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The equation is a conservation law $n_t+J_x=0$ with diffusive flux $J=-D(n_0/n)n_x$. For positive density, insects move down the density gradient, with effective diffusion coefficient $D{n}_0/n$: dispersal is stronger in low-density regions. This is logarithmic diffusion, since the equation is $n_t=D{n}_0(\log n{)}_{xx}$.

Put $\tau =Dt$ and $\xi =x/{\tau }^{\beta }$. A self-similar solution $n=n_0{\tau }^{-\beta }g(\xi )$ conserves total number when ${\int }_{\mathbb{R}}g=1$. Its width shrinks to zero as $t\downarrow 0$; for an integrable nonnegative $g$, changing variables in the integral against a bounded continuous function proves convergence to $n_0$ times the Dirac delta distribution at the origin.

Substitution givesMatching powers for a nonstationary profile forces $\beta =1$. The profile equation integrates to $g^{\prime }/g=-\xi g+C$. For a centred even profile, $C=0$, and integration of $g^{\prime }=-\xi g^2$ gives $g=2/({\xi }^2+a^2)$. Normalization requires $2\pi /a=1$, henceThis is $n_0$ times a Cauchy distribution density with scale $2\pi Dt$. It is positive, has total mass $n_0$, and directly satisfies the nonlinear diffusion equation for $t>0$.

Specify the sign convention for the Lévy characteristic exponent byConditioning on $T_a$ and using the Brownian first-passage Laplace transform givesThereforeThe absolute value is essential for negative $u$. The process is the standard symmetric Cauchy process; for $a>0$, $X_a$ has the Cauchy distribution with location zero and scale $a$. If the exponent convention instead uses $\mathbb{E}{e}^{iu{X}_a}=e^{a\psi (u)}$, the answer is $\psi (u)=-|u|$.

Let $(Y_n)$ be independent and identically distributed random variables with the standard Cauchy distribution, whose probability density function isIts first absolute moment diverges sinceOn the other hand, its characteristic function is $\phi (t)=e^{-|t|}$. The characteristic function of a sum of independent variables therefore givesThus the stability of the Cauchy distribution implies that $S_n/n$ has the standard Cauchy distribution for every $n$. In particular,where $Z$ is standard Cauchy and $\stackrel{d}{\to }$ denotes convergence in distribution.

Differentiate the cumulative distribution function using the chain rule:This is a Cauchy distribution with location zero and scale $x_0$. The probability density function is nonnegative and integrates to $F(\infty )-F(-\infty )=1$.

The probability density is an even function. Its derivative simplifies toIt increases for $z<0$ and decreases for $z>0$, with maximum height $h(0)=2\arctan (1/2)/\pi$. Thus the most probable value, or mode, is $\boxed{z=0}$. The original sketch shows the symmetric, slightly flattened peak and long tails.

There is a genuine qualification to the request for a mean: the mean does not exist. The integral representation shows that $h(z)\sim 1/(\pi z^2)$ as $|z|\to \infty$, soThe two divergent integrals cannot be subtracted to define an expected value. Symmetry does give the Cauchy principal value ${\lim }_{A\to \infty }{\int }_{-A}^{A}zh(z)dz=0$, but that is not the mean. Equivalently, adding the bounded variable $Y$ cannot make the first absolute moment of a Cauchy distribution finite.

For positive standard deviations $a,b$, substitute the two normal densities:The integral is $1/(a^{-2}+v^2{b}^{-2})$, by substituting the exponent's quadratic argument. Thus, with $c=b/a$,This is a centered Cauchy distribution of scale $c$. Since $P(X=0)=0$, the desired probability is

Use the characteristic function to establish convergence in distribution by the Lévy continuity theorem, especially when exponential moments do not exist; a Cauchy distribution, for example, has a characteristic function but no finite moment-generating function on a neighborhood of zero.

Use the moment-generating function for an exponential-tail estimate: for $t>0$ with $M_X(t)<\infty$, Markov inequality givesOptimizing over $t$ produces a Chernoff bound. The modulus-one characteristic function does not directly give this positive exponential bound.

A characteristic function is useful for proving convergence in distribution, including for laws without a finite moment-generating function: the continuity theorem says that pointwise convergence of characteristic functions to a characteristic function continuous at zero implies convergence in distribution to its law. For example, the Cauchy distribution has a characteristic function although its moment-generating function is not finite on any neighbourhood of zero.

A moment-generating function is especially useful for exponential tail bounds: for $t>0$ where it is finite, Markov's inequality givesOptimizing over $t$ yields a Chernoff bound. The positive real exponential in this argument permits the inequality; a unit-modulus complex exponential does not give the corresponding tail bound.

