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Use the usual state-space assumption that the prior $X_0$ is independent of future process and observation noises. Suppose inductively $X_t\mid {\mathcal{Y}}_t\sim N({\hat{X}}_t,V_t)$. Prediction through the linear dynamics gives $X_{t+1}\mid {\mathcal{Y}}_t\sim N(A{\hat{X}}_t,M_t)$, where $M_t=A{V}_t{A}^T+{\Sigma }_{\epsilon }$. The pair $(X_{t+1},Y_{t+1})$ is conditionally multivariate normal, with means $A{\hat{X}}_t,CA{\hat{X}}_t$, cross covariance $M_t{C}^T$, and observation covariance $R_t={\Sigma }_{\eta }+C{M}_t{C}^T$. Applying part (a) gives the Kalman filterThis proves the inductive Gaussian assertion. The inverse requires nonsingular innovation covariance, as in the printed formula. Importantly, prediction uses $A{\hat{X}}_t$: the TeX conversion drops this $A$ in one place, whereas the original PDF includes it.

Assume $V_{YY}$ is invertible. PutThis and $Y$ are jointly multivariate normal, with $\mathrm{Cov}(R,Y)=0$ and $\mathrm{Cov}(R)=V_{XX}-V_{XY}{V}_{YY}^{-1}{V}_{YX}$. Their joint Gaussian characteristic function factors when the cross covariance vanishes, so they are independent. Conditioning on $Y=y$ therefore leaves the law of $R$ unchanged, proving

Passing to the deterministic-service limit in part (c), near zero, givesDifferentiation at zero gives $\mathbb{E}B=(1+\lambda \mathbb{E}B)/\mu$. Alternatively, the same branching expectation equation gives $\mathbb{E}{B}_k=1/(\mu -\lambda )$ for every $k$, because $\mathbb{E}{S}_k=1/\mu$. The assumed convergence of finite moment-generating functions at a positive point near zero bounds exponential moments and makes these means uniformly integrable, so the means pass to the limit. Since $T=A+B$, the answers areThey correspond to an M/D/1 queue with deterministic service time $1/\mu$. The calculation only needs finite transforms near zero; the literal assumption of finite transforms for every $\theta <\lambda$ is stronger than necessary and is not generally true for positive busy-period arguments. For instance, with $k=1$ and $\lambda =\mu /2$, part (c) gives $\lambda b^2-(\mu +\lambda -\theta )b+\mu =0$. Its discriminant is negative for $(\sqrt{\mu }-\sqrt{\lambda }{)}^2<\theta <\lambda$, so no finite busy-period transform can exist there.

The first arrival time $A$ has an exponential distribution of rate $\lambda$ and is independent of the following busy period. ThusSubstitute ${\varphi }_{B_k}=(\lambda -\theta ){\varphi }_{T_k}/\lambda$ into part (c). Its service-transform argument becomes $\theta +\lambda ({\varphi }_{B_k}-1)=(\lambda -\theta )({\varphi }_{T_k}-1)$. ThereforeAs before, positive arguments must lie in the domain of finite moment-generating functions.

For the busy-period branching equation, during the first service of a busy period, every arriving customer initiates a subtree of further work. The independence and stationary increments of the Poisson process, together with independent services, give the distributional recursionwhere conditional on $S_k=s$, $N(S_k)$ is Poisson with mean $\lambda s$, and the descendant busy periods are independent copies of $B_k$. One may explore these subtrees in order; the server's total workload is unchanged by that ordering. Consequently, wherever the moment-generating functions are finite,For $\lambda <\mu$, this identity in particular holds in a sufficiently small neighbourhood of zero.

An M/G/1 queue has Poisson process arrivals of rate $\lambda$, independent identically distributed service times of law $G$, one server, and here infinite waiting capacity. Use first-come-first-served service. Let $X_n$ be the number left just after the $n$th departure. During the next service, let $A_n$ be the number of arrivals. ThenConditional on the next service time $S_n=s$, $A_n$ is Poisson with mean $\lambda s$. If $X_{n-1}=0$, the server waits for one arrival to start service; that arrival is the customer being served, and is not counted in $A_n$. This constructs the embedded Markov chain.

