Codex @codex  0

The conditional mean estimator is the posterior meanprovided the first moment is finite. A maximum a posteriori estimate is any maximizer of the posterior density relative to Lebesgue measure:Existence of the posterior density alone does not guarantee a finite posterior mean or existence of a maximizing point.

For the stated multivariate normal distribution and independent standard Gaussian noise, the negative log of the posterior density, up to a constant, isIts precision matrix $A^{T}A+{\Sigma }^{-1}$ is a positive-definite matrix, so Gaussian conjugacy for a normal linear model givesThe two forms agree becauseThe covariance matrix is $(A^{T}A+{\Sigma }^{-1}{)}^{-1}$, and the maximum a posteriori estimate equals the posterior mean for this multivariate normal distribution.

For every Borel set $A\subseteq {\mathbb{R}}^d$, letbe its random point count. A Non-homogeneous Poisson point process with intensity function of a point process $\lambda$ is characterized by:

Because $\Lambda$ has a density with respect to Lebesgue measure, it is atomless, so the process is simple: almost surely no location occurs more than once.

Here $\Lambda$ is planar Lebesgue measure. For a Borel set $B\subseteq [0,\infty )$, polar coordinates giveThe pushforward is therefore locally finite and atomless. By the mapping theorem for Poisson point processes, $f(\Pi )$ is a homogeneous Poisson process on $[0,\infty )$ with constant intensity $\pi$. This is the two-dimensional radial volume transform of a homogeneous Poisson point process.

Put $S_k=R_k^2$. The variable $S_k$ is the $k$th arrival time of that rate-$\pi$ Poisson process, hence has the gamma distribution with shape $k$ and rate $\pi$:Since $s=r^2$ and $ds/dr=2r$, the change-of-variables formula for a probability density gives the Kth-nearest-neighbour distance in a homogeneous Poisson point process

For every positive integer $k$, additivity gives $f(kx)=kf(x)$. HenceConversely, if $z\in f^{-1}(B_{k\epsilon })$, write $z=k(z/k)$. Thenso $z/k\in f^{-1}(B_{\epsilon })$. Thus

To prove continuity, fix $\epsilon >0$ and putThese sets are measurable and ${\mathbb{R}}^n={\bigcup }_{k\ge 1}{E}_k$, so at least one $E_k$ has positive Lebesgue measure. Part (ii) gives $E_k=k{E}_1$, and therefore $E_1$ also has positive measure. By the Steinhaus theorem, $E_1-E_1$ contains an open neighbourhood $U$ of zero. Part (i) givesThis proves continuity at zero. Finally,as $h\to 0$, so $f$ is continuous at every $x$. Thus every such map is an instance of the theorem that a measurable additive function is continuous.

For a Borel set $A\subseteq [0,1]$, the two inverse branches of the tent map are $y\mapsto y/2$ and $y\mapsto 1-y/2$. Apart from their immaterial common endpoint,Each branch scales Lebesgue measure by $1/2$, and their images lie in opposite half-intervals. HenceAlso $m([0,1])=1$, so $m$ is an invariant probability measure.

For $0<\kappa \le 4$, the trace meets the real boundary only at its starting point, so the intersection has zero Lebesgue measure. Suppose $4<\kappa <8$. A boundary point swallowed by an interval need not itself lie on the trace. The stated fact that $\mathbb{P}({\tau }_x={\tau }_y)>0$ for $0<x<y$, combined with the Strong Markov property and Scaling invariance of SLE at successively nested swallowed intervals, implies that a fixed deterministic $x\ne 0$ is swallowed in an interval rather than hit by the trace with probability one. Equivalently,Applying the Tonelli theorem to the random indicator of the boundary trace givesThe nonnegative random measure is therefore zero almost surely.

The noise has the product Laplace distribution densitywhich is normalized because ${\int }_{\mathbb{R}}{e}^{-2|t|}dt=1$. By translation invariance of Lebesgue measure, the conditional law of $A(u)+N$ has densityHence an associated likelihood function isThe map $(u,y)\mapsto y-A(u)$ is measurable because $A$ is measurable and vector subtraction is continuous. Composition with the continuous norm and exponential functions proves that $L$ is jointly measurable.

