The convex support is the closed convex hull of the canonical sufficient-statistic support. For a minimal exponential family it has nonempty interior. Means at finite interior natural parameters lie in that interior, because a supporting hyperplane at a boundary mean would force the sufficient statistic itself onto that hyperplane.

For each player in a finite bimatrix game, add the positive pure-strategy gains to its current mixed strategy and normalize. This gives a continuous self-map of the product of probability simplices. At a Brouwer fixed-point theorem fixed point, the gain coordinates equal the current strategy times their total gain. If that total were positive, every supported action would beat the current average payoff, contradicting that average. Thus every pure deviation has nonpositive gain for both players, proving a Nash equilibrium.

Define $M_2=K_2$ and $M_{k+1}=\mu (M_k)$ for $k\ge 2$. By the Mycielskian chromatic increase and Mycielskian triangle preservation, each $M_k$ is a finite triangle-free graph with chromatic number $k$. The first steps are $K_2,C_5$, the Grötzsch graph, and then a 23-vertex graph needing five colours. The vertex recurrence $n_{k+1}=2n_k+1$ gives $n_k=3\cdot 2^{k-2}-1$. These explicit examples show how arbitrarily strong graph coloring constraints can accumulate without any three-vertex clique.

A finite simple graph $G$ is a triangle-free graph if and only if its Mycielskian is a triangle-free graph. No triangle in a graph can contain the Mycielski apex, whose neighbours are an independent set, or two Mycielski shadow vertices. A triangle in a graph containing one shadow $u_i$ and two originals $v_j,v_{\ell }$ requires all three original edges $v_i{v}_j,v_i{v}_{\ell },v_j{v}_{\ell }$, hence an original triangle in a graph. Conversely every original triangle in a graph remains. In fact each original triangle in a graph gives four in $\mu (G)$: itself and the three choices of replacing one of its vertices by its shadow. Thus triangle-freeness is preserved, but a positive triangle count is multiplied by four.

For a nonempty finite simple graph, the upper bound copies an optimal proper vertex colouring to its Mycielski shadow vertices and gives the Mycielski apex a fresh colour. For the lower bound, suppose $\mu (G)$ uses $k=\chi (G)$ colours, with $w$ coloured $k$. Shadows avoid $k$. Replace colour $k$ at each original $v_i$ by $u_i$'s colour. Changed originals form an independent set; unchanged neighbours differ because $u_i{v}_j$ is an edge. This proper vertex colouring of $G$ uses $k-1$ colours, contradicting its chromatic number.^[ref]

The nonempty condition means $V(G)\ne \varnothing$, not $E(G)\ne \varnothing$. An edgeless $n$-vertex graph with $n\ge 1$ has chromatic number one; its Mycielskian is a star graph on $w,u_1,\dots ,u_n$ together with $n$ isolated vertices, and needs two colours. If $G$ has no vertices, $\mu (G)$ is a single vertex, so the same formula also holds with the convention $\chi (G)=0$.

The Mycielski apex is the added vertex adjacent to every Mycielski shadow vertex and to no original vertex. Its graph neighbourhood is precisely $\{u_1,\dots ,u_n\}$.

A Mycielski shadow vertex $u_i$ has graph neighbourhood $N_G(v_i)\cup \{w\}$ in $\mu (G)$. Its neighbours other than the Mycielski apex are original vertices. All Mycielski shadow vertices form an independent set.

The Grötzsch graph is the Mycielskian of the five-vertex cycle graph. Take original edges $v_i{v}_{i+1}$, with indices interpreted modulo five; add $u_i{v}_{i-1},u_i{v}_{i+1}$ and $w{u}_i$ for every $i$. There are eleven vertices and twenty edges: five original, ten between originals and Mycielski shadow vertices, and five from the Mycielski apex.

Give $v_1,\dots ,v_5$ colours $(1,2,1,2,3)$, give each $u_i$ the colour of $v_i$, and give $w$ colour 4. This is a proper vertex colouring. The Mycielskian chromatic increase proves that three colours are impossible, so the chromatic number is exactly four. The Mycielskian triangle preservation proves that this is a triangle-free graph. In particular its clique number is only two, even though its chromatic number is four.

The Mycielski apex has an independent set as its graph neighbourhood, yet all three old colours are needed there in any three-colour proper vertex colouring of the subgraph obtained by deleting it. Thus a locally easy group of mutually nonadjacent neighbours can carry globally incompatible colour requirements. Absence of triangles in a graph alone cannot bound the chromatic number.

