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For a normalized two-qubit pure state with coefficient matrix $M$, concurrence is $2|\det M|$. The displayed convex roof extension defines its mixed-state version. Combining decompositions proves convexity: $C({\sum }_j{r}_j{\rho }_j)\le {\sum }_j{r}_{j}C({\rho }_j)$. Every separable quantum state has a decomposition into product vectors, each with zero determinant, so its concurrence is zero.

Use the unary Cantor normal form tree norm, counting one root and recursively counting all repeated exponent subterms. There are only finitely many notations of bounded norm. This finite principle bounds all descending sequences whose norms grow at most linearly with the position. Friedman's finite form of Kruskal's theorem implies it through the natural-sum ordinal rank of a finite rooted tree. The Friedman–Smith miniaturization theorem turns this controlled termination principle for standard epsilon-zero notations into the formal consistency statement for Peano arithmetic; the technical ordinal-analysis result is needed in addition to the tree coding. Section 2.2 of formal.hknu.ac.kr/Publi/kruskalJKMS.pdf records the general miniaturization and soundness theorem.

Let $\mathcal{A}$ be a finite multiset of subsets and let $r_i$ count its occurrences of coordinate $i$. Its nested compression consists of $L_j=\{i:r_i\ge j\}$. Then ${\sum }_{j}H(X_{L_j})\le {\sum }_{A\in \mathcal{A}}H(X_A)$, by repeated entropy submodularity. This is the compression formulation in Balister and Bollobás, "Projections, Entropy and Sumsets".

Let $Y_t={\sum }_{j=1}^{N_t}{X}_j$ for a rate-$\lambda$ Poisson process and independent and identically distributed nonnegative marks independent of that process, with $\varphi (q)=\mathbb{E}{e}^{-q{X}_1}$. The Laplace transform of a nonnegative random variable for the increment over $[s,t]$ is $\exp [\lambda (t-s)(\varphi (q)-1)]$. That increment is independent of the past of $Y$, so the displayed nonnegative process has mean one and satisfies $\mathbb{E}[Z_t\mid \sigma (Y_u:u\le s)]=Z_s$. No finite first moment of the marks is needed for this martingale when $q\ge 0$.

For nondecreasing compound Poisson input $U$, let $W={\sup }_{t\ge 0}(U(t)-Ct)$ and $W_{\delta }={\sup }_{k\ge 0}(U(k\delta )-Ck\delta )$. The second supremum is bounded above by the first. For each $t$, the next grid point is at most $\delta$ later, its input is at least $U(t)$, and it offers at most an extra $C\delta$ service. Hence $W\le W_{\delta }+C\delta$. When the workload is finite almost surely, these inequalities give stationary-tail brackets. If its logarithmic exponential tail rate exists, adding the bounded shift $C\delta$ does not change that rate.

Let $Y$ be a Compound Poisson process with nonzero jump marks and positive rate, independent of a càdlàg process $Z$. Conditional on the path of $Z$, its discontinuities form a countable set. The Poisson jump times have continuous distributions and avoid that set almost surely. At every jump of $Y$, the sum $Y+Z$ therefore has the same nonzero jump. In particular a continuous Lévy process cannot have a positive-rate independent compound Poisson component. Applying this to each truncation of its Lévy measure proves that its jump measure is zero.

Write $m_k=\mathbb{E}{S}^k$, $m_0=1$, and ${\mu }_j=\mathbb{E}{X}^j$. Apply the Poisson size-bias identity with $h(s)=s^{k-1}$, and expand $(S+X{)}^{k-1}$ using the binomial theorem and independence. A finite $k$th severity moment suffices. In particular, the aggregate expected value, variance, and third central moment are $\lambda {\mu }_1$, $\lambda {\mu }_2$, and $\lambda {\mu }_3$, respectively; the latter two use raw severity moments.

Approximate a compound Poisson distribution of positive claims by a shifted gamma distribution whose first three cumulants agree. Here $m_j=\mathbb{E}{X}^j$ are raw severity moments, while the compound cumulants are $\lambda m_j$. Solve $k+\alpha /\nu =\lambda m_1$, $\alpha /{\nu }^2=\lambda m_2$, and $2\alpha /{\nu }^3=\lambda m_3$ to obtain the displayed parameters. The approximation captures positive skewness, but not the atom $e^{-\lambda }$ at zero. Its shift can be negative; with exponential claims of mean $\mu$ it is $-\lambda \mu /3$. Matching three cumulants alone supplies no general tail-error bound.

For a compound Poisson distribution of count parameter $\lambda$ and severity moments $m_1,m_2$, the expected value is $\lambda m_1$ and the variance is $\lambda m_2$. As $\lambda \to \infty$ with the severity law fixed and $m_2<\infty$, the standardized aggregate converges in distribution to the standard normal distribution. Indeed, the logarithm of its characteristic function is $\lambda [{\phi }_X(t/\sqrt{\lambda m_2})-1]-it\lambda m_1/\sqrt{\lambda m_2}$; the second-order expansion of ${\phi }_X$ makes this tend to $-t^2/2$. The approximation loses the zero atom and assigns positive probability to negative losses.

Conditioning on the independent Poisson distribution count in a random sum gives $\mathbb{E}{e}^{tS}=\exp (\lambda (M_X(t)-1))$ wherever finite. Differentiation at zero gives the displayed cumulants when the relevant derivatives exist. The aggregate variance is the count rate times the raw second claim moment, not just the claim variance. Positive moment-generating function arguments need an exponential moment; finite first and second claim moments suffice for the aggregate mean and variance by conditioning or right derivatives at zero of the Laplace transform of a nonnegative random variable.

