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At fixed measured count slope, $N_A^{\mathrm{sed}}\propto a^{-3}$, whereas at fixed measured diffusion coefficient, $N_A^{\mathrm{diff}}\propto a^{-1}$. First-order propagation of uncertainty follows by taking logarithmic derivatives:For independent small errors, the radius contributes $9({\sigma }_a/a{)}^2$ to the relative variance of $N_A^{\mathrm{sed}}$, compared with $({\sigma }_a/a{)}^2$ for $N_A^{\mathrm{diff}}$. Shared radius and temperature calibrations make the two estimates correlated; their agreement is therefore not two wholly independent determinations. As a finite-error example, overestimating $a$ by $5\%$ multiplies the respective answers by $1.05^{-3}\simeq 0.8638$ and $1.05^{-1}\simeq 0.9524$.

This explains the importance of the radius cross-checks in the 1909 memoir: testing the Stokes drag law through an independent radius estimate helps prevent circular inference.^[ref] Convection must also be excluded: a residual horizontal speed $u$ adds $u^2{\tau }^2$ to an uncentred mean squared displacement, making a linear fit misleading.

The predictor values are random, and the regression function is the conditional mean of the response. For independent pairs and uniformly bounded conditional error variances, kernel-window occupancies govern the stochastic part of a local polynomial regression risk bound. This differs from fixed-design nonparametric regression, where the predictor values are deterministic.

For independent study estimates $y_i$ with within-study variances $v_i$, a usual approximate model is $y_i\mid {\delta }_i\sim N({\delta }_i,v_i)$ and ${\delta }_i\sim N(\mu ,{\tau }^2)$. Its marginal distribution is $N(\mu ,v_i+{\tau }^2)$, so for fixed heterogeneity ${\tau }^2$ the inverse-variance weighted mean uses weights $(v_i+{\tau }^2{)}^{-1}$. Here $\mu$ describes the mean effect across comparable studies; ${\tau }^2$ represents between-study variation. Estimation and uncertainty for heterogeneity are essential, especially with few studies.

Randomized response masks which of two complementary binary questions a person answered. Suppose the sensitive property has probability $p$, and independently the direct question is selected with known probability $x$, while the complementary question is selected otherwise. Truthful answers then have yes probability $q=xp+(1-x)(1-p)$. The investigator observes only the answer, preserving uncertainty about an individual's status while permitting population statistical estimation when $x\ne 1/2$.

For $n$ independent responses, $Y\sim \mathrm{Bin}(n,q)$, and the affine estimator $\tilde{p}=(Y/n-1+x)/(2x-1)$ is an unbiased estimator. Its variance is $q(1-q)/[n(2x-1{)}^2]$, by the mean and variance of a binomial distribution. Equivalently the variance is $p(1-p)/n+x(1-x)/[n(2x-1{)}^2]$, exposing the extra sampling uncertainty caused by the masking. At $x=1/2$, all $p$ produce $q=1/2$ and identifiability fails. At $x=0$ or $1$, there is no masking and the extra variance vanishes.

Because $0\le p\le 1$, maximum likelihood estimation gives $\hat{p}=\min (1,\max (0,\tilde{p}))$. This follows by maximizing the binomial likelihood over the interval $q\in [\min (x,1-x),\max (x,1-x)]$ and then inverting its monotone affine parametrization. Clipping can introduce bias: for $n=1$, $x=3/4$ and $p=0$, the clipped estimate has mean $1/4$.

Let $U$ have the uniform distribution on $[0,1)$ and place a point at every $k+U$, $k\in \mathbb{Z}$. This is a simple point process, locally finite, and a stationary point process: a translation only changes the phase modulo one, preserving its uniform distribution. For every Borel set $A$, the Tonelli theorem givesIts consecutive points are exactly one unit apart, despite having the same intensity measure as a unit-rate Poisson point process. This is the one-dimensional translated-grid construction.^[ref]

Every half-open interval of length one contains exactly one point. For $A=[0,1/2)$ and $B=[1/2,1)$, $\Pi (A)+\Pi (B)=1$, each count has the Bernoulli distribution with parameter $1/2$, andMore generally, an interval of length $L=m+r$, with $m$ a nonnegative integer and $0\le r<1$, contains $m$ or $m+1$ points with probabilities $1-r$ and $r$. Its count variance is $r(1-r)$, while a unit-rate Poisson point process has count variance $L$.

A component of measurement error fluctuating between repetitions under specified conditions, as distinguished from a shared systematic error. Independent zero-mean fluctuations of variance ${\sigma }^2$ give a sample-mean variance ${\sigma }^2/N$; correlated fluctuations need not follow this averaging law.

