The final cell is a grouped tail. Thus use twelve probabilitiesMaximize the grouped Grouped Poisson likelihoodthen take $\boxed{{\hat{\pi }}_i=p_i(\hat{\theta })}$ for all twelve cells. For example, the interior likelihood equation isbecause $p_{11}^{\prime }(\theta )=p_{10}(\theta )$. It can be solved numerically, with no numerical work required here.

It would be incorrect to treat the last six observations as exactly eleven. If the underlying ungrouped particle counts were available, their sample mean would be the usual Poisson maximum likelihood estimate. For the data as supplied, the grouped likelihood handles the unknown tail counts directly.

For $t>0$, apply Markov's inequality to the nonnegative random variable $e^{t{S}_n}$. The exponential is then increasing, soAt $t=0$ the right side is one. For $t<0$ it exceeds one because $a>0$, so the same displayed bound remains true trivially; the upper-tail Markov argument itself uses $t>0$.

Complete the square $t^2/2-at=(t-a{)}^2/2-a^2/2$. Choosing $t=a$ gives the strongest bound from this family, $\mathbb{P}(S_n\ge an)\le e^{-n{a}^2/2}$. The distribution is symmetric, so the same bound holds for $S_n\le -an$. A union bound givesOne may also cap this at one. Its exponential decay is stronger than the polynomial Chebyshev bound for fixed $a>0$ and large $n$.

The two-point distribution gives $M(t)=(e^t+e^{-t})/2=\cosh t$. In the power series, $(2j)!={\prod }_{r=1}^j(2r)(2r-1)\ge 2^{j}j!$, so each coefficient satisfiesSumming the nonnegative terms proves $\boxed{M(t)\le e^{t^2/2}}$ for every real $t$. Part (a) therefore givesThis is the sub-Gaussian moment bound for a symmetric random sign.

Chebyshev's inequality states $\mathbb{P}(|Y-\mu |\ge r)\le {\sigma }^2/r^2$ for $r>0$ when $Y$ has mean $\mu$ and finite variance ${\sigma }^2$. Here each summand has mean zero and variance one, so independence gives $\mathbb{E}{S}_n=0$ and $\mathrm{Var}{S}_n=n$. Taking $r=an$ gives

For $n\ge 1$, the unit sphere is nonempty and compact by the Heine-Borel theorem. The function $k(x)=\Vert Ax{\Vert }^2$ is continuous and therefore attains a minimum and a maximum on it by the extreme value theorem. Either point supplies a nonzero eigenvector in part (ii), with a real multiplier. Thus $A^{T}A$ always has a real eigenvalue, indeed a nonnegative one. This gives existence without assuming the spectral theorem first.

On the constraint set, $\Vert x\Vert =1$, so $\nabla f(x)=2x\ne 0$. Part (a)(iv) and the Lagrange multiplier theorem give $2A^{T}Ax=2\lambda x$, henceSince $x$ is nonzero, it is an eigenvector. Moreover $\lambda =x^T{A}^{T}Ax=\Vert Ax{\Vert }^2\ge 0$, so the associated eigenvalue is real and nonnegative.

The Lagrange multiplier theorem for one regular constraint says: if $f,k$ are $C^1$ near $x$, $f(x)=0$, $\nabla f(x)\ne 0$, and $x$ is a local extremum of $k$ restricted to $f^{-1}(0)$, thenFor several constraints with linearly independent gradients, the objective gradient lies in their linear span. The regularity assumption rules out singular points where the usual multiplier conclusion can fail.

For an increment $h$, direct expansion and the chain rule giveFor example, $\Vert x+h{\Vert }^2-\Vert x{\Vert }^2=2x^{T}h+\Vert h{\Vert }^2$, so the quadratic remainder divided by $\Vert h\Vert$ tends to zero. The corresponding gradients are $\nabla g(x)=2x$ and $\nabla k(x)=2A^{T}Ax$; the derivative of $F$ is the constant matrix $A$.

If $F:U\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^m$ is differentiable at $x$, and $G$ is defined on a neighborhood of $F(x)$ and differentiable there, the chain rule statesIn standard coordinates the same statement isThe evaluation point in the first factor is $F(x)$, and the matrix order follows the order in which the linear maps act on increments.

The map is continuously differentiable if it is differentiable at every point of $U$ and the derivative map $x\mapsto D{F}_x$ is continuous. In finite dimensions this is equivalent to all first partial derivatives of all its components existing and being continuous on $U$. The class is denoted $C^1(U,{\mathbb{R}}^m)$.

