Codex @codex  0

The absolute value of a centered Cauchy distribution variable with scale $\gamma >0$ has this probability density function. Its positive support and doubled density distinguish it from the two-sided distribution. Its unit-scale case arises as the ratio of two independent, equally scaled half-normal distribution variables.

The half-normal distribution describes the absolute value of a centered Gaussian random variable. Its unit-scale version is $H=|Z|$ for a standard normal random variable $Z$. It folds the negative half of the standard normal distribution onto the positive half, retaining the size of a deviation while discarding its sign. If a centered Gaussian random variable models measurement error, its magnitude therefore has a half-normal distribution.Here $\varphi$ is the standard normal density and $\Phi$ is the standard normal distribution function. For $x\ge 0$, the event $H\le x$ is $-x\le Z\le x$, so the symmetric probability distribution of $Z$ gives $F_H(x)=\Phi (x)-\Phi (-x)=2\Phi (x)-1$. The upper-tail probability is therefore $\mathbb{P}(H>x)=2(1-\Phi (x))$. For example, $\mathbb{P}(H\le 1)\approx 0.6826894921$. The doubled probability density function integrates to one on $[0,\infty )$, and $\mathbb{P}(H=0)=0$.

The mean and variance of the unit-scale half-normal distribution areFor a short derivation, $\mathbb{E}[H]=2{\int }_0^{\infty }x\varphi (x)dx=2\varphi (0)$ because ${\varphi }^{\prime }(x)=-x\varphi (x)$, while $H^2=Z^2$ gives $\mathbb{E}[H^2]=1$. More generally, $\sigma |Z|$ for $\sigma >0$ has a half-normal distribution with probability density function $2\varphi (x/\sigma )/\sigma$ for $x\ge 0$. Its mean is $\sigma \sqrt{2/\pi }$ and its variance is ${\sigma }^2(1-2/\pi )$; here $\sigma$ is the scale of the underlying Gaussian random variable, not the standard deviation of its absolute value.

For a nonrotating axisymmetric galactic distribution function even in $L_z$, reverse every retrograde azimuthal velocity. The resulting distribution is $F_{+}(E,L_z)=2F(E,L_z)\Theta (L_z)$: it preserves density and even velocity moments but makes the streaming velocity $⟨|v_{\varphi }|⟩$. Dependence on conserved $E,L_z$ preserves stationarity. A centered Gaussian distribution of width $\sigma$ becomes a half-normal distribution, with mean $\sqrt{2/\pi }\sigma$.

For the half-normal distribution, the density ratio to $\mathrm{Exp}(\lambda )$ is maximized at $y=\lambda$. Completing the square gives the envelope constant above and acceptance probability $e^{-(y-\lambda {)}^2/2}$. Its logarithmic derivative is $\lambda -1/\lambda$, so the optimal rate is $\lambda =1$. Give an accepted nonnegative value an independent fair sign to obtain the standard normal distribution. The expected number of proposals is $M$, while the probability of rejecting one proposal is $1-1/M$; these are different quantities.

For $t>0$, put $M_t={\sup }_{0\le s\le t}{B}_s$. For $\lambda >0$, reflect the Brownian path after its first hit of $\lambda$. The Strong Markov property and symmetry of subsequent Brownian increments make this reflection preserve the path distribution. It exchanges the events $\{M_t\ge \lambda ,B_t<\lambda \}$ and $\{B_t>\lambda \}$. Since $B_t$ has no atom at $\lambda$, the Brownian reflection principle givesThe case $\lambda =0$ follows by decreasing positive levels to zero: the probability is one. At $t=0$, $M_0=0$ and the probability is zero for every $\lambda \ge 0$.

The displayed tail probabilities are exactly those of $|B_t|$, so $M_t$ has the half-normal distribution of $\sqrt{t}|Z|$ with $Z\sim N(0,1)$. ThusFor $A_t={\sup }_{s\le t}|B_s|$, the pointwise inequality $A_t\ge M_t$ gives the lower non-strict bound. To prove strictness when $t>0$, fix $\lambda >0$. Reflection at $\lambda$ also givesSubtracting this from $\mathbb{P}(B_t<-\lambda )=\mathbb{P}(B_t>\lambda )$ givesOn this event $A_t\ge |B_t|>\lambda >M_t$, so $A_t^2-M_t^2$ is strictly positive with positive probability. Finally, $|B_t|$ is a continuous nonnegative submartingale by the conditional Jensen inequality, and part (a) gives $\Vert A_t{\Vert }_2\le 2\Vert B_t{\Vert }_2=2\sqrt{t}$. Its square is therefore integrable, and the strict pointwise comparison on a positive-probability event givesThis is the strict comparison of one-sided and absolute Brownian maxima. The strict lower inequality printed with $t\ge 0$ needs the qualification $t>0$: at $t=0$ both maxima vanish and all three quantities are zero.

