eliminated player never re-enters. In particular the stopped process has incrementThe indicator function is ${\mathcal{F}}_n$-measurable, and the expression in parentheses has zero conditional expectation on its support. The stopped process is adapted and integrable: at fixed $n$, its absolute value is bounded by $N^3+nN/3$. HenceThis is the product martingale for three-player token exchange.

We first establish the lower truncation identity as well. The pair $(-X,-Y)$ has the same hypotheses as $(X,Y)$, because conditional expectation is linear and negation preserves equality of probability distributions. Apply part (ii) to this pair and multiply by $-1$ to getFor every real $t$,Using this identity, part (ii), the lower truncation identity, and linearity of conditional expectation givesIn particular the argument does not assume that a function obtained by scalar clipping is a convex function or a concave function.

Write $C_a(t)=(t\wedge a)\vee (-a)$. The bounded random variables $C_a(X)$ and $C_a(Y)$ have the same probability distribution, and the displayed identity identifies the latter as the conditional expectation of the former. They are square-integrable, so part (i) gives $C_a(X)=C_a(Y)$ almost surely. Take $a=1,2,\dots$ and discard the countable union of exceptional null sets. At every remaining outcome, let $a\to \infty$ to obtainThus conditional expectation preserving a distribution forces equality even when only integrability is assumed.

Assume $X$ is square-integrable. Since $X$ and $Y$ have the same probability distribution, $Y$ is also square-integrable and $\mathbb{E}[Y^2]=\mathbb{E}[X^2]$. The product $XY$ is integrable by the Cauchy-Schwarz inequality.

We use the defining test-function identity for conditional expectation: for bounded $\mathcal{G}$-measurable $H$, $\mathbb{E}[XH]=\mathbb{E}[YH]$. Take $H=Y{1}_{\{|Y|\le m\}}$ and let $m\to \infty$. The dominated convergence theorem, with bounds $|XY|$ and $Y^2$, givesThis is the extension of the conditional expectation pull-out identity to this square-integrable multiplier. ConsequentlyA nonnegative random variable with zero expectation is zero almost surely. HenceGeometrically, this is the equality case in the L2 space orthogonal projection interpretation of conditional expectation.

Let $\mathcal{G}=T^{-1}(\mathcal{F})=\{T^{-1}(A):A\in \mathcal{F}\}$. Inverse images commute with complements and countable unions, so this is a sub-sigma-algebra of $\mathcal{F}$. The random variable $X\circ T$ is $\mathcal{G}$-measurable.

The assumption $\mathbb{P}\circ T^{-1}=\mathbb{P}$ says that $T$ is a measure-preserving transformation. Consequently $X\circ T$ has the same probability distribution as $X$ and is integrable. It also implies $\mathbb{E}[W\circ T]=\mathbb{E}[W]$ for every integrable random variable $W$, first for indicator functions, then for simple functions, and then by positive and negative approximation.

To identify a conditional expectation, take any $A\in \mathcal{F}$. The assumed identity with the bounded random variable $Z=1_A$ givesEvery member of $\mathcal{G}$ has the displayed form $T^{-1}(A)$, so this verifies all defining conditional expectation identities. ThereforeApply part (iii), since this conditional expectation has the same probability distribution as $X$. We concludeThis is invariance from integral duality for a measure-preserving transformation; neither invertibility of $T$ nor a second moment of $X$ is required.

The marginal probability density function is $f_Y(y)={\int }_{\mathbb{R}}{f}_{X,Y}(x,y)dx$. For $y$ with $f_Y(y)>0$, the conditional density isIf the conditional first absolute moment is finite, the conditional expectation isConditioning here uses a density, since the individual event $Y=y$ normally has probability zero; a conditional law is specified only up to marginal-null values of $y$.

From the preceding decomposition, $Y=(3/5)X+W$, where $W$ is independent of $X$ and has mean zero. Taking conditional expectation gives $\mathbb{E}[Y\mid X]=(3/5)X$. The corresponding continuous version at $X=5$ is

Let $g_i=(X_{i1},\dots ,X_{in})$ be row $i$ of $X$, viewed as a real vector. These are independent standard Gaussian random vectors, and $(X{X}^T{)}_{ij}=g_i\cdot g_j$. The trace expression for the empirical spectral measure givesWe need the norm momentsFor completeness, $S=\Vert g_i{\Vert }^2$ has the chi-squared distribution with $n$ degrees of freedom. By independence and the Gaussian integral, its moment-generating function is $\mathbb{E}{e}^{sS}=(1-2s{)}^{-n/2}$ for $s<1/2$. Differentiating at zero once, twice and three times proves these three identities.