In an 1880 paper, John William Strutt, known as Lord Rayleigh, studied the resultant of many oscillations with a common frequency and arbitrary oscillation phases. His large-number calculation distinguished a mean intensity from a definite resultant amplitude and used planar coordinates followed by a polar integral. The resulting amplitude density is the Rayleigh distribution in a different scale convention. This is a large-number description, not an exact law for every finite collection of oscillations.^[ref]

To see the connection in a chosen model, take independent oscillation phases ${\Theta }_j$ with uniform distribution on $[0,2\pi )$ and equal component amplitude $a$. The phasor sum has coordinatesBoth have mean zero, variance $n{a}^2/2$ and zero covariance. The multivariate central limit theorem gives an approximately circular Gaussian random vector for large $n$, with $\sigma =a\sqrt{n/2}$. Its amplitude $\sqrt{X_n^2+Y_n^2}$ is therefore approximately Rayleigh. For every finite $n$, that amplitude is bounded by $na$, whereas a Rayleigh distribution has an unbounded tail. At $n=1$ it is exactly $a$. This simple endpoint check prevents the approximation from becoming a false exact identity.

For independent claim sizes with common expected value $m$ and variance $v$, independent of the nonnegative integer count $N$, the aggregate satisfies $\mathbb{E}S=m\mathbb{E}N$ and $\mathrm{Var}(S)=v\mathbb{E}N+m^2\mathrm{Var}(N)$. Its moment-generating function is $G_N(M_X(t))$ wherever finite. These identities follow from the law of total expectation and law of total variance.

A real-valued random variable is a measurable function $X$ from a probability space $(\Omega ,\mathcal{F},\mathbb{P})$ to the real numbers with their Borel sigma-algebra. More general targets give random elements. For every Borel set $A$, measurability ensures that $\{\omega :X(\omega )\in A\}$ is an event, so the question "does the observed value lie in $A$?" has a defined probability.

The function $X$ is fixed; its input outcome is uncertain. This separates what we measure from how the uncertainty is modeled. A probability distribution is the resulting pushforward measure, ${\mathbb{P}}_X(A)=\mathbb{P}(X\in A)$. A cumulative distribution function, probability mass function, or probability density function describes that measure in an appropriate setting. The random variable and its probability distribution are different objects: the former also retains its relationship to other measurements on the same sample space.

For two independent uniform dice, let $U,V$ record their results and define $S=U+V$ and $M=\max (U,V)$. The 36 ordered pairs are equally likely, but the resulting probability mass functions differ:For the sum, count the diagonal of the outcome grid having total $s$. For the maximum, subtract the $(m-1{)}^2$ pairs with both results below $m$ from the $m^2$ pairs with both results at most $m$. These examples show how a deterministic function transforms a simple input probability distribution into a nonuniform output one.

The corresponding expected values are $\mathbb{E}S=7$ and $\mathbb{E}M=161/36$. The first follows immediately from linearity of expectation; the second follows from ${\sum }_{m=1}^{6}m(2m-1)/36$. Neither number must be an attainable result. An expected value describes an average under the probability measure, not a prediction that the next observation will equal it.

The distinction between a random variable and its probability distribution matters for dependence. $U$ and $V$ have the same discrete uniform distribution and are independent random variables. $U$ and a second copy of $U$ also have that same marginal probability distribution, but they always agree. Their joint probability distribution is concentrated on the diagonal; the marginals alone cannot say which relationship holds. This is why a model for several measurements must specify their joint law or an appropriate independence assumption.

Indicator random variables turn events into numbers: $1_A$ equals one on $A$ and zero elsewhere, and $\mathbb{E}{1}_A=\mathbb{P}(A)$. This connects event counting with expected value and variance, and lets tools for random variables answer questions about probabilities. More complicated functions of observations can describe totals, waiting times, or estimation errors using the same framework.

For a random walk with centered independent identically distributed increments of variance one, the maximum of its polygonal interpolation is the maximum at its grid vertices. The maximum functional is Lipschitz for the supremum norm, so the continuous mapping theorem applied to the Donsker invariance principle proves the displayed convergence in distribution. The Brownian reflection principle then gives upper-tail limits $2(1-\Phi (a))$ at all $a>0$, since the limiting maximum has no atom there.

The exact range is a separation beyond which the semivariogram reaches its sill. A Gaussian semivariogram with nonzero structured variance approaches its sill asymptotically and therefore has infinite exact range. Software range parameters may instead specify a scale.

For independent uniform random variables on $(\theta ,3\theta )$ with $\theta >0$, the sample minimum and maximum form a sufficient statistic. Conditional on these endpoints, exchangeability and the uniform law of the remaining observations give $E[X_1\mid X_{(1)},X_{(n)}]=(X_{(1)}+X_{(n)})/2$. Applying the Rao-Blackwell theorem to $X_1/2$ gives the displayed unbiased estimator. Its variance is ${\theta }^2/[2(n+1)(n+2)]$, compared with ${\theta }^2/12$ for $X_1/2$.

For $n\ge 3$ independent labels with three type probabilities, conditioning the indicator of an ordered $(0,1,2)$ initial triple on the multinomial distribution type counts gives the displayed unbiased estimator. The Rao-Blackwell theorem reduces its variance strictly when all three probabilities are positive. At a boundary parameter with a missing type, both variances vanish.