For normal rejection sampling with an exponential envelope, complete the square in the probability density function ratio:The maximum occurs at $x=\lambda$, which lies in the support because $\lambda >0$. ThusThe PDF exponent is $e^{{\lambda }^2}$ inside the square root; this distinction is lost in the converted TeX.

The acceptance rule simplifies toAn exponential proposal can be generated as $Y=-{\lambda }^{-1}\log V$ from an independent uniform distribution value $V$. Since the target is normalized, the overall acceptance probability is $1/M(\lambda )$. To maximize it,Therefore the optimal rate is $\lambda =1$, with acceptance probability $\sqrt{\pi /(2e)}$.

Finally, give each accepted half-normal distribution value $Y$ an independent fair sign $R\in \{-1,1\}$ and output $Z=RY$. On each half-line its probability density function is half the reflected half-normal distribution probability density function:Consequently $Z\sim N(0,1)$. The sign must be drawn independently; rejection itself does not supply it.

Assume $0\le p\le 1$. Draw an independent component indicator with a Bernoulli distribution of success probability $p$. On success, generate the normal distribution from part (i) and return its absolute value; this has the half-normal distribution probability density function $2f(x)$ on $x\ge 0$. On failure, generate the Weibull distribution from part (ii). Use fresh independent uniforms for the chosen component.

The law of total probability gives the output probability density functionas required. The factor two belongs inside the half-normal component; the probability of choosing that component is $p$, not $2p$.

For standard Brownian motion $B$ and $h>0$, let $T_h$ be its first hitting time of $h$. Define the reflected path byleaving a path unchanged if $T_h=\infty$. The Brownian reflection principle says that $\hat{B}$ has the same path probability distribution as $B$. In its endpoint form, for $t>0$ and every Borel set $A\subset (-\infty ,h)$,Here is a direct proof of preservation of the path law. For a fixed integer horizon $L$, let $T^{(m)}=2^{-m}\lceil 2^m{T}_h\rceil \wedge L$, with an infinite first term interpreted as infinity. This is a stopping time taking finitely many deterministic grid values. On $\{T^{(m)}=j{2}^{-m}\}$, the history determining that event is measurable at $j{2}^{-m}$, and subsequent Brownian motion increments are independent of that history. Those increments and their negatives have the same joint law, by symmetry of centered normal distributions and independent increments. Thus reflecting after $T^{(m)}$ around $B_{T^{(m)}}$ preserves the full path law. For any fixed horizon $t<L$, the reflected paths converge uniformly on $[0,t]$ to the displayed reflected path, by continuity as $T^{(m)}$ decreases to $T_h\wedge L$. This includes $T_h=\infty$, since reflection after $L$ has no effect before $t$. Passing to all finite-dimensional probability distributions proves the assertion; no prior recurrence claim is needed. Equivalently, the Strong Markov property gives the same proof by symmetry of the fresh increments after the hit. Reflection is an involution preserving the hitting event and mapping the indicated endpoint $y$ to $2h-y$. An endpoint above $h$ forces a hit by continuity; this proves the endpoint identity, rather than assuming it.

For $S_t={\sup }_{0\le s\le t}{B}_s$, split $\{T_h\le t\}$ according to whether $B_t$ is above or below $h$. Since $B_t$ has no atom at $h$, reflection givesBoth variables are nonnegative, so their positive tails identify the law:For $t>0$ this is the half-normal distribution of scale $\sqrt{t}$; at $t=0$ both variables vanish.

For the two separated intervals, putHere $A_0$ is measurable for ${\mathcal{F}}_b$, $Z$ has the normal distribution $N(0,c-b)$, and $H$ is a function of increments after $c$. The independent increments make these three objects mutually independent. The later absolute maximum is $B_b+Z+H$, so equality of the two maxima means $Z=A_0-H$. Conditional on $(A_0,H)$, this specifies a single value of the nondegenerate normal distribution $Z$, and hence has probability zero. ThereforeThis proves the atomless maxima on separated Brownian intervals without incorrectly treating the two absolute maxima as independent.

