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Put $r=1-\alpha \in (0,1)$. The variable $Y=X_1-\theta$ has density $r{y}^{r-1}$ on $(0,1)$, so $\mathbb{E}{Y}^j=r/(r+j)$. ThusThe unbiased estimator isTaking expectations verifies that its bias is zero.

Assume the zero set $A=\{s\in [0,t]:g(s)=0\}$ is measurable, as is ensured by measurability of $g$. This is an implicit necessary hypothesis: for an arbitrary real function the printed integral and event need not be measurable. Put $a={\int }_0^t{1}_A(s)ds$. Conditional on $N_t=n$, the uniform order statistics description makes the probability of no jump in $A$ equal to $(1-a/t{)}^n$. Averaging over the Poisson distribution givesConsequentlyIn particular a measure-zero zero set is almost surely missed, despite possibly containing infinitely many points.

A rate-$\lambda$ Poisson process has $N_0=0$, right-continuous paths with upward jumps of size one, stationary independent increments, and $N_t-N_s\sim \mathrm{Poisson}(\lambda (t-s))$ for $t\ge s$. Assume $\lambda >0$. Its successive waiting times are independent exponential variables of rate $\lambda$. For $0<t_1<\cdots <t_n<t$, the product of their densities and the probability of no further jump before $t$ isDivide by $\mathbb{P}(N_t=n)=e^{-\lambda t}(\lambda t{)}^n/n!$ to obtain the conditional joint densityBoundary inequalities do not affect a density. Equivalently these are the ordered values of $n$ independent uniform points of $[0,t]$. For $n=0$, the conditional tuple is empty.

No, under the absolute-integral convention. Direct integration givesOn a fixed positive-length portion of every period, the numerator is bounded below. Summing the resulting lower bounds gives a divergent harmonic series, so the Fourier transform is not in $L^1$. Also there is no continuous almost-everywhere representative: continuity would force it to be $1$ on $(0,1)$ and $0$ on $(-\infty ,0)$, giving incompatible limits at $0$. Symmetrically truncated conditional inversion does hold at interior continuity points and gives half the jump value at the endpoints; this does not contradict the answer.

Define $g$ by the inverse integral. Its absolute convergence gives $|g(x)|\le (2\pi {)}^{-d}\Vert \hat{f}{\Vert }_1$. If $x_n\to x$, the exponential factors converge pointwise and their differences are bounded by $2|\hat{f}|$. The dominated convergence theorem proves $g(x_n)\to g(x)$. Therefore $g$ is bounded and continuous and equals $f$ almost everywhere.

First ${\int }_{\mathbb{R}}{e}^{-x^2/2}dx=\sqrt{2\pi }$: square the positive integral and use polar coordinates to obtain $2\pi {\int }_0^{\infty }r{e}^{-r^2/2}dr=2\pi$. Scaling therefore normalizes the one-dimensional Gaussian function $g_t$.

Let $I(\xi )={\int }_{\mathbb{R}}(2\pi t{)}^{-1/2}{e}^{-x^2/(2t)}{e}^{-i\xi x}dx$. Differentiation under the integral is justified by the integrable majorant $|x|g_t(x)$. Since $g_t^{\prime }(x)=-x{g}_t(x)/t$, integration by parts, with zero endpoint terms, gives $I^{\prime }(\xi )=-t\xi I(\xi )$. With $I(0)=1$, this yields $I(\xi )=e^{-t{\xi }^2/2}$. Multiplication of the one-dimensional integrals givesApplying the same computation with reciprocal scale $1/t$ evaluates its inverse integral:For $f\in L^1$, Fubini's theorem gives $\widehat{f\ast g_t}=\hat{f}{e}^{-t|\xi {|}^2/2}$, which is integrable because $\hat{f}$ is bounded. Its inverse integral isAbsolute integrability of $|f(y)|e^{-t|\xi {|}^2/2}$ justifies this second application of Fubini's theorem. Thus Fourier inversion holds for every Gaussian convolution, in fact pointwise for this continuous representative.

