Unbiasedness of an estimator means unbiasedness for every parameter value, not just one chosen $\theta$. Thus $DX\theta =\theta$ for every $\theta \in {\mathbb{R}}^p$. Evaluating this identity on a basis, or identifying the two linear maps, gives

All entries must lie between zero and one. The tightest restriction comes from $p_{01}=\lambda$, so together with the given positivity the parameter range is

The interarrival times are independent and identically distributed with exponential distribution of rate $\lambda$: for any finite $k$ their joint density is ${\lambda }^k\exp (-\lambda {\sum }_{j=1}^k{x}_j)$ on $x_j>0$. In particular

A Poisson process of rate $\lambda >0$ is a counting process with $N_0=0$, independent stationary increments, and $N_t-N_s\sim \mathrm{Poisson}(\lambda (t-s))$ for $0\le s<t$, with the usual right-continuous unit-jump sample paths.

The moment-generating function is $M_X(\theta )=\mathbb{E}(e^{\theta X})$ for those real $\theta$ where this expectation is finite. When it exists on a neighbourhood of zero it determines the distribution, and its derivatives at zero give the corresponding moments.

For $x\ne y$, apply the scalar mean value theorem to $h(t)=⟨f(x+t(y-x)),z⟩$ on $[0,1]$. It is continuous on the closed interval and differentiable inside by the stated segment and regularity assumptions. The chain rule givesThus $h(1)-h(0)=h^{\prime }(\theta )$ for some $\theta \in (0,1)$, precisely the asserted identity. If $x=y$, both sides are zero when the zero-direction derivative is naturally taken as $Df(x)0$.

For the final claim, all partial derivatives tending to zero imply $\Vert \nabla g(x)\Vert \to 0$ away from the origin. For a nonzero small vector $h$, the one-variable function $t\mapsto g(th)$ is continuous on $[0,1]$ by continuity at zero and differentiable on $(0,1)$. Thereforefor some $\theta \in (0,1)$. For any $\epsilon >0$, if $\Vert h\Vert$ is sufficiently small, the entire punctured segment has gradient norm below $\epsilon$. Hence $|g(h)-g(0)|\le \epsilon \Vert h\Vert$, provingThis argument does not presuppose differentiability at the origin.

The $i$th partial derivative is ${\partial }_{i}f(x_0)={\lim }_{t\to 0}[f(x_0+t{e}_i)-f(x_0)]/t$. The directional derivative in direction $v$ is the same limit with $v$ in place of $e_i$, using the specified vector rather than normalizing it. Differentiability means that there is a linear map $L:{\mathbb{R}}^n\to \mathbb{R}$ withFor $v=(v_1,v_2)$ with $v_1\ne 0$, the difference quotient at the origin is $t{v}_2^3/v_1$, tending to zero. If $v_1=0$, the function is identically zero on the directional line, again giving derivative zero. Thus every directional derivative exists and is zero.

Along the curved path $(x,y)=(t^3,t)$, however, $f(t^3,t)=1$ for $t\ne 0$, while $f(0,0)=0$. Hence the function is not continuous at the origin. Differentiability implies continuity, so it is not differentiable there either. Direction-by-direction limits alone do not control general approaches to a point.

The difference map $(z,w)\mapsto z-w$ is continuous, and $X$ is its preimage of the open set $\mathbb{C}\setminus \{0\}$. Thus $X$ is open in the product.

The class of $(z,w)$ is $C_z=\{z\}\times (\mathbb{C}\setminus \{z\})$. A punctured plane is path-connected: two points can be joined by a polygonal path avoiding the deleted point. Consequently every class is connected.

The projection $\pi$ is onto, since $(z,z+1)\in X$ for every $z$. It is continuous as a restricted coordinate projection. If $U$ is open in $X$, it is open in the product because $X$ itself is open. For a point $(z,w)\in U$, the product-topology basis gives open sets $V,W$ with $(z,w)\in V\times W\subseteq U$. Then $V\subseteq \pi (U)$. Hence $\pi (U)$ is open.

The map is constant exactly on the equivalence classes, so it induces a bijection $\overline{\pi }:X/R\to \mathbb{C}$. It is continuous by the defining property of the quotient topology. It is also open: for quotient-open $V$, $\overline{\pi }(V)=\pi (q^{-1}(V))$ is open. ThusThe plane is connected, and the classes are connected, so part (b) now yields $X$ is connected.