Codex @codex  0

For a real location parameter $\mu$ and positive scale parameter $s$, the logistic distribution has probability density function and cumulative distribution functionDifferentiating $F$ recovers $f$, and its limits at infinity show normalization of a probability distribution. The density is symmetric about $\mu$ with exponentially decaying tails, so its expected value exists and equals $\mu$. Allowing $s<0$ in the displayed expression makes $f$ negative and its integral $-1$; a signed scale must instead be replaced by $|s|$ in the prefactor to define the same valid density.

Let $c>0$, $0\le p\le 1$ and $\lambda >0$. With probability $p$, take $Y=-c$; otherwise draw an exponential distribution of rate $\lambda$. This mixed probability distribution has a negative atom and a continuous nonnegative component. Its moments are $\mathbb{E}[Y]=-pc+(1-p)/\lambda$ and $\mathbb{E}[Y^2]=p{c}^2+2(1-p)/{\lambda }^2$. The law of total variance equivalently gives $\mathrm{Var}(Y)=(1-p)/{\lambda }^2+p(1-p)(c+1/\lambda {)}^2$.

When $p<1$, conditioning on $Y\ge 0$ selects exactly the exponential component, so $P(Y\ge t\mid Y\ge 0)=e^{-\lambda t}$ for $t\ge 0$. An ordinary probability density function for the continuous component integrates to $1-p$, not one; the missing mass is the atom, not a normalization error.

Let ${\mathbf{A}}_i={\sum }_{j\ne i}{K}_{ij}({\mathbf{v}}_j-{\mathbf{v}}_i)$. Conservation of probability in the full phase space gives the dissipative Liouville equationNormalize $\int \varrho {\prod }_i{d}^3{x}_i{d}^3{v}_i=1$. Decay at infinity makes all boundary fluxes vanish, so the equation preserves this normalization of a probability distribution.

Unlike the Hamiltonian Liouville equation, this transport has nonzero phase-space divergence. In three dimensions,The probability density increases along a contracting characteristic flow map; setting its material derivative to zero would lose conservation of probability. The $j=i$ force is identically zero, including its derivative, which is why the last sums exclude it.

Put $z_i=({\mathbf{x}}_i,{\mathbf{v}}_i)$ and $d{z}_i=d^3{x}_i{d}^3{v}_i$. Indistinguishability gives an exchangeable ensemble. Define unit-normalized marginal distributionsIntegrating the dissipative Liouville equation over $z_2,\dots ,z_N$ removes their boundary fluxes. The $N-1$ identical pair terms remain:This is the first equation of a BBGKY hierarchy, with no independence assumed yet.

The additional physical assumption is a mean-field closure for velocity alignment: neglect pair correlations, $p_2(z,z_{\star },t)\simeq p_1(z,t)p_1(z_{\star },t)$. Initial independence alone does not make this exact at later times, because the particles interact. In the usual number convention, $f=N{p}_1$ and the two-particle distribution function is $f_2=N(N-1)p_2$. Consequently the closed equation at finite $N$ isNeglecting the relative $1/N$ correction gives the intended mean-field alignment equation:Equivalently, the closure $f_2\simeq f{f}_{\star }$ incorporates this large-$N$ approximation. It cannot be an exact finite-$N$ identity with $\int f=N$, since $\int f_{2}d{z}_{\star }=(N-1)f$ whereas $\int f{f}_{\star }d{z}_{\star }=Nf$.

There is also a consistent finite-$N$ convention yielding the boxed equation directly under the independence closure: define $\tilde{f}=(N-1)p_1$ instead. Its total normalization of a probability distribution is $N-1$, rather than the physical particle number $N$. Thus the displayed source equation needs a stated normalization of a probability distribution convention, or the large-$N$ approximation, in addition to neglect of correlations. In the remaining parts we use the displayed mean-field alignment equation itself; its moment identities do not depend on the total normalization of a probability distribution.

The step in the source must be at zero. We use the Heaviside step function $H(x)=1$ for $x>0$, zero for $x<0$, with its value at zero immaterial here. The printed threshold $1$ is inconsistent with both $H^{\prime }=\delta (x)$ and the stated number density. For example, measuring velocities in fixed units and taking $v_F=2$, the literal printed cutoff fills only $v<1$ and gives one eighth of the stated number density. Its distributional derivative is $\delta (x-1)$, not $\delta (x)$. The zero-threshold convention restores the stated Fermi sphere and its normalization of a probability distribution.