Here Fourier inversion means the absolute-integral version: $\hat{f}\in L^1({\mathbb{R}}^d)$ andThis convention is necessary for the bounded continuous version in part (d). Conditional inversion by symmetric truncation is a different statement and need not have that conclusion.

For $f\in L^1({\mathbb{R}}^d)$, use the Fourier transform conventionThe integral is absolutely convergent for every $\xi$, and $|\hat{f}(\xi )|\le \Vert f{\Vert }_1$.

Homogenization gives $F=X^4+X^2{Y}^2+Y^2{Z}^2+Z^4$, withIf the characteristic is not $2$, a singular point with $Y=0$ would force $X=Z=0$, impossible. If $Y\ne 0$, then $X^2+Z^2=0$. If either $X$ or $Z$ vanishes, both vanish, giving $[0:1:0]$. Otherwise the other two derivative equations require $Y^2=-2X^2=-2Z^2$, whence $X^2=Z^2$ and $2X^2=0$, again impossible. ThusIn the chart $Y=1$, the quadratic term is $X^2+Z^2$, the product of two distinct linear factors over $k$, so this is an ordinary double point.

In characteristic $2$,All first derivatives of this equation vanish. The equation defines a nonreduced scheme, singular everywhere; it no longer satisfies the irreducible/reduced hypersurface premise. If “curve” means the associated reduced variety, it is the union of two distinct lines, and only their intersection $\boxed{[1:0:1]}$ is singular. These two interpretations must not be conflated.

The Jacobian criterion on an affine chart $x_j\ne 0$ says that a point of the projective hypersurface is singular exactly when all derivatives of its dehomogenized equation vanish there. The Euler homogeneous function theorem identity ${\sum }_i{x}_i{f}_{x_i}=df$ then recovers the omitted $j$th derivative on $X$, without dividing by $d$. Thus the correct statement in every characteristic isIf the printed $Z(I)$ denotes a zero locus inside $X$, this proves it as written. If it means the ambient projective zero locus, the assertion is false when the characteristic divides $d$. When the characteristic does not divide $d$, the Euler homogeneous function theorem forces $f=0$ at every common zero of the derivatives, and the intersection with $X$ is redundant.

For a genuine counterexample to the unrestricted ambient interpretation, in characteristic $p$ takeThis is an irreducible homogeneous polynomial: it is primitive and linear in $X_2$ over $k[X_0,X_1]$, with coprime coefficients. At $[1:0:0]$ all partial derivatives vanish, including when $p=2$, but $f=1$. This point is outside $X$. Hence the ambient-locus version needs the displayed intersection or a restriction on the degree.

A point $P$ belongs to the smooth locus of a variety if its local ring ${\mathcal{O}}_{X,P}$ is a regular local ring, meaningEquivalently, the Zariski tangent space has the expected local dimension. Over the algebraically closed field, this agrees with smoothness for a reduced variety. In affine coordinates, the Jacobian criterion expresses it by the rank of the matrix of first derivatives of defining equations.

Suppose $(x_k{)}_{k\ge 1}$ were a countable Hamel basis. Put $E_n=\mathrm{span}(x_1,\dots ,x_n)$. Each $E_n$ is a finite-dimensional, hence closed, linear subspace of $X$. It is proper because $X$ is infinite dimensional, and therefore a nowhere dense set. A Hamel basis means that every vector is a finite linear combination, so $X={\bigcup }_n{E}_n$. This contradicts the Baire category theorem for the Banach space $X$. Hence an infinite-dimensional Banach space has no countable Hamel basis. The same proof excludes any countable algebraic spanning set.

One form of the Baire category theorem is: a nonempty complete metric space cannot be a countable union of closed nowhere dense subsets. Equivalently, a countable intersection of open dense subsets is dense. A Banach space is a complete metric space for its norm metric, so this form applies directly to the next part.

Because $Y$ is closed, it is a nowhere dense set exactly when its interior is empty. If a ball about $y\in Y$ lay inside $Y$, translating it by $-y$ would put a ball about $0$ inside $Y$. Every vector of $X$ has a small scalar multiple in that ball, and closure under scalar multiplication would force $Y=X$. This contradiction proves a proper closed linear subspace is nowhere dense.