Write $Z_0=r_0{e}^{i{\alpha }_0}$ with $0<r_0<1$ and choose its initial argument once. The exponential map in the source is a covering of the punctured disc, rather than a bijective map onto the full disc. Its pathwise inverse is the continuous liftFor $t<T_{\mathbb{D}}$ this lies in the upper half-plane, and at the disc exit it lies on the real line. Local conformal invariance of planar Brownian motion, or the quadratic variation of its analytic derivative $-i/Z$, shows that $W_t$ is planar Brownian motion run at the clockThe disc exit time is finite almost surely. The path avoids zero on the compact interval through its exit, so its minimum modulus there is positive and $A_{T_{\mathbb{D}}}<\infty$. Before this terminal clock time the lifted imaginary coordinate is positive, and at it the coordinate is zero. Hence it is exactly the upper-half-plane exit time of the time-changed Brownian motion, not an additional arbitrary stopping time.

The starting point of the lift is ${\alpha }_0+ib$ with $b=-\log r_0>0$. Its real exit coordinate has Cauchy distribution $b/[\pi ((s-{\alpha }_0{)}^2+b^2)]$. Since that coordinate is ${\alpha }_0+{\theta }_{T_{\mathbb{D}}}$, we obtain the stronger exact lawIts first absolute moment diverges:There is no contradiction with the terminal point lying on the compact unit circle: reducing its argument modulo $2\pi$ loses the number and direction of turns accumulated before exit.

Using the Cauchy distribution from the preceding part gives one common probability space for all starting points:As $(x,y)\to (0,0)$ with $y>0$, $x+yC\to 0$ almost surely, since $C$ is a finite real random variable almost surely. Continuity of $f$ at zero gives $f(x+yC)\to f(0)$ almost surely, and boundedness supplies an integrable constant bound. Dominated convergence theorem therefore provesThe argument applies to every sequence approaching zero, including tangential approaches. Along the real boundary the extension is already $f$, which is continuous. Hence the extension is continuous at zero in the closed half-plane. No finite moment of the Cauchy distribution is required.

Take $\lambda >0$ and let ${\sigma }_{\lambda }$ be the first time $X$ hits $\lambda$. Reflect the horizontal coordinate after ${\sigma }_{\lambda }$, leaving the vertical coordinate unchanged. The Strong Markov property and symmetry preserve the law of planar Brownian motion and leave its real-axis exit time unchanged. On paths with ${\sigma }_{\lambda }<\tau$, the reflection swaps the two terminal events $X_{\tau }<\lambda$ and $X_{\tau }>\lambda$. There is no terminal atom at $\lambda$. Hence the Brownian reflection principle givesHorizontal symmetry gives the same probability for $-\inf X>\lambda$. A union bound therefore yieldsAlso $X^{\ast }\ge |X_{\tau }|$, whose Cauchy distribution is $2{\pi }^{-1}{\int }_{\lambda }^{\infty }(1+x^2{)}^{-1}dx$. Thus the maximum tail is bounded above and below by positive constants times ${\lambda }^{-1}$ for large $\lambda$. The tail integral formula for moments for positive moments givesThe large-$\lambda$ integral is finite exactly when $0<p<1$; it diverges for $p\ge 1$. Near zero its upper bound by ${\int }_0^{1}p{\lambda }^{p-1}d\lambda$ is finite.

The vertical coordinate remains nonnegative up to exit and starts at one. The strip calculation, now with upper boundary $\lambda \ge 1$, givesThus $Y^{\ast }$ has density ${\lambda }^{-2}$ on $[1,\infty )$, and in factFor completeness, if the source's $p<1$ includes negative powers, $X^{\ast }$ also has all negative moments. Let ${\sigma }_{\eta }$ be the first horizontal exit from $(-\eta ,\eta )$. ThenThe first bound follows from the Gaussian reflection bound for the vertical hit of zero; the second follows by splitting time into blocks of length ${\eta }^2$, each having a fixed positive chance of exiting the horizontal interval. This small-value estimate integrates against every negative power. At $p=0$ both moments equal one.

The long-time Spitzer winding theorem states that, for the continuous unwrapped angle ${\Theta }_t$ of planar Brownian motion started at $1$,Here $C$ has the standard symmetric Cauchy distribution. The notation means convergence in distribution: for every real $x$,Equivalently, expectations converge for every bounded continuous test function. This is a statement about the laws at large deterministic times, without asserting almost-sure convergence of sample paths. The angle is unwrapped; a principal argument is bounded and would instead give zero after division by $\log t$.

The probable error is the half-width of an interval around a specified central value containing probability $1/2$. For a normal distribution $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, its quantile function giveswhere $\Phi$ is the standard normal cumulative distribution function and $\sigma$ is the standard deviation. The factor is specific to the normal distribution. The centered Cauchy distribution with parameter $\gamma >0$ has probable error $\gamma$, despite having no finite variance. Thus a central probability width need not require a finite second moment.

Take two stipulated oppositely charged daughters with equal rest mass $m=0.14\mathrm{GeV}/c^2$ and exactly opposite momenta of magnitude $p=0.30\mathrm{GeV}/c$. Their total momentum is zero, soThis is a constructed event, not an identification of a historical particle. Boosting both four-momenta by $\beta =0.60$ along the first daughter's direction gives total energy $0.82765\mathrm{GeV}$ and total momentum $0.49659\mathrm{GeV}/c$. Substitution recovers the same invariant mass, although the laboratory energy has increased.

