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Combining the independent standard normal variable $\sqrt{n}(\overline{X}-\mu )/\sigma$ with the independent chi-squared variable $S_{XX}/{\sigma }^2$ givesthe Student's t-distribution with $n-1$ degrees of freedom.

The sample mean isWrite the Gaussian sample vector as its orthogonal projection onto the span of $(1,\dots ,1)$ plus its projection onto the orthogonal complement. These two Gaussian projections are independent. Their squared standardized lengths give Cochran's theorem:

Let $x_1,\dots ,x_N$ be the distinct real roots of ${\mathrm{He}}_N$. The Gauss-Hermite quadrature rule iswhere $w_j$ is the weighted integral of the $j$th Lagrange interpolation polynomial. Equivalently,The orthogonality of ${\mathrm{He}}_N$ makes the rule exact for every polynomial of degree at most

The probabilists' Hermite polynomials satisfywhich follow by differentiating the Rodrigues formula. Thusso ${\alpha }_n=0$ and ${\beta }_n=n$.

For the squared norm, substitute the Rodrigues formula and integrate by parts $n$ times. All boundary terms vanish under the Gaussian weight, and ${\mathrm{He}}_n^{(n)}=n!$, giving

Inductively, $p_n$ is monic of degree $n$, because the leading term of $p_{n+1}$ comes only from $x{p}_n$. Assume $p_0,\dots ,p_n$ are mutually orthogonal. The definitions giveUsing the preceding recurrence in the formgivesso $⟨p_{n+1},p_{n-1}⟩=0$ by the definition of ${\beta }_n$. If $j\le n-2$, then $x{p}_j$ lies in the span of $p_{j-1},p_j,p_{j+1}$, all orthogonal to $p_n$, and the remaining recurrence terms are also orthogonal to $p_j$. Hence $⟨p_{n+1},p_j⟩=0$. Induction proves that the recurrence defines monic orthogonal polynomials.

Bilinearity and symmetry follow from the corresponding properties of the integral. Since $w(x)>0$,If this integral is zero, the nonnegative continuous integrand must vanish everywhere, so $f=0$. Therefore the displayed formula is a real inner product on $C[a,b]$.

If $K<0$ everywhere, compactness makes the integral negative, so $\chi (S)<0$. For every triangulation, $3F=2E$ again yields the general identitySince $V\ge 1$ and $\chi (S)<0$,This supplies bounds depending only on the surface.

A geodesic triangulation of a closed smooth surface decomposes it into finitely many triangular discs whose edges are geodesic arcs and whose intersections are common faces. Its Euler number isThe Gauss-Bonnet theorem for a closed surface statesFor any triangulation of a closed surface, each triangular face has three edges and every edge belongs to two faces, so $3F=2E$. The sum of all vertex valencies is $2E$. Hence the average valency isThe torus has Euler characteristic $\chi =V-E+F=0$. Using $F=2E/3$ gives $V-E/3=0$, so $E=3V$ and

A local form of the Picard-Lindelöf theorem is as follows. Let $F(t,y)$ be continuous on a rectangle about $(t_0,y_0)$ and uniformly Lipschitz in $y$ there. Then some interval about $t_0$ admits a unique continuously differentiable solution ofThe solution is the unique fixed point of the associated integral operator. If $F$ is continuous on $[a,b]\times {\mathbb{R}}^n$ and globally Lipschitz in $y$ with one constant, the solution exists uniquely throughout $[a,b]$.

Factor the equation in the Eisenstein integers:Let $\pi =1-\omega$. Its norm is $3$, andso $3$ is a unit times ${\pi }^2$. Suppose $x\equiv 2(\mathrm{mod}3)$, say $x=3k+2$. Since $\omega =1-\pi$,The first term is divisible by ${\pi }^2$ and the second has discrete valuation exactly one, so $v_{\pi }(x+\omega )=1$. Conjugation gives $v_{\pi }(x+{\omega }^2)=1$ as well. ConsequentlyBut $v_{\pi }(y^3)=3v_{\pi }(y)$ is divisible by three, a contradiction. Hence

Every Euclidean domain is a principal ideal domain: in a nonzero ideal choose an element of least positive Euclidean value and divide every other element by it; minimality forces every remainder to vanish. Every principal ideal domain is a unique factorization domain: the ascending-chain condition supplies factorizations into irreducibles, and Euclid lemma follows because every irreducible generates a prime ideal. Therefore

The Eisenstein integers are $\mathbb{Z}[\omega ]$, where ${\omega }^2+\omega +1=0$. Their multiplicative algebraic norm isGiven $\alpha ,\beta \ne 0$, choose an Eisenstein integer $q$ nearest to the complex number $\alpha /\beta$. The hexagonal lattice has covering radius $1/\sqrt{3}$, soWith $r=\alpha -q\beta$, multiplicativity givesThus division with remainder decreases the nonnegative integer norm, proving that $\boxed{\mathbb{Z}[\omega ]\text{ is a Euclidean domain}}$.

For any action $a$, decompose the posterior expected quadratic loss around the posterior mean:The unique minimizer is therefore the Bayes estimator under squared error loss. Using the mean of the posterior Beta distribution givesUnder repeated sampling at parameter value $\theta$, $S$ has the binomial distribution with expected value $n\theta$. Consequentlyso its bias of an estimator is $(1-2\theta )/(n+2)$.

The quadratic loss, also called squared-error loss, for reporting an action $a$ when the parameter is $\theta$ is

Let $S={\sum }_{i=1}^n{X}_i$ be the number of heads. The likelihood function for the Bernoulli distribution sample is proportional to ${\theta }^S(1-\theta {)}^{n-S}$. A uniform distribution prior on $[0,1]$ is the $\mathrm{Beta}(1,1)$ distribution, so Beta-binomial conjugacy gives the posterior distribution