The total derivative of $F$ at $x$ is the linear map $D{F}_x:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe Jacobian matrix is the $m\times n$ matrix $J_F(x)=(\partial F_i/\partial x_j)(x)$. When the total derivative exists, this is its matrix in standard bases: $D{F}_x(h)=J_F(x)h$. Existence of partial derivatives alone need not imply existence of the total derivative.

If $A$ is disconnected, take a separation $A=A_1\cup A_2$. Each piece is also relatively closed, being the complement of the other. They are compact as closed subsets of the compact space $A$.

Part (b), using the ambient Hausdorff property, gives disjoint ambient open neighborhoods of $A_1$ and $A_2$. They meet $A$, cover it, and are disjoint everywhere, as required. Conversely, such ambient open sets give a separation on intersecting with $A$, by part (ii). Hence compactness and the Hausdorff property permit fully disjoint ambient neighborhoods.

If $A$ is disconnected, a separation writes it as $A=A_1\cup A_2$ with $A_1,A_2$ disjoint, nonempty and open in its subspace topology. By definition there are ambient open sets $U,V$ with $A_1=U\cap A$ and $A_2=V\cap A$. These give exactly the four conditions stated.

Conversely, those conditions make $U\cap A,V\cap A$ two nonempty disjoint relatively open sets whose union is $A$. They are therefore a separation. This proves the equivalence. The ambient sets may overlap outside $A$, which is why the condition here is $U\cap V\cap A=\varnothing$.

A topological space is connected if it cannot be written as the union of two disjoint nonempty open subsets. Equivalently, its only subsets that are both open and closed are the empty set and the whole space. Under this definition the empty space is connected.

Suppose no $F_N$ is contained in $U$. Then $K_i=F_i\cap (X\setminus U)$ are nested nonempty closed subsets of the compact space. Part (ii) gives a point in their intersection. Such a point lies both in ${\bigcap }_i{F}_i$ and outside $U$, contradicting the hypothesis. Thus some $F_N\subseteq U$, and nesting givesCompactness upgrades containment of the limiting intersection to eventual containment of the whole nested sets.

If the intersection were empty, the open complements $X\setminus F_i$ would cover the compact space. A finite subcover would give an empty finite intersection of the corresponding $F_i$. But the sets are nested, so any finite intersection is the set with largest index, which is nonempty. This contradiction provesThis is the nested-closed-set form of the finite intersection property for compact spaces.

A topological space is compact if every open cover of it has a finite subcover. That is, if $X={\bigcup }_{i\in I}{U}_i$ with all $U_i$ open, finitely many of those same open sets already cover $X$.

Show that the subspace $C$ of convergent real sequences is closed in $X$. Suppose $a^{(m)}\in C$ converges uniformly in the coordinate index to $a\in X$. Given $ϵ>0$, choose $m$ so ${\sup }_j|a_j-a_j^{(m)}|<ϵ/3$. Since the fixed sequence $a^{(m)}$ converges, it is Cauchy, so for sufficiently large $j,k$ its coordinate difference is below $ϵ/3$. ThenThus $a$ is a Cauchy sequence of real numbers and converges by completeness of $\mathbb{R}$. Hence $a\in C$, so $C$ is closed. Part (a)(iii) proves the convergent-sequence subspace is complete.

Let $a^{(m)}=(1,1/2,\dots ,1/m,0,0,\dots )$. Each has finite support, and in the supremum norm metric it converges to $a=(1,1/2,1/3,\dots )$, because $d(a^{(m)},a)=1/(m+1)$.

The limit is bounded but does not have finite support. Thus the finite-support subspace is not closed in $X$, and by part (a)(iii) it is not complete. Equivalently, the displayed sequence is Cauchy in that subspace but has no limit there.

If $A$ is closed in the complete ambient metric space, any Cauchy sequence in $A$ converges in $X$ by completeness, and its limit belongs to $A$ by part (i). Thus $A$ is complete with its restricted metric.

Conversely, suppose $A$ is complete and a sequence $a_n\in A$ converges in $X$ to $x$. A convergent sequence is Cauchy, also in the restricted metric, so completeness of $A$ gives $a_n\to a$ for some $a\in A$. Uniqueness of a metric limit gives $x=a\in A$. Part (i) now makes $A$ closed. Hence