The following independent simulation stipulates 700 signal combinations from a Cauchy distribution centered at $0.80\mathrm{GeV}/c^2$ with scale $0.020\mathrm{GeV}/c^2$, truncated to $0.50<M<1.10\mathrm{GeV}/c^2$, plus 2400 background combinations whose density decreases linearly across that interval. Each combination is converted to a back-to-back two-daughter event with $m=0.14\mathrm{GeV}/c^2$. Independent isotropic directions are chosen, each Cartesian momentum component receives normal distribution noise with standard deviation $0.012\mathrm{GeV}/c$, and invariant mass is reconstructed from the perturbed momenta. All inputs and counts are invented. No historical experimental points or fitted historical parameters appear.

For planar Brownian motion started at $1$, let ${\Theta }_t$ be its continuous unwrapped angle, initialized at zero. Then $2{\Theta }_t/\log t$ converges in distribution to the standard symmetric Cauchy distribution, with density $1/[\pi (1+x^2)]$. The angle records all completed turns rather than a bounded principal argument. The assertion is a convergence in distribution at deterministic times; it does not assert convergence of sample paths. A precise statement appears as Theorem 7.31 in Mörters and Peres, Brownian Motion.

A probability distribution is stable when every linear combination of independent copies has the same distribution up to a location and scale change. The Cauchy distribution is stable with stability parameter one.

If $s>0$ and $\Theta$ has the continuous uniform distribution on $(-\pi /2,\pi /2)$, then $Y=s\tan \Theta$ has the Cauchy distribution with location zero and scale $s$. The tangent function is strictly increasing on this interval, so its cumulative distribution function isDifferentiating gives the probability density function $s/[\pi (s^2+y^2)]$. The probability of either endpoint is zero, so including endpoints in the uniform interval does not alter this probability distribution. The slow quadratic decay makes the expected value undefined: its positive and negative parts both have infinite integrals. Its second moment is also infinite, so it has no finite standard deviation.

Temporal optical coherence compares a field at one position with a delayed copy. In a Michelson interferometer, unequal round-trip lengths supply this delay. Even a source with rapidly changing wave phase can give high interferometric visibility close to zero delay, because both arms then deliver nearly the same source history. A source narrowed in optical spectrum generally retains correlation over larger delays; a single width does not describe sources with several separated lines. ^[ref].

For a stationary scalar field, choose the carrier convention $e^{-i2\pi \nu t}$, with frequency $\nu$ in cycles per second. Let $S(\nu )\ge 0$ be its spectral density per unit frequency, with $0<\int S(\nu )d\nu <\infty$. Its normalized field correlation is the Fourier transformThe optical spectrum weights the rotating contributions; a broad spread of frequencies allows them to cancel at short delays. This is the stationary field correlation–spectrum relation. The relevant spectrum is that of the optical field in the detected mode, rather than the low-frequency fluctuations of a detector's light intensity. ^[ref].

Here are two stipulated line profiles, each centered at ${\nu }_0$ with full width at half maximum $\Delta \nu >0$. Write $f=\nu -{\nu }_0$ and use normalized profiles on the whole detuning line:The first is a Lorentzian, mathematically a Cauchy distribution; the second is a Gaussian function. Taking their Fourier transforms gives, with the carrier factor $e^{-i2\pi {\nu }_0\tau }$ omitted from the magnitudes,These are calculated model results, not measured line profiles. The detuning approximation treats the narrow line as extending to arbitrarily large positive and negative $f$; for optical applications require $\Delta \nu \ll {\nu }_0$. Exact spectra of the complex positive-frequency field have no negative absolute frequencies. Changing to angular frequency replaces $\Delta \nu$ by $\Delta \omega /(2\pi )$; it must not silently preserve the same numerical constant.

Equal detected single-arm light intensities give interferometric visibility equal to the correlation magnitude. Doubling the Gaussian full width at half maximum from $1$ to $2\mathrm{GHz}$ halves every fixed-contrast delay. At $\tau =0.5\mathrm{ns}$ the model gives Gaussian interferometric visibility approximately $0.410686$ at $1\mathrm{GHz}$ and $0.0284471$ at $2\mathrm{GHz}$; the $1\mathrm{GHz}$ Lorentzian gives $0.207880$. The dashed curve instead keeps the $1\mathrm{GHz}$ Gaussian correlation and changes $I_2/I_1$ to $1/4$, multiplying all interferometric visibilities by $0.8$. Its lower contrast at zero delay therefore comes entirely from intensity imbalance.

A Voigt profile is the convolution of a Gaussian distribution and a Cauchy distribution. For an absorption line, thermal Doppler broadening contributes the Gaussian core while finite excited-state lifetime contributes the Lorentzian wings. Instrumental broadening and unresolved motions must also be modeled when inferring column density and temperature.