Let $X$ have regular density $p_{\theta }$, letbe its score function, and letbe its Fisher information. Regularity gives the mean-zero score identity ${\mathbb{E}}_{\theta }{S}_{\theta }=0$.

Suppose an estimator $T$ has mean ${\mathbb{E}}_{\theta }T=\psi (\theta )$. Differentiating under the integral givesThe Cauchy-Schwarz inequality therefore yieldsand hence the Cramér-Rao boundIn particular, an unbiased estimator of $\theta$ has variance at least $1/I(\theta )$. For an independent sample, the information is the sum of the individual informations.

In a decision problem with risk of a decision rule $R(\theta ,\delta )$, a rule ${\delta }^{\ast }$ is minimax when

Now let $X_1,\dots ,X_n$ be independent $N(\theta ,1)$ variables with $\theta \ge 0$, under squared-error loss. The sample mean is unbiased with variance $1/n$, sofor every $\theta \ge 0$. Thus the minimax value is at most $1/n$.

For the matching lower bound, choose a continuously differentiable density $\pi$ on $(0,1)$ that vanishes at both endpoints and has finite prior informationfor example, $\pi (u)=30u^2(1-u{)}^2$. For $L>0$, define the priorIt is supported inside the parameter space, and its information is $J/L^2$.

Here the likelihood score iswhose Fisher information is $n$. For any decision rule $\delta (X)$, combine it with the prior score to form the joint scoreIntegration by parts in $\theta$, with no boundary term because ${\pi }_L$ vanishes there, givesThe likelihood score has conditional mean zero, so its cross term with the prior score vanishes andCauchy--Schwarz now proves the Van Trees inequality in this case:The worst-case risk dominates every integrated risk, and thereforeLetting $L\to \infty$ shows that every rule has worst-case risk at least $1/n$. Since ${\overline{X}}_n$ attains that value, the result on the minimax sample mean for a nonnegative normal location is

Suppose first that the affine variety $V$ is irreducible. If $fg\in I(V)$, thenBoth pieces are Zariski closed in $V$. Irreducibility forces one to equal $V$, so either $f\in I(V)$ or $g\in I(V)$. Thus $I(V)$ is a prime ideal.

Conversely, suppose that $I(V)$ is prime and that $V=A\cup B$ with $A,B$ closed in $V$. If both were proper, the Hilbert Nullstellensatz would giveChoose $f\in I(A)\setminus I(V)$ and $g\in I(B)\setminus I(V)$. Their product vanishes on $A\cup B=V$, so $fg\in I(V)$, contradicting primality. Hence one piece equals $V$. This proves the criterion of irreducibility and prime vanishing ideal:

For the semicubical parabolaconsiderCertainly $(y^2-x^3)\subseteq \ker \theta$. Divide any polynomial by the monic polynomial $y^2-x^3$ as a polynomial in $y$; its remainder is $a(x)y+b(x)$. If this remainder lies in $\ker \theta$, thenThe first summand contains only odd powers of $t$ and the second only even powers, so both vanish and $a=b=0$. ThereforeThe latter is a subring of the integral domain $k[t]$, hence is a domain. Thus $(y^2-x^3)$ is prime and

An affine variety of dimension $d$ is smooth at $p$ when its Zariski tangent space has dimension $d$; equivalently, the Jacobian criterion requires the Jacobian of its defining ideal to have rank $n-d$. For this plane curve, put $F(x,y)=y^2-x^3$. Its gradient isAt $(0,0)\in C$ the gradient vanishes, so the tangent space is all of $k^2$ and has dimension two rather than the curve dimension one. Hence the result on the coordinate ring and singularity of the semicubical parabola gives

Let $F:X\to Y$ be a nonconstant holomorphic map of degree $d$ between compact connected Riemann surfaces. At $p\in X$, let $e_p$ be the ramification index of a holomorphic map. Choose a triangulation of $Y$ containing every branch value among its vertices, and lift it to $X$.

If the target triangulation has $V$ vertices, $E$ edges, and $T$ triangles, then the lifted triangulation has $dE$ edges and $dT$ triangles, because no branch value lies in an edge or face interior. For a target vertex $v$, local degree counting givesConsequently the number of vertices above $v$ isWriting $R={\sum }_{p\in X}(e_p-1)$, the Euler characteristic is thereforeSince a compact orientable surface of genus $g$ has Euler characteristic $2-2g$, this Triangulation proof of the Riemann-Hurwitz formula givesor equivalently

Now extend a cubic polynomial $f:\mathbb{C}\to \mathbb{C}$ to a degree-three holomorphic mapof the Riemann sphere. Both genera are zero, so Riemann--Hurwitz givesThe point at infinity is totally ramified with $e_{\infty }=3$, contributing two. Thus the finite points contribute exactly two. There are consequently two cases.

If there is one finite ramification point $\alpha$, it has index three. Choose an affine source map $h$ sending zero to $\alpha$. Thenfor some $c\ne 0$, and integration givesAn affine target map subtracts $d$ and rescales by $3/c$, yielding $z^3$.

Otherwise there are two distinct finite ramification points, each of index two. Choose an affine source map $h$ sending $-1$ and $1$ to them. The derivative of $f\circ h$ is then a nonzero multiple of $z^2-1$:HenceThe affine target map $g(w)=(w-d)/c$ givesTherefore the affine normal forms of a complex cubic polynomial are precisely