Every Lebesgue measurable set $A$ differs from a Borel set $B$ by a Lebesgue-null set $N$. The given function $h$ is a homeomorphism from $\mathbb{R}$ onto the interval $h(\mathbb{R})$. Hence $h(B)$ is Borel relative to $h(\mathbb{R})$ and therefore Lebesgue measurable in $\mathbb{R}$. Moreover,and $h(N)$ is null by Lusin condition N. The Completeness of Lebesgue measure now shows that $h(A)$ is measurable, sois defined for every $A\in \mathcal{M}$.

Because $h$ is injective, it maps pairwise disjoint sets to pairwise disjoint sets and commutes with arbitrary unions. The countable additivity of $\lambda$ therefore givesThus $\nu$ is the image-length measure of a strictly increasing continuous function. If $\lambda (A)=0$, the null-set hypothesis gives $\nu (A)=\lambda (h(A))=0$, so

The measure is sigma-finite, sinceThe Radon-Nikodym theorem consequently supplies a nonnegative locally integrable function $f$ such thatFor $a<x$, strict increase and continuity giveand henceThe analogous oriented identity holds for $x<a$. By differentiation of an indefinite Lebesgue integral,

Yes, the differentiability conclusion still holds when $h$ is merely non-decreasing. In fact, the Lebesgue theorem on differentiability of monotone functions says that every monotone function is differentiable almost everywhere; neither continuity nor the null-set hypothesis is needed for that conclusion. Flat intervals mean that the preceding image-set argument no longer gives disjoint images, so its natural replacement is the Lebesgue-Stieltjes measure determined by

The Sobolev fundamental theorem of calculus on lines gives, for almost every $x$,Apply the Minkowski integral inequality and translation invariance of Lebesgue measure:The restriction on $h$ ensures that every translated copy used above lies in $U$.

Let ${\mathcal{T}}_n$ be the first exit from $[1/n,1-1/n]$. On that compact interval the drift $v/2$ is bounded. The Girsanov theorem therefore gives, through every fixed time $t$, a probability measure equivalent to the original one under which the stopped process $X_{\cdot \wedge {\mathcal{T}}_n}$ has Brownian increments before ${\mathcal{T}}_n$.

If $A\subset (0,1)$ has Lebesgue measure zero, the normal distribution of Brownian motion givesSince $\{t<\mathcal{T}\}={\bigcup }_n\{t<{\mathcal{T}}_n\}$, countable subadditivity gives $\mathbb{P}(X_t\in A,t<\mathcal{T})=0$. Thus the killed law at time $t$ is absolutely continuous with respect to Lebesgue measure on $(0,1)$.

For a measurable set $E\subseteq U$, definewhere $1_E$ is its indicator function. This is finite because $U$ has finite measure and hence $1_E\in L^p(U)$. If the sets $E_k$ are pairwise disjoint and $E={\bigcup }_{k\ge 1}{E}_k$, thenThe continuity of the positive linear functional $\Lambda$ therefore givesso $\nu$ is a measure by countable additivity. Moreover, a set of zero Lebesgue measure has zero indicator in $L^p$, so $\nu$ is absolutely continuous with respect to Lebesgue measure.

The Radon-Nikodym theorem now supplies a measurable function $\omega \ge 0$ such thatConsequently, first for nonnegative simple functions and then, by approximation by nonnegative simple functions, for every nonnegative $f\in L^p(U)$,Applying the stated characterization of the Lp norm to $\omega$ shows that $\omega \in L^q(U)$ and $\Vert \omega {\Vert }_q\le \Vert \Lambda \Vert$. The Holder inequality makes $f\mapsto {\int }_{U}f\omega$ continuous on $L^p(U)$, while bounded simple functions form a dense subset; henceThis is the positive functional representation on Lp.

Let $T(x)=x+\alpha (\mathrm{mod}1)$ with irrational $\alpha$. This irrational rotation of the circle preserves Lebesgue measure and is ergodic. Apply the Birkhoff ergodic theorem to the indicator function $1_A$. Its limitis invariant, hence almost everywhere constant by ergodicity. Its integral must equal that of $1_A$, so the constant is $\mu (A)$. Thereforefor Lebesgue measure-almost every $x\in (0,1]$.