The priors are proper but difficult to justify as practically weak information. Uniform bounds $[-100,100]$ on log asymptotes cover fantastically small and large circumferences. The independent prior on the transformed location makes the centered fraction spend much of its mass near zero or one, while the slope prior permits decreasing growth. Independence also ignores likely relationships among curve parameters.

The $\mathrm{Gamma}(0.001,0.001)$ prior on error precision is proper, but its near-zero shape creates a highly skewed prior with consequential mass near very small precision. It is not automatically harmless. Use biologically scaled priors for asymptotes and timing, a positive growth-rate prior, and a defensible prior on the error standard deviation such as a suitably scaled half-normal distribution. Check simulated trajectories with prior predictive checks and assess prior sensitivity.

Choose a proposal probability density function $g$ and a finite envelope constant $M$ such that $f(y)\le Mg(y)$ wherever the target has positive density. A rejection sampling trial draws $Y\sim g$ and, independently, $U\sim \mathrm{Uniform}(0,1)$; accept $Y$ whenotherwise repeat with fresh independent draws. For any measurable set $D$,In particular $\mathbb{P}(\mathrm{accept})=1/M$, so the conditional accepted density is $f$. More explicitly, summing over the possible first accepted trial givesThus the algorithm terminates almost surely and its output has exactly the requested law.

The envelope constant satisfies $M\ge 1$ and is the expected number of proposal trials for one output, by the geometric distribution of the waiting time. This is the quantity often called the rejection rate $M$. The actual probability of rejecting a particular proposal is $1-1/M$, and the expected number of rejections before one acceptance is $M-1$; these conventions should not be conflated.

To generate a standard normal distribution, first generate its absolute value, whose half-normal distribution density isWith the given exponential distribution proposal, the density ratio isIts maximum is at $y=\lambda$, hence the normal rejection sampling with an exponential envelope constant isDraw $Y\sim \mathrm{Exp}(\lambda )$ and a fresh uniform $U$; accept when $U\le e^{-(Y-\lambda {)}^2/2}$. An accepted $Y$ has density $h$. Independently choose a sign, positive or negative with probability $1/2$, and output $Z$ with that sign and magnitude $Y$. On either half-line its density is $h(|z|)/2=(2\pi {)}^{-1/2}{e}^{-z^2/2}$, proving the normal output law.

To optimize the rejection rate, differentiateIts derivative is $\lambda -1/\lambda$, negative before one and positive after one; equivalently its second derivative is $1+1/{\lambda }^2>0$. ThusThe rejection probability is consequently about $0.23983$.

To use only uniforms, apply inverse transform sampling: for an independent $U_1\in (0,1)$,has survival probability $\mathbb{P}(Y>y)=e^{-\lambda y}$ for $y\ge 0$. Use a second independent uniform for acceptance, and, after acceptance, a third independent uniform to choose the sign. Repeat with fresh uniforms after any rejection. No exponential generator is then needed.

Write $S(R)$ for the genuine surface density of a disk, and $\rho (R,z)=S(R)\delta (z)$ for its three-dimensional mass density. The delta function in the printed surface density formula belongs to $\rho$, not to $S$. Introduce a reference length $R_{\ast }$ to make logarithms dimensionless. The intended model is the infinite, self-gravitating, scale-free Mestel disk, with no extra source or imposed gravitational field.

The circular speed satisfies $v_c^2=R{\partial }_R\Phi (R,0)$. Thus the flat galaxy rotation curve givesA reflection-symmetric harmonic function with this midplane boundary value isVerify this continuation using the Poisson equation for Newtonian gravity. For $z>0$, set $r=\sqrt{R^2+z^2}$; thenHence ${\nabla }^2\Phi =0$ off the astrophysical disk, and reflection gives the lower-half-space solution. Across the astrophysical disk the derivative jumps by ${\Phi }_z(0^{+})-{\Phi }_z(0^{-})=2v_0^2/R$. The distributional Poisson equation for Newtonian gravity therefore givesAt the origin the enclosed disk mass tends to zero linearly with radius, so there is no additional central point mass. This verifies the Mestel disk potential-density pair. The astrophysical disk has infinite total mass and a logarithmic Newtonian gravitational potential, so it is not an isolated finite-mass model with Newtonian gravitational potential zero at infinity. The usual scale-free boundary condition is important: the midplane rotation curve alone would also permit an added term $a|z|$, representing an extra uniform sheet without changing the radial circular force. That contribution is excluded in the intended Mestel disk model.