If zero-sized claims are allowed, a compound Poisson distribution count parameter need not be unique. Poisson thinning of zero claims gives the displayed positive-claim count parameter and the conditional claim size probability distribution given $X>0$, without changing the aggregate. If no positive claims occur, the aggregate is identically zero. This matters when representing a finite weighted sum of Poisson distribution counts whose coefficients include zero.

Let $N\sim \mathrm{Pois}(\lambda )$ be independent of independent and identically distributed random variables $(Y_i,Z_i)$ with finite second moments. Conditioning on $N$ gives covariance $N\mathrm{Cov}(Y_1,Z_1)$ and conditional expectations $N\mathbb{E}{Y}_1,N\mathbb{E}{Z}_1$. Law of total covariance and $\mathrm{Var}N=\lambda$ yield the formula. For per-claim excess of loss reinsurance payouts $Y=\min (X,M)$ and $Z=(X-M{)}_{+}$, it becomes $\lambda M\mathbb{E}[(X-M{)}_{+}]$. The insurer and reinsurer aggregates share their claim count and are generally dependent.

If a Poisson distribution count $M$ of clusters has mean $\nu$, and each cluster independently contains a Poisson number of claims of mean $\lambda$, then the total count is a compound Poisson distribution with the displayed probability generating function. Conditioning gives $\mathbb{E}N=\nu \lambda$ and $\mathrm{Var}N=\nu \lambda (1+\lambda )$, so it is not Poisson for positive parameters. With independent severities of mean $\mu$ and variance ${\sigma }^2$, the aggregate payment has mean $\nu \lambda \mu$ and variance $\nu [\lambda ({\sigma }^2+{\mu }^2)+{\lambda }^2{\mu }^2]$. independent validity thinning replaces the per-cluster count mean by $\lambda (1-p)$ and multiplies expected payment by $1-p$.

If independent summands have exponential distribution of mean $\mu$ and the independent count has a zero-based geometric distribution of parameter $p$, their random sum of independent claims has an atom $p$ at zero and an exponential positive component of rate $p/\mu$. Its Laplace transform of a nonnegative random variable is $p/[1-(1-p)/(1+\mu s)]=p+(1-p)p/(p+\mu s)$. Thus $\mathbb{P}(S>x)=(1-p)e^{-px/\mu }$ for $x>0$. Unlike the geometric sum of exponential variables with a positive count, this law is not purely exponential.

For independent factories with accident probability $p$, and independent Poisson distribution claim counts of mean $\lambda$ per accident, the accident count is binomial and each accident supplies a cluster. The random-sum transform identity gives the displayed probability generating function. In particular the no-claim chance is $(1-p+p{e}^{-\lambda }{)}^m$; a factory with an accident can still have zero claims. The binomial accident law has Panjer parameters $a=-p/(1-p)$ and $b=(m+1)p/(1-p)$.

For an even-rank real vector bundle $\xi$, a stable complex structure is a complex structure on $\xi \oplus {\mathbb{R}}^{2s}$, considered up to adding trivial complex bundles and homotopy. This gives a stable complex class but does not necessarily make $\xi$ itself a complex vector bundle. Stable Chern classes therefore need not recover the unstable Euler class: $T{S}^4\oplus {\mathbb{R}}^2$ can be made trivial complex, while $e(T{S}^4)$ evaluates to $2$.

This is a smoothly varying positive-definite Hermitian form on each fiber, without requiring a holomorphic base or bundle. A partition of unity glues local positive metrics into a global one. Applying smooth Gram-Schmidt to local frames then makes the transitions unitary, reducing the structure group of a vector bundle to $U(k)$. For a line bundle its unit vectors form a principal bundle for the circle group.

For a crack on the real axis under applied shear traction $p$, write $G=\mu F^{\prime }$ for the additional complex stress potential for antiplane shear derivative. Reflection of the induced displacement gives $G^{-}=\overline{G^{+}}$. Traction-free crack faces therefore require $G^{+}+G^{-}=-2p$; intact portions have $G^{+}=G^{-}$. This is a scalar Riemann-Hilbert problem. Endpoint regularity, single-valued displacement and the absence of additional remote loads fix its homogeneous terms.

Substituting the free complex scalar field modes into $i\int ({\varphi }^{†}\dot{\varphi }-{\dot{\varphi }}^{†}\varphi )d^{3}x$ gives the formally ordered expression $Q_{\mathrm{bare}}=\int d{\Pi }_p(a^{†}a-b{b}^{†})$. Its divergent vacuum constant is removed by normal ordering, equivalently by choosing $Q|0⟩=0$. The resulting complex scalar charge operator has $[Q,a^{†}]=a^{†}$ and $[Q,b^{†}]=-b^{†}$. Subtracting the constant changes neither its action on fields nor current conservation, but it is essential when identifying the vacuum charge and the charges of one-particle states.

Every scalar holomorphic function on a compact connected complex manifold is constant by the maximum modulus principle. Apply this to each entry of the holomorphic Adjoint representation of a Lie group; its value is constantly the identity. Differentiation makes the complex Lie algebra abelian, hence connectedness makes the group abelian. The holomorphic Lie exponential map identifies it with ${\mathbb{C}}^n$ modulo a discrete subgroup. Compactness forces that subgroup to span the underlying real vector space, giving a full-rank Euclidean lattice and a complex torus.