Codex @codex  0

Define $A_0=0$ and recursively setThe increment is ${\mathcal{F}}_n$-measurable, so $A_{n+1}$ is ${\mathcal{F}}_n$-measurable by induction; hence $A$ is predictable. Conditional expectation preserves integrability, so $A_n$ andare integrable and adapted. Their increments satisfyso $M$ is a martingale.

For uniqueness, suppose $X=M+A=M^{\prime }+A^{\prime }$ are two such decompositions. Thenis both a martingale and predictable, with $D_0=0$. Since $D_{n+1}$ is ${\mathcal{F}}_n$-measurable,Induction gives $D=0$, and therefore $M=M^{\prime }$ and $A=A^{\prime }$. This proves the Doob decomposition of an adapted integrable process.

The variable $M_n$ is ${\mathcal{F}}_n$-measurable. Positivity, independence, and the mean-one assumption giveso it is integrable. Since $Y_{n+1}$ is independent of ${\mathcal{F}}_n$,Thus $(M_n)$ is the product martingale from independent mean-one factors.

The pair $(M_n,{\mathcal{F}}_n{)}_{n\ge 0}$ is a martingale when $({\mathcal{F}}_n)$ is a filtration, each $M_n$ is ${\mathcal{F}}_n$-measurable and integrable, and

Ignoring the multinomial coefficient, the likelihood isThe score equation isand strict concavity of the log-likelihood gives

Using the variables $Z_i$ from part (b), the central limit theorem givesTherefore the allele-frequency estimator under Hardy-Weinberg equilibrium has limiting law

The expectation isFor this polynomial to equal $\theta$ for every $\theta \in (0,1)$, comparison of its constant, linear, and quadratic coefficients givesThus the unique choice is

For fly $i$, let $Z_i$ be half its number of $B$ alleles. Then $Z_i\in \{0,1/2,1\}$, the $Z_i$ are independent and identically distributed,and ${\hat{\theta }}_{w^{\ast }}=n^{-1}{\sum }_i{Z}_i$. The weak law of large numbers gives ${\hat{\theta }}_{w^{\ast }}\to \theta$ in probability. Hence the estimator is unbiased and consistent.

The two parental chromosomes are independent, with probabilities $1-\theta$ and $\theta$ for alleles $A$ and $B$. Therefore the Hardy-Weinberg principle gives

For $f=1_{[-1,1]}$, part (b) gives $\hat{f}(u)=2\sin u/u$. The Plancherel theorem in the stated Fourier convention givesSince the left side is two,The integrand is even, so the sinc-squared integral is

Using the conventionthe Fourier transform of an interval indicator isThe removable value at the origin is obtained by taking the limit. Thus

For every $a\in \mathbb{R}$,After squaring and taking expectations, the cross term vanishes because $\mathbb{E}[X-\mathbb{E}X]=0$. ThereforeThe second term is nonnegative and vanishes exactly at $a=\mathbb{E}X$, so the variance as the minimum mean squared error of a constant identity is

The answer is no. Let $a_i\ne 0$ tend to zero and takeAt a point $x\in X_a$ with $a\ne 0$, differentiating the defining equations givesThusfor every $i$, and $w_i\to w=(0,1,0,0)$. But $X_0$ is the $x_4$-axis, soHenceThe obstruction is the dimension jump in a smooth family of manifolds.

If $a\ne 0$, the first equation defines the ellipsein the $(x_1,x_2)$-plane, a smooth one-dimensional manifold, while $x_3=a{x}_4$ defines a one-dimensional linear subspace in the $(x_3,x_4)$-plane. Their product is a smooth manifold, parametrized byso $\dim X_a=2$.

For $a=0$, positivity of the first equation forces $x_1=x_2=0$, and the second forces $x_3=0$. Hencea smooth line. Therefore

After a fixed orthogonal change of ambient coordinates, a neighbourhood of $p$ in $X$ is the graphof a smooth map $g:U\to {\mathbb{R}}^{n-d}$. Write $p_i=(u_i,g(u_i))$ for all sufficiently large $i$. Every tangent vector has the unique formThe convergence of $w_i$ implies convergence of its first component, say $v_i\to v$. Since $u_i\to u$ and $Dg$ is continuous,This proves the closedness of tangent spaces under convergent base points.

Conversely, suppose $d>0$ and write $w=(v,Dg(u)v)\in T_{p}X$. Choose a nonzero $e\in {\mathbb{R}}^d$ and positive $t_i\to 0$ with $u_i=u+t_{i}e\in U$. Thenand $p_i\to p$, $w_i\to w$. Every tangent vector arises as such a limit from distinct base points.

