The two possible ways to observe one ball of each colour have probabilitiesConditioning on their union therefore gives

The prior probabilities of choosing the two urns are equal. Their white-ball probabilities areBy Bayes' theorem,

The covariance matrix isIf $\mathrm{Cov}(X_1,X_2)=0$, the moment-generating function in part (ii) factors into the product of the two marginal moment-generating functions. Thus $X_1$ and $X_2$ are independent random variables; this is the independence of uncorrelated jointly normal variables.

By bilinearity of covariance,ChooseThen $(X_1,Y)$ is bivariate normal by part (i) and has zero covariance, so $Y$ is a normal random variable independent of $X_1$. Hence $X_2=a{X}_1+Y$.

For every $\lambda \in {\mathbb{R}}^2$, the scalar ${\lambda }^{T}X$ is normal with mean ${\lambda }^T\mu$ and variance ${\lambda }^{T}V\lambda$. The moment-generating function of a normal distribution therefore givesThus $\mu$ and $V$ determine the moment-generating function on all of ${\mathbb{R}}^2$. By the uniqueness theorem for moment-generating functions, they uniquely determine the probability distribution of $X$.

On the first and third coordinates, the identity $SJS=J$ from part (b) gives $e^{xB}C{e}^{xB}=C$; the second coordinate is also unchanged. Thus

The rotation matrix $e^{xA}$ commutes with $C$ and is orthogonal, soSince $C^n=\mathrm{diag}(1,(-1{)}^n,(-1{)}^n)$,

The matrix $B$ fixes the second coordinate and satisfies the hyperbolic-generator relation on the first and third coordinates. Therefore

The matrix $A$ fixes the first coordinate and restricts to one half of the rotation generator on the last two coordinates. Hence

The vector triple product identity givesTaking the dot product with $\mathbf{a}\times \mathbf{b}$ therefore yields

Using the Levi-Civita symbol and the epsilon-delta identity,This proves the vector triple product identityApplying it inside the scalar triple product gives Lagrange's identity

Here ${\omega }_4=i$. Splitting into even and odd pairs giveswhose two-point inverse discrete Fourier transforms are $(2,0)$ and $(-2,0)$. The butterfly therefore yields

Put $\Delta r=r_2-r_1$. Expanding the quotient $S/T$ consistently to first order givesThusBecause $\log (r_2/r_1)/(r_2-r_1)\to 1/r_2$ as $r_1\to r_2$,This local limit is the expected speed of a massless particle measured in a freely falling local inertial frame; the finite-interval effective velocity combines nonlocal length and clock comparisons.

A radial null geodesic satisfiesso for the outgoing rayThe coordinate-time interval is thereforeThe stationary observer at $r_2$ measures proper time $dT=\sqrt{1-2M/r_2}dt$. Hence

Along the radial curve at fixed $t,\theta ,\phi$, the spatial line element is $dS=\sqrt{1+2M/r}dr$. To first order,so

At $B=0$,with eigenvalues ${\lambda }_{+}=a+b=2\cosh (\beta J)$ and ${\lambda }_{-}=a-b=2\sinh (\beta J)$. Diagonalizing givesSubstitution in part (i), together with $Z={\lambda }_{+}^N+{\lambda }_{-}^N$, givesSince ${\lambda }_{-}/{\lambda }_{+}=\tanh (\beta J)$, the required One-dimensional Ising correlation function is

Insert $s_1{s}_i$ into the spin-basis expansion of the trace. Summing the spins along the first path from site $1$ to site $i$ produces $(M^{i-1}{)}_{st}$, while summing along the complementary path around the periodic chain produces $(M^{N-i+1}{)}_{ts}$. Therefore the two-point correlation function is

Assuming $\mu$ is $T$-invariant, $T$ is an ergodic transformation when every measurable set invariant modulo null sets is trivial:Equivalently, the invariant sigma-algebra is trivial modulo null sets.

A probability measure $\mu$ is $T$-invariant whenEquivalently, the pushforward measure satisfies $T_{\ast }\mu =\mu$, so $T$ is a measure-preserving transformation.

For an example satisfying $(\ast )$, consider the incomplete surface z equals r to three halvesA meridian of a surface of revolution reaches the missing origin in the finite lengthso $S_{\ast }$ is geodesically incomplete. The curvatures of a parametrized surface of revolution giveBy constant speed of an affinely parametrized geodesic, a geodesic with a finite endpoint has finite length and converges in ${\mathbb{R}}^3$. The finite-dimensional continuation criterion rules out an endpoint in a compact subset of $S_{\ast }$, and infinity is infinitely far away. It must approach the origin, so $(\ast )$ holds.

The answer to the second question is no. The punctured circular coneis incomplete because a generator reaches the missing vertex in finite time. It is inextendible: a proper connected smooth extension would have to add the origin, but the tangent planes approaching the origin depend on $\theta$ and therefore cannot be the continuous tangent planes of a smooth surface. Yetby the Gaussian curvature of a cone away from its vertex. Thus this incomplete inextendible surface does not satisfy $(\ast )$.

Suppose that a connected smooth surface $\tilde{S}\subset {\mathbb{R}}^3$ properly contains $S$. Since $S$ and $\tilde{S}$ have the same dimension, $S$ is open in $\tilde{S}$. Choose a point of its relative boundary. Inside a sufficiently small compact normal neighbourhood in $\tilde{S}$, choose $q\in S$ and a closest point $p\in \tilde{S}\setminus S$. A minimizing $\tilde{S}$-geodesic from $q$ to $p$ stays in $S$ before its endpoint, since encountering the complement earlier would contradict the choice of $p$.

The restriction of this curve is consequently a geodesic of $S$ with a finite maximal endpoint at $p$. On the other hand, the Gaussian curvature of $S$ is the restriction of the smooth Gaussian curvature of $\tilde{S}$, so it remains bounded near $p$. This contradicts condition $(\ast )$. Henceso $S$ is an inextendible embedded surface. This is the curvature-blowup criterion for inextendibility of a surface.