A mechanism that waits for ownership by repeatedly trying while consuming CPU execution time. A spinlock built from test-and-set can enforce mutual exclusion without guaranteeing starvation freedom. Progress also requires the owner to run and release: if a waiting software thread monopolizes the only CPU while the owner is suspended, exclusive entry alone cannot help.^[ref]

A weak charged current decay described at energies much below the W boson mass by a four-fermion interaction with coefficient given by the Fermi constant. At leading order with negligible final-state masses, its width is $G_F^2{m}_{\mu }^5/(192{\pi }^3)$.

For $Z_t=W_t+\Theta W_{t-1}$ and vector white noise of covariance matrix $D$, the only nonzero covariance lags are $\Gamma (0)=D+\Theta D{\Theta }^T$, $\Gamma (1)=D{\Theta }^T$ and $\Gamma (-1)=\Theta D$. This follows by retaining only matching noise times in $\mathbb{E}[Z_t{Z}_{t+h}^T]$. Its spectral matrix, with this lag convention, is $(I+\Theta e^{i\lambda })D(I+{\Theta }^T{e}^{-i\lambda })/(2\pi )$. In two dimensions, with $D=\mathrm{diag}(a,b)$, its cross-spectrum is $[a{\theta }_{11}{\theta }_{21}+b{\theta }_{12}{\theta }_{22}+a{\theta }_{21}{e}^{-i\lambda }+b{\theta }_{12}{e}^{i\lambda }]/(2\pi )$.

For centered jointly Gaussian random variables, an odd product has zero expectation, and an even product's expectation is the sum over all pairings of the products of the pair covariances. In particular,For a proof, the moment-generating function is $\exp (t^T\Sigma t/2)$. Differentiating once in each variable at zero leaves precisely products of covariance entries indexed by complete pairings. No independence assumption between the coordinates is needed.

This criterion holds even for reducible nonnegative matrices. If a fixed point $p\ne 0$ exists, $1-e^{-z}<z$ for $z>0$ gives $(\Lambda p{)}_i>p_i$ on its support. Some $\alpha >1$ therefore satisfies $\Lambda p\ge \alpha p$, forcing the spectral radius above one by the Perron–Frobenius theorem. Conversely take $v\ge 0$, $v\ne 0$, with $\Lambda v=\rho v$ and $\rho >1$. For sufficiently small $\eta >0$, $1-e^{-\eta \rho v_i}\ge \eta v_i$. Thus $\eta v$ is a subsolution; monotone iteration under $T$ gives a positive fixed point below $1$. Strictness of $1-e^{-z}<z$ excludes survival at the critical value for this Poisson offspring law.

If a region is invariant under one coordinate reflection and an integrable function changes sign under that reflection, its multiple integral is zero. Apply the change of variables formula to the reflection, whose absolute Jacobian determinant is one: the integral equals its negative. This applies to odd coordinate powers in symmetric volumes and to the off-diagonal entries of an inertia tensor when the mass density shares the coordinate-reflection symmetries.

Counts of several categories before a fixed number of stopping-category outcomes have a negative multinomial distribution. For one stopping outcome of probability $r=1-\alpha -\beta$, the two-count mass function is $r\left(\genfrac{}{}{0ex}{}{x+y}{x}\right){\alpha }^x{\beta }^y$. Its probability generating function is $r/(1-\alpha s-\beta t)$ and its means are $\alpha /r$ and $\beta /r$.

For cell means ${\mu }_j$ with ${\sum }_j{\mu }_j=n$, the independent Poisson distribution likelihood and the multinomial distribution likelihood with probabilities ${\mu }_j/n$ differ by a parameter-independent factor. Thus a maximum-likelihood estimator for the unrestricted Poisson model whose fitted total is $n$ is also an estimator for the constrained multinomial model. Equality of named parameter estimates presumes identifiability and uniqueness; without these, the conclusion concerns the sets of maximizers. A free intercept in a log-linear Poisson model enforces the fitted-total equation.

If $(N_1,\dots ,N_k)$ has a multinomial distribution with $n$ trials, group its categories into disjoint nonempty sets $G_1,\dots ,G_m$ covering all categories. The grouped counts in the displayed equation have the multinomial distributionEach original trial belongs to exactly one group, its group probability is the sum of its category probabilities, and grouping preserves independence between trials. This gives the rule without summing the original probability mass function over every possible internal count pattern.

For the four-trial example with probabilities $(1/2,3/10,1/5)$, merge the last two categories. Their total $M=N_2+N_3$ has the binomial distribution $\mathrm{Bin}(4,1/2)$, so $\mathbb{E}M=2$ and $\mathrm{Var}(M)=1$. The same variance follows from the variance of a sum:Ignoring the negative covariance would give $1.48$, although the two counts share a fixed total. Multinomial aggregation lets the resolution of category labels change while retaining a consistent model for the counts.