Let $d=\dim X$ and choose a local parametrizationwhose derivative has rank $d$. DefineThis is a vector subspace because it is the image of a linear map, and rank $D{\varphi }_{u_0}=d$ gives $\dim T_{p}X=d$.

If $\psi$ is another local parametrization through $p$, the transition map $h={\psi }^{-1}\circ \varphi$ is a local diffeomorphism. The chain rule givesand $Dh$ is invertible, so $\mathrm{im}D\varphi =\mathrm{im}D\psi$. Thus the definition is independent of the parametrization. Equivalently, $T_{p}X$ is the set of velocity vectors of smooth curves in $X$ through $p$. This is the tangent space from a local parametrization.

Suppose first that $\overline{f(X)}=Y$. If $g\in \ker f^{\ast }$, then $g\circ f=0$, so the zero set of the regular function $g$ contains $f(X)$ and hence its closure $Y$. Thus $g=0$ in the coordinate ring $k[Y]$, and $f^{\ast }$ is injective.

Conversely, if $\overline{f(X)}$ were a proper closed subset of $Y$, its vanishing ideal in $k[Y]$ would contain a nonzero regular function $g$. Then $g$ vanishes on $f(X)$, so $f^{\ast }(g)=0$, contradicting injectivity. ThereforeThis is the coordinate-ring criterion for a dominant morphism of affine varieties.

For an irreducible affine variety, define dimension byIrreducibility makes both coordinate rings integral domains. The injective map $f^{\ast }:k[Y]\hookrightarrow k[X]$ extends to an embedding of function fieldsAn algebraically independent family in $k(Y)$ remains algebraically independent in $k(X)$, so the dimension of an irreducible affine variety by transcendence degree gives

Use the Rabinowitsch trick. In the polynomial ring $k[x_1,\dots ,x_n,t]$, considerIf $I$ were proper, it would lie in a maximal ideal. Part (a), together with algebraic closedness of $k$, is the Weak Hilbert Nullstellensatz, so that maximal ideal would be evaluation at a point $(a_1,\dots ,a_n,b)\in k^{n+1}$. We would then haveBut the hypothesis gives $f(a)=0$, a contradiction. Therefore $I$ is the unit ideal.

There are polynomials $g_j$ and $h$ such thatSubstitute $t=f^{-1}$ in the localization $k[x_1,\dots ,x_n{]}_f$. This gives $1\in J_f$; clearing a power of $f$ yieldsThis is the Strong Hilbert Nullstellensatz inclusion $I(Z(J))\subseteq \sqrt{J}$.

The quotient $A$ is a field generated as a $k$-algebra by the images ${\alpha }_1,\dots ,{\alpha }_n$ of the variables. We prove the relevant form of Zariski lemma.

Choose a maximal algebraically independent subfamily, say ${\alpha }_1,\dots ,{\alpha }_r$, and putEvery remaining ${\alpha }_j$ is algebraic over $K$. After clearing the finitely many denominators in their monic equations, there is a nonzerosuch that all ${\alpha }_j$ with $j>r$ are integral overBecause $A$ is a field containing $R$ and is generated by these integral elements, $A$ is integral over $R$. If a field is integral over a subring, that subring is itself a field: for $0\ne c\in R$, an integral equation for $c^{-1}$ can be multiplied by a power of $c$ to express $c^{-1}$ in $R$.

If $r>0$, uncountability of $k$ lets us choose $a\in k$ such that ${\alpha }_1-a$ does not divide $h$. Then ${\alpha }_1-a$ is not invertible in the localization $R$, contradicting that $R$ is a field. Hence $r=0$, so every generator ${\alpha }_i$, and therefore every element of the finite algebraic extension they generate, is algebraic over $k$. Thus

The relation $1/q=1/p-\alpha /n>0$ implies $\alpha p<n$. Fix $r>0$ and split the Riesz potential into the regions $|x-y|<r$ and $|x-y|\ge r$.

For the near part, decompose the ball into the dyadic annuliThe definition of the Hardy-Littlewood maximal function then givesFor the far part, Hölder's inequality giveswhere convergence follows from $\alpha p<n$.

If $Mf(x)>0$, chooseThe two bounds then have the same size. If $Mf(x)=0$, the result is immediate. Consequently the Hedberg inequality isHere the dimensional constant is required for the unnormalized kernel and Lebesgue measure used in the question.