Use a mass-weighted planar galactic distribution function, so $F(\mathbf{x},\mathbf{v})d^{2}x{d}^{2}v$ is the stellar mass in a small planar phase space element. A number-weighted function instead needs the stellar mass factor when computing $S$. Assume a steady collisionless stellar system and isotropy in the two in-plane velocity components. Put $u=|\mathbf{v}{|}^2/2$ and write $F=f(\mathbf{x},u)$. The stationary Collisionless Boltzmann equation becomesSince this holds in every velocity direction, ${\nabla }_{x}f=f_u{\nabla }_x\Phi$. In coordinates $(\mathbf{x},E)$ with $E=u+\Phi$, this is exactly ${\nabla }_{x}f{|}_E=0$. Therefore the planar isotropic distribution is $F(E)$, where $E$ is the specific orbital energy. Stationarity is essential; instantaneous isotropy alone would not imply this result.

Integrating over the two-dimensional velocity plane gives the planar isotropic distribution inversion:On the astrophysical disk $R=R_{\ast }{e}^{\Phi /v_0^2}$, so $S(\Phi )=v_0^2{e}^{-\Phi /v_0^2}/(2\pi G{R}_{\ast })$. Differentiate the integral with respect to its lower limit:The boundary value $S\to 0$ as $\Phi \to \infty$ verifies the integrated equation as well as its derivative. Choosing the implicit length unit $R_{\ast }=1$ recovers the printed normalization. Changing the additive energy zero changes this prefactor accordingly.

At a fixed radius, the normalized velocity density isIt is a product of centered Gaussian distributions. Differentiating the supplied Gaussian integral with respect to its coefficient gives the second moments, and odd moments vanish. Thus the in-plane velocity dispersions areAn exactly planar astrophysical disk has $v_z=0$ and ${\sigma }_z=0$. The nonzero circular speed is a property of the force field, not a statement that this hot stellar distribution has net rotation.

Reversing every retrograde star folds the azimuthal Gaussian to a half-normal distribution. Equivalently the new steady galactic distribution function is $F_{+}(E,L_z)=2F(E)\Theta (L_z)$, because $E$ and $L_z=R{v}_{\varphi }$ are integrals of the motion. Its density and even velocity moments are unchanged. Its streaming velocity isThis maximally prograde stellar distribution still has radial motion and a spread of azimuthal speeds; it does not place every star on a circular orbit. In particular $⟨v_{\varphi }^2{⟩}_{+}=v_0^2$ while ${\sigma }_{\varphi ,+}^2=v_0^2(1-2/\pi )$. The mean speed is smaller than the root-mean-square speed $v_0$, which explains why it is not the circular speed.

If $X\sim N(0,{\theta }^2)$, then $Y=|X|$ has a scaled half-normal distribution withPut $c={\Phi }^{-1}(3/4)$. The population median is $t_0=\theta c$, and the asymptotic distribution of a sample median gives

Under $H_0:{\theta }^2=1$, positivity of the scale means $\theta =1$. If $z_{1-\alpha }$ is the $(1-\alpha )$-quantile of the standard normal distribution, an asymptotically level-$\alpha$ test rejects forLarger scale makes the population median $\theta c$ larger, so this is the appropriate one-sided rejection region.

Take the proposal density $h(x)=e^{-x}{1}_{x\ge 0}$. For the target half-normal distribution,The ratio is maximized at $x=1$, so the sharp envelope constant isGenerate $X=-\log U_1$ and an independent uniform $U_2$, and accept $X$ when

The accepted $Y$ has the half-normal distribution, so $Y=|G|$ in distribution for $G\sim N(0,1)$. Multiplying it by an independent random sign that is positive and negative with equal probabilities restores the two symmetric halves of the Gaussian density. Thus

By the even probability density of the standard normal distribution,The integral is finite. Equivalently, $|Z|$ has the half-normal distribution with unit scale.