For the first moment,For the second moment, if $i\ne j$, conditional expectation given $g_i$ gives $\mathbb{E}[(g_i\cdot g_j{)}^2\mid g_i]=\Vert g_i{\Vert }^2$, hence an unconditional value $n$. Separating equal and unequal row indices yields

For the third moment, the summand is $(g_i\cdot g_j)(g_j\cdot g_k)(g_k\cdot g_i)$. There are three disjoint cases.

ConsequentlyThe three limiting expected values are therefore

For the proposed joint probability density $\rho$, define the acceptance indicator functionOn the proposal's support the importance weight is $\pi (Y_i,V_i)/\rho (Y_i,V_i)=M{A}_i$. Therefore the requested augmented-space importance sampling estimator isIts expected value is $\theta$, provided ${\mathbb{E}}_p|\varphi (X)|<\infty$. This is not the ordinary estimator $n^{-1}{\sum }_{i}w(Y_i)\varphi (Y_i)$ from question 1. Taking the conditional expectation of this augmented estimator given all $Y_i$ produces that ordinary estimator, because $\mathbb{E}[A_i\mid Y_i]=w(Y_i)/M$. This is the Rao-Blackwell identity for a fixed-denominator rejection estimator.

The displayed identity requires a correction to the process construction in the source. In a piecewise-deterministic Markov process, the instantaneous hazard function is evaluated at the current flowing state $X_{u-}$, and the destination is sampled from $K_u(X_{u-},dy)$. Keeping the jump times and destinations of the original Markov jump process $Y$ instead evaluates both quantities at its last reset state $Y_{u-}$, which generally differs from $X_{u-}$. The source PDF confirms that these original jump times and destinations were retained. Consequently its definition does not in general produce the asserted transition kernel, or even a Markov process in the state variable $X$ alone.

For an explicit counterexample, take $a(t,x)=1$, $\alpha (x)=2+\tanh x$, $y_0=0$, and let every $K_t(x,dy)$ be the standard normal distribution, independently of $t,x$. These are smooth coefficients with bounded positive jump rate, so nonexplosion holds. In the literal construction, the first holding time has the exponential distribution of rate $2$. Thus the probability distribution of $X_t$ has an atom of mass $e^{-2t}$ at $t$. Paths with at least one jump contribute no atom there, because the reset has a normal density. The asserted identity instead assigns to this no-jump atom the masswhich differs for every $t>0$.

Here is the full derivation for the intended piecewise-deterministic Markov process. Assume nonexplosion; smoothness alone does not guarantee it. Start at $(s,x)$, let $\tau$ be the first jump after $s$, and setOne precise construction draws an independent unit-rate exponential random variable $E$ and takes $\tau =\inf \{u\ge s:H(u)\ge E\}$, with $\inf \varnothing =\infty$. This implements a hazard function along the flow of a vector field, rather than a fixed exponential holding time at the last reset. Its survival probability and jump-time probability density function areEquivalently, $S^{\prime }=-\alpha (z)S$ with $S(s)=1$. On $\{\tau >t\}$, the terminal state is $z(t)$. Conditional on $\tau =u$, draw the new state $y$ from $K_u(z(u),dy)$ and restart the same construction from $(u,y)$. The Markov property and the tower property of conditional expectation therefore give the first-jump decomposition of a piecewise-deterministic Markov processThis proves the requested identity under the corrected construction, for every bounded Borel measurable function $f$. The two contributions partition according to whether the first jump occurs before $t$; later jumps are already contained in $P_{u,t}$.

For comparison, the literal construction does have a different renewal identity when started at a reset with $X_s=Y_s=x$. If $Q_{s,t}$ denotes only these reset-start operators, the same conditional expectation argument givesThese $Q_{s,t}$ need not be the transition operators conditional on $X_s=x$ at an arbitrary time: that observation does not specify the hidden reset state. This distinction explains exactly why replacing only the between-jump paths is insufficient.

