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Because $X$ is a martingale,Writingand using ${\xi }_n\in \{-1,1\}$ givesThus

For signs $e_1,\dots ,e_m\in \{-1,1\}$, repeated use of the tower property of conditional expectation givesHence the increments are IID symmetric signs, proving the symmetric signs forced by the martingale property result. Thus $X_n-X_0$ is a simple symmetric random walk.

A process $(X_n{)}_{n\ge 0}$ is a martingale with respect to a filtration $({\mathcal{F}}_n{)}_{n\ge 0}$ when:

Equivalently, every increment has conditional mean zero given the past.

The model factors into the normal distribution densitiesand, conditionally on $X_{i-1}$,Ignoring constants independent of $\theta$, the log likelihood isIts score isIn terms of the independent innovations,Variance additivity for independent random variables therefore yields

Since $I_1(\theta )=1$, the equality $I_n=n{I}_1$ holds for all $n$ exactly whennamely whenThus the information tensorizes only in the independent case.

Finally, the Cramér-Rao bound gives every unbiased estimator $\hat{\theta }$ the lower boundThis is the Fisher information in a stationary Gaussian autoregressive location model.

Write $f_{\theta }(x)$ for the joint density and assume regularity permits differentiation under the integral. Sinceone differentiation givesDifferentiating the score itself,Taking expectation givesThus the information identity is

Letbe the log-likelihood. The score function and scalar Fisher information areUnder the usual regularity conditions the score has mean zero, so this is also its variance.

The information tensorizes when it is additive over observations. For identically distributed observations,For an independent sample the joint score is the sum of the individual scores, and the mean-zero score identity makes all cross covariances vanish. This is the Tensorization of Fisher information.

LetThe compact support of $g$ ensures that this sum has only finitely many nonzero terms almost surely. The intensity measure of the ideal gas isApplying the first-moment identity in Campbell theorem gives

For the second moment, separate equal and distinct particles:Subtracting $(\mathbb{E}{S}_g{)}^2$ yieldsThese are the mean and variance formulas in Campbell's theorem for a homogeneous Poisson point process.

Put ${\lambda }_i=z|V_i|$. Thenand ${\lambda }_i\to \infty$. The Poisson central limit theorem givesSince $\sqrt{{\lambda }_i}=\sqrt{z}\sqrt{|V_i|}$,

Equivalently, the characteristic function of the left-hand side iswhich is the characteristic function of that centered normal distribution.

Let $N_j(A)$ be the number of particles of gas $j$ in a bounded Lebesgue measurable set $A$. Because Lebesgue measure is diffuse, two independent spatial Poisson point processes have no common particle almost surely. Hence the mixture count isWriting ${\lambda }_j=z_j|A|$, independence gives the probability generating functionThus

If $A_1,\dots ,A_m$ are pairwise disjoint, the familyis independent: counts in disjoint sets are independent within each gas, and the two gases are independent. Therefore the sums $N(A_r)$ are independent. These two facts prove directly that the mixture is a Poisson point process. By the Superposition theorem for Poisson point processes, its activity is

Yes. Almost-sure convergence givesalmost surely. Since the random variables are nonnegative, Fatou lemma applies and givesThe inequalities remain valid in the extended nonnegative reals, so no integrability assumption beyond nonnegativity is required.

On the probability space $([0,1],\mathcal{B},\lambda )$, letFor every $x\in [0,1]$, eventually $x\notin (0,1/n)$, so $f_n(x)\to 0$. Nevertheless,for every $n$. ConsequentlyThis gives strict inequality in Fatou lemma.

Fatou lemma states that for any sequence $(f_n)$ of nonnegative measurable functions on a measure space,

For the proof, defineThen $(g_n)$ is a nonnegative increasing sequence andThe monotone convergence theorem therefore givesFor each $k\ge n$, we have $g_n\le f_k$, so monotonicity of the Lebesgue integral yieldsTaking the limit in $n$ proves the asserted inequality and the proof of Fatou lemma.

For a chosen unit normal field $N$, the second fundamental form isIn a local parametrization $\varphi$, its coefficients, up to the equivalent simultaneous sign convention, areThe Gaussian curvature is the determinant of the shape operator, henceTheorema Egregium states that $K$ is determined entirely by the first fundamental form. In particular, it is intrinsic and is preserved by every local isometry.

Since $z>0$, the cone has the parametrizationIts coordinate derivatives satisfyso its first fundamental form is

On a sufficiently narrow angular patch, defineand map the cone to the plane point with polar coordinates $(R,\Theta )$. The Euclidean metric pulls back asThe map is therefore a local isometry from a circular cone to the plane, proving that $\Sigma$ is locally isometric to the Euclidean plane.

For each $p\in S$, the first fundamental form is defined without coordinates bythe restriction of the Euclidean inner product to the tangent space of $S$.

If $\varphi (u,v)$ is a local parametrization andthenTaking their dot product giveswhere

The three coordinate axes are the three irreducible components ofTheir common intersection is the origin. Likewise, if ${\ell }_1,{\ell }_2,{\ell }_3$ are distinct linear forms defining the three lines in ${\mathbb{A}}_k^2$, thenand its three irreducible components have only the origin in common. Any isomorphism of algebraic varieties would permute irreducible components and therefore map the origin of $V$ to the origin of $W$.

Every defining polynomial $xy,xz,yz$ has zero differential at the origin, soThe defining polynomial of $W$ is homogeneous of degree three, so its differential also vanishes at the origin, givingAn isomorphism induces an isomorphism of Zariski tangent spaces, which is impossible here. Hence $V$ and $W$ are not isomorphic. This is the tangent-space obstruction between two unions of three lines.

Since $I_1\cap I_2\subseteq I_j$, every point of either $V(I_j)$ lies in $V(I_1\cap I_2)$. Conversely, supposeChoose $f_j\in I_j$ with $f_j(p)\ne 0$. Thenbut $(f_1{f}_2)(p)\ne 0$, so $p\notin V(I_1\cap I_2)$. Therefore

A point $p$ belongs to $V(I_1+I_2)$ exactly when every polynomial $f_1+f_2$, with $f_j\in I_j$, vanishes at $p$. This implies that every member of each ideal vanishes by taking the other summand to be zero, and the converse follows by addition. HenceThis is the first vanishing loci of ideal sums and intersections identity.

Subtracting the two defining equations givesConsequently the variety isEach member is irreducible. Indeed, after imposing its linear equation, its coordinate ring isThe invertible linear substitutionturns this intowhich is an integral domain. The two curves are distinct and neither contains the other, so they are precisely the irreducible components of two intersecting complex hyperbolas.