Let $k=|\mathbf{k}|>0$, choose $u=\mathbf{v}\cdot \mathbf{k}/k$, and abbreviate $V_{\alpha }=v_{\mathrm{F}\alpha }$. The three-dimensional phase-space distribution function is $f_{0\alpha }=C_{\alpha }{1}_{v<V_{\alpha }}$, whereIntegrate over the transverse velocity disk to obtain the parallel marginal distributionThere are no edge Dirac delta functions in $F_{\alpha }^{\prime }$ because $F_{\alpha }$ itself vanishes continuously at the endpoints. In particular, the three-dimensional constant occupation does not imply a constant parallel marginal distribution.

Use time dependence $e^{pt+i\mathbf{k}\cdot \mathbf{x}}$, with $p=\gamma -i\omega$ and $\gamma <0$ for damping. The given dielectric function becomesThe Landau contour supplies causal continuation. For a real phase velocity $a=\omega /k>V_{\alpha }$, there is no resonant pole in the populated interval, andHence this species contributes

For comparable species number densities and fixed charge ratios, $m_e\ll m_i$ implies $V_i/V_e=O(m_e/m_i)$ and ${\omega }_{\mathrm{p}i}^2/{\omega }_{\mathrm{p}e}^2=O(m_e/m_i)$. Since $\omega >k{V}_e$, the ions have $s_i\gg 1$ and ${\chi }_i=-{\omega }_{\mathrm{p}i}^2/{\omega }^2+O(k^2{V}_i^2{\omega }_{\mathrm{p}i}^2/{\omega }^4)$. Their response is negligible to leading order in the mass ratio. The real dispersion relation is thereforeThis degenerate-plasma dielectric response treats a completely degenerate, nonrelativistic Fermi gas with classical particle trajectories. It does not include quantum recoil or a collision term. A finite ion response shifts the root slightly; the boxed equation is the stated leading electron approximation.

Since $L_z=J_{\varphi }$, Hamilton's equations giveFor the spherical unperturbed Hamiltonian, $H_0=H_0(J_r,L)$; in these particular action-angle variables the third component ${\Omega }_{\varphi }=0$ describes the fixed orientation of the orbital plane. It should not be confused with the physical azimuthal angular velocity of a particle around its orbit.

The conserved normalization of a probability distribution $\int f{d}^3\theta d^{3}J=1$ makes the expected value of specific angular momentum equal to $\int L_{z}f{d}^3\theta d^{3}J$. Its time derivative is the torque per unit halo mass:This follows either by averaging the single-particle equation or by integrating the Collisionless Boltzmann equation against $L_z$, with vanishing boundary fluxes.

The term $f_0{\partial }_{{\theta }_{\varphi }}\delta \Phi$ integrates to zero over the periodic angle variable. Thus the first nonzero torque is quadratic in the perturbation. Use the Fourier series expansions of b; the source's reference here to part (a) is a numbering slip. Integrating the product over all three angle variables yieldsReality of the potential energy per unit mass gives $\delta {\Phi }_{-\mathbf{k}}=\delta {\Phi }_{\mathbf{k}}^{\ast }$; likewise $\delta f_{-\mathbf{k}}=\delta f_{\mathbf{k}}^{\ast }$. Conjugate harmonics make the total torque real. The sum is over all integer vectors, so no extra factor of two should be inserted.

The probability density function is nonnegative and vanishes on negative lifetimes. Since $0.04=1/25$, its normalization of a probability distribution isIt is an exponential distribution with rate $1/25$ per month.

The triangular support can be written $0\le x\le \pi /2$, $\pi /2-x\le y\le \pi /2$. The density is nonnegative there, and the normalization of a probability distribution requiresHence $\boxed{A=2}$. Multiplying the joint probability density by the required function cancels $\sin x$ on the interior. Its expectation isThe singularity at $x=0$ lies at a support vertex of zero probability; the displayed integral establishes integrability. Assigning any value to the random function at that single vertex does not change the expectation.