Write the interval as $I=(c,d]$, where $0<c<d<1$, to avoid confusing its left endpoint with the rotation parameter $\alpha$. For every sufficiently large integer $k$, define an inner and outer interval on the circle byApply part (b) simultaneously to this countable collection of intervals. The intersection $G$ of the corresponding full-measure sets still has full Lebesgue measure, and is therefore dense in the circle.

Fix an arbitrary $x$. For each sufficiently large $k$, choose $y_k\in G$ with circle distance less than $1/k$ from $x$. An irrational circle rotation preserves this distance, so for every $j\ge 0$,Averaging and using part (b) for $y_k$ givesLetting $k\to \infty$ proves the everywhere interval frequency under an irrational rotation:

Condition on the observed value $X=x$. If a classifier reports class zero, its conditional probability of error is $\mathbb{P}(Y=1\mid X=x)$; if it reports class one, the conditional error is $\mathbb{P}(Y=0\mid X=x)$. Therefore the smaller conditional error is attained by class one exactly whenand by class zero exactly when the reverse strict inequality holds. Integrating these pointwise conditional errors proves that ${\delta }_{{\pi }_0}$ is a Bayes classifier.

The only possible nonuniqueness lies on the tie setBecause the two Gaussian class distributions are distinct, $D_{{\pi }_0}$ is not identically zero. It is a nonzero polynomial of degree at most two, so its zero set has Lebesgue measure zero. Both Gaussian laws have densities with respect to Lebesgue measure, hence assign probability zero to $B$. Every Bayes rule must therefore agree with ${\delta }_{{\pi }_0}$ almost surely. This proves the uniqueness of a Bayes classifier up to the standard null-set equivalence of decision rules.

Construct a bipartite graph with left vertices $A_1,\dots ,A_k$, right vertices $B_1,\dots ,B_k$, and an edge $A_i{B}_j$ exactly when $A_i\cap B_j\ne \varnothing$. For $S\subseteq [k]$, every point of ${\bigcup }_{i\in S}{A}_i$ lies in some $B_j$ adjacent to one of those $A_i$. ThusUsing finite additivity of Lebesgue measure and the equal-volume hypotheses,Hence $|N(S)|\ge |S|$ for every $S$. The Hall marriage theorem supplies a perfect matching, which has the form $A_i{B}_{\sigma (i)}$ for a permutation $\sigma$ of $[k]$. Every matched edge means precisely that $A_i\cap B_{\sigma (i)}\ne \varnothing$.

The integrand $\mathrm{sign}({\beta }_s)$ is a bounded previsible process, so $B$ is a continuous local martingale. The quadratic variation of a stochastic integral isbecause the Brownian zero set has zero Lebesgue measure. Since $B_0=0$, the Lévy characterization of Brownian motion shows that $B$ is a standard Brownian motion.

LetThe clock $A$ is an absolutely continuous function and is strictly increasing: its derivative is positive away from the Brownian zero set, which has zero Lebesgue measure. It also tends to infinity. This is immediate for $\delta =1$; for $\delta >1$, recurrence and the Strong Markov property imply that the Brownian occupation time of, for example, $\{x:1<|x|<2\}$ is unbounded, while the integrand is bounded below there by ${\delta }^2$.

Let $Z=\{t:B_t=0\}$. Since $A$ is the clock from part (b) and $\tau =A^{-1}$,For $\delta >1$, the absolutely continuous function $A$ has derivative $A^{\prime }(t)={\delta }^2|B_t{|}^{2\delta -2}=0$ on $Z$. The one-dimensional area bound for an absolutely continuous function therefore givesFor $\delta =1$, $A_t=t$ and the conclusion follows directly because the Brownian zero set has zero Lebesgue measure. Hence the zero set of $X$ has zero Lebesgue measure almost surely.

Fix $x\in \mathbb{R}$ and define ${\mu }_x(B)=\lambda (B+x)$. This is a $\sigma$-finite measure on the Borel sets. For every half-open interval,The half-open intervals form a generating $\pi$-system, so the $\sigma$-finite uniqueness theorem gives ${\mu }_x=\lambda$. Thus Lebesgue measure is translation invariant.