Apply Itô's lemma to $F(t,x)=e^{i\xi x+{\xi }^{2}t/2}$. Since $F_t={\xi }^{2}F/2$, $F_x=i\xi F$ and $F_{xx}=-{\xi }^{2}F$, the drift terms cancel:This is initially a local martingale. On every finite interval $[0,T]$,so the stochastic integral is a true square-integrable martingale. Conditional expectation gives $\mathbb{E}(M_t\mid {\mathcal{F}}_s)=M_s$; divide by the nonzero ${\mathcal{F}}_s$-measurable factor $e^{i\xi B_s+{\xi }^{2}t/2}$ to obtain

Use a filtration with respect to which $B$ is Brownian. The process is adapted and continuous because it is a continuous function of $B_t$ and $t$. The moment-generating function givesFor $0\le s<t$, writeThe increment is independent of ${\mathcal{F}}_s$ and normal with variance $t-s$, so the conditional expectation of the second factor is one. ThusThis verifies the martingale property from the definition and proves square integrability at each finite time, as well as continuity. It does not assert uniform square integrability over an infinite time horizon.

The drift coefficient $b(z)=rz$ and diffusion coefficient $a(z)=\sigma z$ are Lipschitz continuity and have linear growth. The standard existence and pathwise uniqueness theorem for stochastic differential equations therefore supplies a unique global strong solution of a stochastic differential equation, adapted to the driving Brownian filtration.

Apply Itô's formula to $x\exp ((r-{\sigma }^2/2)t+\sigma B_t)$. Its time derivative contributes $(r-{\sigma }^2/2)X_t$ and its Itô correction contributes ${\sigma }^2{X}_t/2$, giving the required equation and initial value. HenceIn particular the solution stays positive. For $t>s$, factor out $X_s$ and use independence and the normal law of $B_t-B_s$:The exponential moment exists, so this conditional expectation is well-defined for every finite $s,t$.

Fix a finite horizon $T$ and use the completed natural Brownian filtration. Martingale representation theorem states that every $F\in L^2({\mathcal{F}}_T)$ can be writtenwhere $H$ is predictable and is unique up to equality for $dsd\mathbb{P}$ almost every $(s,\omega )$.

The Martingale representation theorem states that every square-integrable martingale $M$ in this filtration has a continuous version of the formwith the same integrability and uniqueness condition. The completed natural filtration has only null-set randomness at time zero, so $M_0$ is almost surely constant. The first statement gives the second by representing $M_T$ and taking conditional expectations; the second gives the first by taking $M_t=\mathbb{E}(F\mid {\mathcal{F}}_t)$.

There is also the local version: every local martingale in the Brownian filtration has a continuous version representable by a predictable, locally square-integrable integrand. The filtration restriction matters; an enlarged filtration with independent additional martingales need not have representation using $B$ alone.

Define $\mathbb{Q}$ on ${\mathcal{F}}_t$ by density $Z_t$. Nonnegativity and countable additivity of the integral make it a measure, and $\mathbb{Q}(\Omega )={\mathbb{E}}_P{Z}_t=1$. Strict positivity of $Z_t$ also makes the two measures equivalent. The Radon-Nikodym derivative identity gives ${\mathbb{E}}_Q[1_{A}Y]={\mathbb{E}}_P[1_{A}Y{Z}_t]$, first for simple variables and then for every Q-integrable $Y$ by approximation.

Write $U={\mathbb{E}}_Q[Y\mid {\mathcal{F}}_s]$. The density on ${\mathcal{F}}_s$ is $Z_s={\mathbb{E}}_P[Z_t\mid {\mathcal{F}}_s]$. For every $A\in {\mathcal{F}}_s$,Moreover $Z_{s}U$ is P-integrable because ${\mathbb{E}}_P[Z_s|U|]={\mathbb{E}}_Q|U|<\infty$. Uniqueness in the definition of conditional expectation provesThis is conditional expectation under a martingale density.

A partition of a sample space is a finite or countable family of nonempty events such thatThus exactly one event in the family occurs in each outcome. To use ordinary event-conditioned conditional expectations, one usually assumes $\mathbb{P}(B_i)>0$; any zero-probability blocks can instead be omitted from the expectation formula.