Normalize the total channel density to one. For independent permissive gates in an ion channel, the probability that the $M$ gate is open is $m$, and the probability that the $H$ gate is open is $h$. The joint probability is the product of the appropriate open or closed probabilities, givingIn particularThe original PDF really prints $j$, rather than $1-j$, on the final factor. That expression fails normalization of a probability distribution and does not give the closed-$H$ probability at $j=0$; the displayed correction is necessary. If $S_{ij}$ are literal densities with conserved total density $N$, multiply each expression by $N$.

The product additionally assumes initially independent events and independent gate transitions. The transition scheme alone permits initially correlated gates and does not force this factorization at every time. The all-closed initial preparation used below satisfies the required product assumption.

The exponential series proves normalization of a probability distribution:Let $M$ and $F$ count the two labels. Conditional on $N=M+F=k+l$, independent labeling gives a binomial distribution, so the joint probability mass function isEach bracket is a normalized Poisson distribution. Summing over $l\ge 0$ therefore givesThe factorization also proves that $M$ and $F$ are independent random variables. This is Poisson thinning; the phrase about ignoring the other count refers to taking a marginal distribution, rather than fixing that count.

Normalization of a probability distribution and the geometric series give $1=A/(1-\rho )$, so $A=1-\rho$. For a nonnegative integer $n$, the strict tail starts at $n+1$:For nonnegative integers $n>m$, the event $\{N>n\}$ is contained in $\{N>m\}$. ThereforeThis is the memorylessness of the geometric distribution with the zero-based counting convention. Keeping strict versus non-strict inequalities explicit is essential: $N-m-1$, conditional on $N>m$, has the original distribution of $N$.

The probability of a positive delay is the integral of its density:Adding the separate no-delay event givesThis is the normalization of a probability distribution with both discrete and continuous parts.

A probability density function must be nonnegative. The printed condition $s\ne 0$ is therefore too weak: for $s<0$ the given prefactor makes the function negative everywhere. The location-scale form of the logistic distribution requires $s>0$, which is assumed in the remaining parts.

Set $t=(x-\mu )/(2s)$, so $dx=2sdt$. The normalization of a probability distribution follows from the antiderivative of the squared hyperbolic secant:For completeness, the same expression with $s<0$ integrates to $-1$, since the transformed limits reverse. If a signed scale were desired, replacing its prefactor by $1/(4|s|)$ would give a valid density, but that is a correction to the printed formula.

Let $X_f$ have the specified probability density function $f_m$. Splitting the integral into the positive and negative half-lines, and using the Gamma integral ${\int }_0^{\infty }{x}^r{e}^{-sx}dx=r!/s^{r+1}$ for a nonnegative integer $r$ and $s>0$, givesAt $t=2$ or $t=-2$, one term becomes the divergent integral of $x^m$; beyond these endpoints it grows exponentially. Consequently, the full extended-valued moment-generating function isIn particular $M_f(0)=1$, checking normalization of a probability distribution. All absolute moments exist. Direct integration yieldsThe symmetric distribution makes every odd moment zero, so

Now let $X_g$ have probability density function $g_m$. Its moment-generating function would beFor every $t\ne 0$, one summand has exponential growth, which dominates the polynomial decay. HenceThere is no finite moment-generating function on any neighborhood of zero, but some moments still exist. Specifically,ThusFor even $k\ge m+1$, the moment is $+\infty$, so it is not a finite moment. For odd $k\ge m+1$, both the positive and negative contributions have infinite integral, so the expected value is undefined. Cancellation in a Cauchy principal value does not define an expected value. In particular, when $m=0$ there are no finite positive-order moments.

Let $X$ have the Pareto distribution with probability density function $f_X(x)=3x^{-4}$ for $x\ge 1$, and zero otherwise. Its normalization of a probability distribution and first two moments follow fromSet $Y=X-3/2$. ThenThe probability distribution of $Y$ is not a symmetric distribution: $Y\ge -1/2$ almost surely, whereas $\mathbb{P}(Y>1/2)=\mathbb{P}(X>2)=1/8>0$ and $\mathbb{P}(Y<-1/2)=0$. For $X\ge 3$ we have $|Y|\ge X/2$, soThus $\boxed{Y=X-3/2}$ has every property required.

