An estimator $T$ of a parameter or estimand $g(\theta )$ is an unbiased estimator if its expectation exists andUnbiasedness concerns the sampling mean of the estimator over repetitions, rather than equality to the target for any particular observed sample.

For any fixed $p>1/4$ below 1, the ratio to the null isIt increases with $y$. Its ordering is therefore the same for every such alternative, while the size is determined only by the null. The very same rule, reject when $Y\ge 4$, is most powerful against each alternative and hence is a uniformly most powerful test. At $p=1$ its power is 1, also optimal.

More generally, for $0<p_0<1$, the binomial likelihood ratio against any $p>p_0$ increases with $y$. Choose an integer cutoff $c$ so thatand reject for $Y>c$, randomize with probabilitywhen $Y=c$, and accept below. This gives exact size $\alpha$ and is a uniformly most powerful test for every positive integer $n$. Exact prescribed size may require this boundary randomization; a deterministic cutoff alone need not attain it. For $p_0=0$, reject all $Y>0$ and reject at $Y=0$ with probability $\alpha$, which is likewise optimal. At $p_0=1$ the stated one-sided alternative is empty. The cutoff generally depends on $p_0,n,\alpha$; there is no claim that one fixed cutoff serves every null parameter.

The likelihood ratio for the two binomial probabilities isstrictly increasing in $y$. By the Neyman-Pearson lemma, reject in the highest upper tail. Under the null,Hence reject exactly when $Y\ge 4$, with no randomization required. Its statistical power at the stated alternative is

Reading positive entries as directed edges, the communicating class decomposition isFor the first class, paths $1\to 5\to 8\to 9\to 1$ establish mutual reachability, and every transition from one of these states stays in this set. States 2 and 3 communicate, but state 2 can enter the first class and never return. States 4,6,7 communicate through $4\to 6\to 7\to 4$, but state 6 can enter state 2 with no route back. These paths also exclude any larger communicating class. In a finite Markov chain, the closed class is recurrent and the two open classes are transient.

For sufficiently large $n$, $p_n=\lambda /n$ lies in $(0,1]$. The waiting time has the geometric distribution on $1,2,\dots$, so summing its geometric series giveswhenever $(1-\lambda /n)e^{t/n}<1$. Fix $t<\lambda$. The expansions $e^{t/n}=1+t/n+O(n^{-2})$ show that the numerator is $\lambda /n+O(n^{-2})$ and the denominator is $(\lambda -t)/n+O(n^{-2})$. ThereforeThe convergence theorem for moment-generating functions says that pointwise convergence of finite moment-generating functions on a fixed open interval about zero to the moment-generating function of a probability distribution implies convergence in distribution to that distribution. For example take $|t|<\lambda /2$; all large $n$ satisfy the convergence condition throughout that interval. The limit is the moment-generating function from part (a), proving

The independent identically distributed central limit theorem states that for variables of common finite mean $\mu$ and finite positive variance ${\sigma }^2$,For the exponential distribution, $\mu =1/\lambda$ and $\sigma =1/\lambda$. Thus the event in question is exactlyThe limiting normal distribution has a continuous distribution function at every real $x$, so convergence in distribution givesHere $\Phi (x)=(2\pi {)}^{-1/2}{\int }_{-\infty }^x{e}^{-u^2/2}du$, including the strict-inequality convention because the limiting distribution is continuous.

The moment-generating function of a random variable $X$ is $M_X(t)=\mathbb{E}[e^{tX}]$ wherever this expectation is finite. Existence on an open interval containing zero is the condition needed for the usual uniqueness and convergence theorems. For the exponential distribution with rate $\lambda >0$,For $t\ge \lambda$ the integral diverges. Differentiating at zero gives mean $1/\lambda$ and variance $1/{\lambda }^2$.

Positive definiteness gives a positive smallest eigenvalue $c$ with $x^{t}Ax\ge c|x{|}^2$. Hence the set $x^{t}Ax=y^{t}Ay=1$ is compact, and $h$ attains a maximum there. Its two constraint gradients are $(2Ax,0)$ and $(0,2Ay)$; they are linearly independent because $x,y$ are nonzero and $A$ is invertible.

Put $s=x^{t}Ay$. At any constrained extremum the multiplier equations areMultiplication by $x^t$ and $y^t$ gives $\lambda =\mu =s^2$. If $s=0$ the objective is zero. Otherwise $Ay=sAx$, so $y=sx$ and normalization gives $s^2=1$. Thus every maximum is at most one, and the value one is attained at $x=y$. This proves $\boxed{(x^{t}Ay{)}^2\le 1}$ on the normalized constraints.

For nonzero arbitrary vectors apply that result to $x/\sqrt{x^{t}Ax}$ and $y/\sqrt{y^{t}Ay}$. If either vector is zero the conclusion is immediate. ThereforeEquality for nonzero vectors holds precisely when they are linearly dependent. This is the Cauchy-Schwarz inequality for the inner product induced by $A$.

Write the variables as $(u,v)\in {\mathbb{R}}^n\times {\mathbb{R}}^k$, with $a=(u_0,v_0)$. The hypothesis makes $D_{u}f(a)$ invertible. The implicit function theorem supplies neighbourhoods $U,V$ of $u_0,v_0$ and a unique smooth map $\varphi :V\to U$ such thatIts derivative is $D\varphi (v_0)=-(D_{u}f(a){)}^{-1}{D}_{v}f(a)$. Thus the derivative of this parametrization has image exactly $\ker Df(a)$.

Since $h$ has a constrained maximum at $a$, differentiating $v\mapsto h(\varphi (v),v)$ gives zero in every direction. Therefore $Dh(a)$ vanishes on $\ker Df(a)$. In finite-dimensional linear algebra, $(\ker Df(a){)}^{\perp }$ is the row space of $Df(a)$. ConsequentlyThe full-rank hypothesis is what permits this Lagrange multiplier conclusion; it is not a consequence of constrained maximality alone at a singular constraint.

A space is compact if every open cover has a finite subcover. The finite-intersection formulation says that every family of closed sets with the finite intersection property has nonempty total intersection.

A space is Hausdorff if distinct points have disjoint open neighbourhoods. In symbols, for $x\ne y$ there are open $U\ni x$ and $V\ni y$ with $U\cap V=\varnothing$.

True in both cases. For every open $U$, continuity of $f$ makes $f^{-1}(U)$ open. Thenis Lebesgue measurable by measurability of $g$. The value $f(0)$ has no bearing on this preimage argument.

False in both cases. Without the zero-at-zero condition, $g=0$ and $f\equiv 1$ already give an integrable input and a non-integrable composition on $\mathbb{R}$.

Even imposing $f(0)=0$ does not control how fast $f(t)$ approaches zero. Take $g(x)=(1+x^2{)}^{-1}$ and the continuous function $f(t)=\sqrt{|t|}$. The input has integral $\pi$, buthas a divergent integral, because its tails are comparable to $1/|x|$. Thus continuity at zero is weaker than the linear growth estimate that would preserve integrability.

Under the bounded-support convention, false in general, true if $f(0)=0$. For the first claim, take $g=0$ and $f\equiv 1$: the composition is one on all of $\mathbb{R}$, and cannot have bounded support. For the second, refine the finitely many defining intervals into disjoint intervals on which $g$ is constant. The composition is constant on each of them and vanishes outside their bounded union because $f(0)=0$. Hence it is another step function.

Under the unrestricted piecewise-constant convention, true in both cases: composition with any function preserves constancy on the same partition, including its unbounded cells. This also applies to the broader measurable-partition convention. Continuity of $f$ is not needed for this particular assertion.