NCERT's teaching experiment uses a moving-coil galvanometer, a 0–6 V supply, a 0–10 kΩ series electrical resistance box $R$, a 0–200 Ω parallel electrical resistance box $S$, two switches and a voltmeter. With the parallel path open, close the supply switch and record the pointer deflection. Keep $R$ fixed, close the parallel path, and adjust $S$ until the deflection halves. The guide requests repeated settings, clean tight contacts and constant supply electromotive force; it introduces high $R$ before energizing to protect the winding. ^[ref]

Write the winding electrical resistance as $R_g$, the supply electromotive force as $E$, and assume negligible supply internal electrical resistance. Ohm's law gives the two winding electric currentsIf proportional deflection gives $I_2=I_1/2$, solving yields $R_g=RS/(R-S)$, requiring $R>S$. The shortcut $R_g\simeq S$ requires $R\gg R_g$; closing the parallel path generally changes the total supply electric current. For chosen values $R=9900\Omega$ and $S=99.0\Omega$, the exact result is $R_g=100\Omega$, rather than $99\Omega$. The manual gives this exact relation and asks how finite supply internal electrical resistance affects it. ^[ref]

As an additional calibration control, record the supply-off zero, apply several small positive and negative electric currents, and recheck zero after each sequence. With $R_g$ known, calculate each electric current from the measured supply electric potential difference and the total series electrical resistance; plot settled deflection against that electric current. Equal-magnitude reversed readings should lie equally far from a stable zero. A fitted relation ${\theta }_{\mathrm{read}}={\theta }_0+S_{I}I$ separates zero offset from galvanometer current sensitivity. The manual's “figure of merit” is electric current per scale division, not the mechanical torsional spring constant $k$ used above. These reversal and zero checks extend the documented procedure; they are proposed controls, not reported measurements.

The galvanometer current sensitivity is the change in settled deflection angle per change in electric current. In the linear model its derivative is constant, $S_I=NAB/k$. The illustrative design has $S_I=3000rad/A$, or $0.1719^{\circ }$ per $\mu \mathrm{A}$. A chosen full-scale angle ${\theta }_{\max }=0.180\mathrm{rad}$ corresponds to $|I|=60.0\mu \mathrm{A}$; a straight-line extrapolation beyond that range does not establish real instrument behavior.

Finite sensitivity and finite readable angle set different limits. Under the stipulated $0.10^{\circ }$ tolerance, a reading indistinguishable from zero means $|I|\lesssim 0.582\mu \mathrm{A}$ only if the zero is stable and the model applies. More turns can increase galvanometer current sensitivity while also increasing winding electrical resistance; the response to an applied electric potential difference depends on both. Friction, zero drift, departures from the radial magnetic field, and spring nonlinearity can spoil the calibration. Damping sets how long one must wait for a settled reading. A rapidly reversing electric current can produce little mean pointer deflection even when its instantaneous magnitude is large.

For a rectangular winding with $N$ turns, active side length $\ell$, side separation $w$, and per-turn area $A=w\ell$, the magnetic force on a current-carrying wire gives equal and opposite side forces of magnitude $N|I|\ell B$. Their force couple has lever arm $w$. Choosing the positive rotation to match positive electric current in the illustration gives the signed torque and spring balanceHere $B$ is the working-gap magnetic flux density, $k>0$ is the torsional spring constant in $\mathrm{N}m/rad$, and $\theta$ is the deflection angle in radians from the no-current zero. The balance holds after motion has settled, within the radial-field and linear elasticity ranges. Do not identify $\theta$ with the angle between a magnetic dipole moment and a fixed uniform magnetic field: that different geometry has torque magnitude $NIAB|\sin \alpha |$. ^[ref]

For an original illustrative design, choose $N=120$, $A=3.00\times 10^{-4}{\mathrm{m}}^2$, $B=0.250\mathrm{T}$, and $k=3.00\times 10^{-6}\mathrm{N}m/rad$. Then $NAB=9.00\times 10^{-3}\mathrm{N}m/A$. At $I=+40.0\mu \mathrm{A}$, the magnetic torque is $+3.60\times 10^{-7}\mathrm{N}\mathrm{m}$ and $\theta =+0.120\mathrm{rad}\simeq +6.88^{\circ }$. Reversing the electric current reverses both the torque and the deflection, giving $-6.88^{\circ }$ without changing the magnetic field.