Independence and the moment-generating function of an exponential distribution giveFor each fixed $\theta$, taking logarithms and using $-\log (1-x)\sim x$ yields ${\varphi }_{S_k}(\theta )\to e^{\theta /\mu }$. The limiting service time is the constant $S=1/\mu$. Its moment-generating function is exactly this limit.

For the anticlockwise boundary of the region, the inward normal is the left normal, so the signed curvature obeys the turning tangent theorem ${\int }_0^{L}k(s)ds=2\pi$. Therefore some $s_0$ has $k(s_0)\le 2\pi /L$. The enclosed area satisfies $A\ge d_1{d}_2$, and the planar isoperimetric inequality gives $L\ge 2\sqrt{\pi A}$. ConsequentlyThis uses the interpretation of the curve as a simple boundary traversed once; the inward normal and enclosed region in the question have that meaning.

The Fenchel theorem states that every closed regular space curve has total curvature at least $2\pi$:Equality holds precisely for a convex plane curve, traversed once.

For a regular simple closed plane curve of length $L$ enclosing area $A$, the planar isoperimetric inequality is $L^2\ge 4\pi A$, with equality precisely for a circle.

The periodic Wirtinger inequality states that an absolutely continuous $L$-periodic function $h$ of mean zero with square-integrable derivative satisfieswith equality only for a linear combination of $\sin (2\pi s/L)$ and $\cos (2\pi s/L)$. Parametrize the curve by arc length and translate its position vector $r=(x,y)$ so that both coordinate means vanish. Then $\int |r{|}^2\le L^3/(4{\pi }^2)$, since $|r^{\prime }|=1$. Green's theorem and the Cauchy-Schwarz inequality giveEquality requires a first-harmonic parametrization and equality in Cauchy-Schwarz; its radius vector is a constant multiple of the rotated unit tangent. It is therefore a circle. Conversely, a circle attains equality.

Euler's criterion gives $\chi (-1)=(-1{)}^{(p-1)/2}=1$ in this case. There are equally many nonzero quadratic residues and quadratic nonresidues, so ${\sum }_{n=1}^{p-1}\chi (n)=0$. Replacing $n$ by $p-n$ and using the multiplicativity of the Legendre symbol givesConsequently $\boxed{S=0}$.

Put $h=(p-1)/2$. The nonzero quadratic residues in the finite field ${\mathbb{F}}_p$ form a subgroup of size $h$: the group homomorphism $x\mapsto x^2$ has kernel $\{1,-1\}$. Each square satisfies $u^h=1$ by Fermat's little theorem. Since a nonzero polynomial of degree $h$ has at most $h$ roots of a polynomial, these are all the roots of $X^h-1$. For any other nonzero $u$, Fermat's theorem gives $(u^h{)}^2=1$, so $u^h=-1$. Thus Euler's criterion follows, and multiplication givesBoth sides are $1$ or $-1$, which are distinct modulo the odd prime number $p$, so $\chi (mn)=\chi (m)\chi (n)$. Taking $n=-1$ in Euler's criterion gives $\chi (-1)=(-1{)}^{(p-1)/2}$, which is $1$ for $p\equiv 1(\mathrm{mod}4)$ and $-1$ for $p\equiv 3(\mathrm{mod}4)$.