For $0<q\le 1$, put probability mass $1-q$ at zero and $q/2$ at each of $-1$ and $1$. This symmetric probability distribution has mean zero. Its second and fourth moments are both $q$, because the nonzero outcomes have square and fourth power equal to one:For $q=1/4$ the kurtosis is $4$; for $q=1/100$ it is $100$. In the latter case $99\%$ of the probability mass is at zero, yet all of the fourth central moment comes from the remaining $1\%$. The standard score $Z=X/\sqrt{q}$ takes values $0$ and $\pm q^{-1/2}$, with variance one. Each nonzero outcome contributesto kurtosis. At $q=1/100$, the two rare outcomes are ten standard deviations from the mean, and their contributions are $50+50=100$.

Every member has finite support of a probability distribution, so all its moments are finite. Nevertheless, kurtosis becomes arbitrarily large as $q$ approaches zero: high kurtosis does not require an unbounded support of a probability distribution or a heavy-tailed distribution. At $q=0$ the variance vanishes and kurtosis is undefined. At $q=1/3$ this discrete probability distribution even has the normal distribution value $\kappa =3$ and excess kurtosis zero, so those values do not establish a normal distribution.

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.

The Rayleigh distribution is the probability distribution of the Euclidean norm of a two-dimensional random vector with independent zero-mean normal random variables of equal variance as components. If $X,Y\sim N(0,{\sigma }^2)$ are independent, thenHere $\sigma$ is the standard deviation of each component, not of the radius $R$. At $\sigma =1$ this gives the unit-scale probability density function $r{e}^{-r^2/2}$.^[ref]

This geometry turns two signed errors into one nonnegative distance. For example, suppose horizontal and vertical positioning errors are independent $N(0,4{\mathrm{cm}}^2)$ random variables. Their means both vanish, but the distance from the intended point is usually positive. The Rayleigh distribution with $\sigma =2\mathrm{cm}$ gives the probability of landing inside any centered disk. These are chosen model assumptions, rather than measurements of a particular instrument.

An apparently paradoxical feature is that the most likely location per unit area is the origin, while the radial mode is $\sigma$. A narrow annulus farther out has more area. Near zero that growing circumference outweighs the falling planar density; farther out the exponential function wins. Distinguishing densities measured against different elements of area or length resolves the paradox.

The joint probability density of $X,Y$ isIn polar coordinates, $x=r\cos \theta$, $y=r\sin \theta$, and the absolute Jacobian determinant is $r$. Thus the joint density of radius and angle factors asIntegrating over angle gives the Rayleigh distribution. The same factorization shows that the angle has uniform distribution and is independent of the radius. This is a short change of variables argument: an annulus of width $dr$ has area $2\pi rdr$ to first order.

The assumptions are substantive. Unequal component variances give elliptical rather than circular density contours; a nonzero mean displaces the center. Even zero covariance and equal normal distributions do not suffice without independence. For an original counterexample, let $X\sim N(0,{\sigma }^2)$ and let $S$ be an independent Rademacher random variable. Put $Y=SX$. Then $Y$ has the same normal distribution as $X$, and $\mathrm{Cov}(X,Y)=\mathbb{E}[S{X}^2]=0$. Nevertheless, all points lie on $y=x$ or $y=-x$, and $R=\sqrt{2}|X|$ has densitywhich is positive at zero, unlike the Rayleigh distribution. This pair is not a Gaussian random vector.

Let $r$ be an unnormalized density, $Z=\int r$ with $0<Z<\infty$, and $\pi =r/Z$. Given independent draws ${\theta }_t\sim \pi$ and an evaluable normalized density $q$ supported where $r>0$, reciprocal importance sampling estimates $Z^{-1}$ by $T^{-1}{\sum }_{t}q({\theta }_t)/r({\theta }_t)$. Its expectation is $Z^{-1}\int q=Z^{-1}$ and its variance is $(T{Z}^2{)}^{-1}(\int q^2/\pi -1)$. Broad tails of $q$ relative to $\pi$ can make this infinite; reversing the importance sampling ratio reverses the dangerous tail mismatch. Taking the reciprocal of the estimate is generally biased for $Z$.

If a set partition refines a rectangle $U\times V$ of a graph into rectangles $U_a\times V_b$, write $d=d(U,V)$ for its edge density of a bipartite graph. The weighted variance identity isIt follows by expanding the square and using ${\sum }_{a,b}|U_a||V_b|d(U_a,V_b)=|U||V|d$. In particular, equitable regularity energy cannot decrease under refinement. If a witness rectangle has relative sizes at least $\epsilon$ and edge density of a bipartite graph discrepancy greater than $\epsilon$, its contribution, after division by the squared total vertex count, exceeds ${\epsilon }^4|U||V|/N^2$.

For air approximated by index one, $n=\sin i/\sin r$, so $dn/n=\cot idi-\cot rdr$. Taking the variance gives the displayed first-order error formula, with angles and their errors in radians. Independence removes the covariance term; hard error bounds require a different combination rule. Common angular-zero errors can produce covariance and should not automatically be added in quadrature.