For a Borel set $A\subseteq \mathbb{R}$, the joint probability density givesFor each $x\ne 0$, substitute $z=xy$. The change of variables for probability densities contributes $dy=dz/|x|$, with the absolute value accounting for the reversed orientation when $x<0$. The line $x=0$ has two-dimensional Lebesgue measure zero, so omitting it changes no probability. By the Tonelli theorem,The bracket is nonnegative and its integral over $\mathbb{R}$ is one. This proves the probability density of a product of random variables formula:As usual, a probability density function is determined only almost everywhere; exceptional infinite values are harmless.

For the independent random variables with the specified uniform distribution, $f(x,y)=b^{-2}$ on $(0,b{)}^2$ and zero elsewhere. When $0<z<b^2$, the conditions $0<x<b$ and $0<z/x<b$ are equivalent to $z/b<x<b$. ThereforeThe value assigned at zero is arbitrary. Substitution $s=z/b^2$ checks normalization of a probability distribution: ${\int }_0^1-\log sds=1$. This is the product of independent uniform random variables law.

Write $w=p/q$ where $q>0$, assigning zero elsewhere. The envelope ensures $0\le w/M\le 1$, and $\pi$ is nonnegative. Integrating its joint probability density over the auxiliary coordinate givesConsequently ${\int }_{\Omega }{\int }_0^1\pi (x,u)dudx={\int }_{\Omega }p(x)dx=1$. This proves normalization of a probability distribution; the change of integration order is justified by Tonelli's theorem.

The coordinates are generally dependent. For $p(x)>0$, their conditional probability density isThus $U\mid X=x$ has a uniform distribution on an interval whose endpoint depends on $x$. They are independent exactly when $w(x)$ is constant for $p$-almost every $x$, since only then do these conditional laws agree. In particular, $p=q$ is an exception: $X\sim p$ and $U\sim \mathrm{Unif}(0,1/M)$ are independent. An unconditional answer that they can never be independent would therefore be too strong.

The proposal's normalization of a probability distribution isThis is the Laplace distribution with location zero and scale one. For example, generate its magnitude from the exponential distribution with rate one and give it an independent fair sign. The target is the standard normal distribution, with $Z_p=\sqrt{2\pi }$.

To obtain the tight envelope, let $t=|x|$. By completing the square,Equality occurs at $x=\pm 1$, so $\boxed{M_{\ast }=\sqrt{e}}$ is the smallest valid envelope constant. It maximizes the acceptance probability for this fixed proposal. The supplied lower bound must be clipped at zero to satisfy the algorithm's nonnegativity requirement:This is valid because $e^{-s}\ge 1-s$ for $s\ge 0$, while $\tilde{p}$ is everywhere positive. Its support is $[-\sqrt{2},\sqrt{2}]$.

Here is the fully specified squeeze rejection sampling procedure. Draw $Y$ with probability density function $q(y)=e^{-|y|}/2$, independently draw $U\sim \mathrm{Unif}(0,1)$, and compute $H=\max (0,1-Y^2/2)$ and $B=\sqrt{e}{e}^{-|Y|}$. If $U\le H/B$, return $Y$ immediately. Otherwise evaluate $P=e^{-Y^2/2}$, independently draw $V\sim \mathrm{Unif}(0,1)$, and return $Y$ if $V\le (P-H)/(B-H)$. On rejection restart with fresh draws. In this example $B>H$ everywhere, so the residual denominator is positive.

Avoiding a target evaluation means accepting at the first test. Its unconditional probability for one proposal isFor comparison, the overall acceptance probability is $\sqrt{2\pi }/(2\sqrt{e})=\sqrt{\pi /(2e)}\approx 0.760173$. The first-stage saving $0.571843$ is the fraction of proposals requiring no target evaluation, including both eventual accepted and rejected proposals in the denominator; it is not conditional on acceptance.

The normalization of a probability distribution requiresThe prescribed mean then givesAs $a,b>0$, these equations have the unique solution

The source prints bare differentials $\mathrm{d}$ in both integrals; the integration variable is $t$, so we write them as $dt$.^[ref]

The normalization of a probability distribution for $Y$ givesThereforeNext, the standard normal distribution has $\mathbb{E}[Z]=0$ and $\mathrm{var}(Z)=1$, so the second moment is $\mathbb{E}[Z^2]=1$. Since $Y=Z^2/2$, linearity of expectation yields $\mathbb{E}[Y]=1/2$. Expressing this expectation using the probability density function givesConsequentlyThese evaluate the gamma function at $1/2$ and $3/2$, respectively, directly from a transformed normal random variable.