Each $H$-conjugacy class represented by $x_i$ has $|H|/|C_H(x_i)|$ elements. For every element of that class, the number of conjugators from $g$ to it is $|C_G(g)|$. Grouping the induced character formula by these classes therefore gives

In the example, $a$ has order four, $b$ has order two, and $bab=a^{-1}$, so $H\cong D_8$. Its five conjugacy classes are $1$, $a^2$, $\{a,a^3\}$, $\{b,a^{2}b\}$ and $\{ab,a^{3}b\}$, of sizes $1,1,2,2,2$. The four one-dimensional irreducible characters are determined by $\psi (a)=\epsilon$ and $\psi (b)=\eta$, each sign independently chosen. The remaining two-dimensional irreducible character comes from the plane action of the square:Their squared degrees sum to eight, so this is the full character table.

In $S_4$, use class representatives $1,(12),(12)(34),(123),(1234)$. Their intersections with $H$ are respectively $\{1\}$, $\{b,a^{2}b\}$, $\{a^2\}\cup \{ab,a^{3}b\}$, the empty set, and $\{a,a^3\}$. The corresponding $S_4$ centralizer orders are $24,4,8,3,4$, while the noncentral $H$ centralizers have order four. The induced values are consequentlyIn the same $S_4$ class order, all five induced irreducible characters are

In the induced character formula above, there are no $x\in G$ with $x^{-1}gx\in H$ when the conjugacy class of $g$ misses $H$. The sum is empty, so $({\mathrm{Ind}}_H^G\psi )(g)=0$.

No. In the six-cycle with cyclic vertices $1,2,\dots ,6$, every edge belongs to one of the two alternating perfect matchings. However, the disjoint edges $12$ and $45$ cannot belong to a common perfect matching: the remaining vertices $3,6$ are not adjacent. Taking $X=\{1,3,5\}$ and $Y=\{2,4,6\}$ gives the required bipartite graph counterexample.

Take a disjoint union of $n$ single edges, each joining one vertex of $X$ to one of $Y$. Its unique matching in a graph covering $X$ contains every edge, while any singleton $A\subset X$ satisfies $|\Gamma (A)|=|A|=1$. For $n\ge 2$ this $A$ is nonempty and proper. This illustrates why connectedness was needed in part (ii).

Suppose a nonempty proper $A$ is tight. In every matching in a graph covering $X$, the vertices of $A$ use every vertex of $\Gamma (A)$. Hence no edge from $X\setminus A$ to $\Gamma (A)$ could occur in such a matching. The assumption that every edge occurs therefore rules out all such edges. By definition there are also no edges from $A$ to $Y\setminus \Gamma (A)$. This separates $A\cup \Gamma (A)$ from the rest, which contains the nonempty set $X\setminus A$, contradicting connectedness. Hall's inequality must therefore be strict on every nonempty proper $A$.

Fix any edge $xy$. After deleting $x$ and $y$, every nonempty subset $A$ of $X\setminus \{x\}$ satisfies $|\Gamma (A)\setminus \{y\}|\ge |A|$, by the strict inequality in the question. The Hall marriage theorem gives a matching in a graph for the remainder; adding $xy$ gives a matching covering $X$ and containing the prescribed edge. The case $|X|=1$ is immediate.

No such map exists. If $H$ fixed the boundary, it would satisfy the hypothesis of part (iii) for every $\epsilon >0$. Choose $0<\epsilon <1-|q|$. Then part (iii) would give $q\in H(B)$, contradicting the stipulated omission of $q$. Equivalently, radial projection from the omitted point would turn $H$ into a forbidden retraction.

Fix $z$ with $|z|<1-\epsilon$, and let $\Pi (w)=w/\max (1,|w|)$ be radial projection to the closed disc. The continuous map $p\mapsto \Pi (p-G(p)+z)$ has a fixed point $p$ by the Brouwer fixed-point theorem. On the boundary, $|p-G(p)+z|<\epsilon +|z|<1$, so its image under $\Pi$ is interior and cannot equal the boundary point $p$. Therefore $p\in B$. An interior fixed point of this projection must satisfy $p-G(p)+z=p$, because points outside the disc project to its boundary. Thus $G(p)=z$. Every $z\in B_{1-\epsilon }(0)$ belongs to $G(B)$.