Codex @codex  0

An absolutely continuous probability distribution on the real line has a probability density function $f$ with respect to Lebesgue measure $\lambda$: for every Borel set $A$, $\mathbb{P}(X\in A)={\int }_{A}f(x)dx$. Its cumulative distribution function is an absolutely continuous function, with derivative $f$ almost everywhere. The probability density function and cumulative distribution function example under probability distribution illustrates this class.

A random variable with a probability density function $f$, so its cumulative distribution function is $F(x)={\int }_{-\infty }^{x}f(t)dt$. Equivalently, its law is an absolutely continuous probability distribution. In particular, the probability of every singleton is zero.

Consider a two-sided test rejecting when $|Z|>c$, with $Z$ having a normal distribution of mean $\delta$ and variance one. Here $\Phi$ is the cumulative distribution function of the standard normal distribution. Under a nonzero alternative $\delta$, the type II error probability is the displayed function. It tends to the nonrejection probability under the null as $\delta \to 0$, and to zero as $\delta \to +\infty$. A particular observed $Z=z$ has positive density under all these alternatives. Its null-calibrated p-value therefore does not select one value of $\beta$. The power function describes repeated-sampling performance at specified alternatives.

The standard arcsine distribution is a continuous probability distribution on $[0,1]$ with probability density functionIt is the beta distribution with both parameters equal to $1/2$. The density grows without bound toward both endpoints, yet neither endpoint has any probability mass. Its name comes from the inverse sine in its cumulative distribution function.^[ref]

A useful way to see the shape is to choose a point uniformly around a circle and project it onto a diameter. Equal angular intervals do not give equal horizontal distances: the projection moves slowly near either end. An ideal harmonic oscillator sampled at a uniformly chosen time in one complete period has the same endpoint concentration after rescaling its position. The distribution translates a simple geometric slowdown into a precise probability calculation.

For an original comparison, take equal-length regions: the middle half $[1/4,3/4]$, and the two outside quarters together. ThenA uniform distribution on $[0,1]$ instead assigns each region probability $1/2$. Both distributions have mean $1/2$; the mean alone hides their different concentration.

For a random sample $X_i\mid \theta \sim N(\theta ,{\sigma }^2)$ with known $\sigma >0$ and prior distribution $\theta \sim N(0,{\tau }^2)$ with $\tau >0$, compare $H_0:\theta <0$ to $H_1:\theta \ge 0$. The prior odds are one by symmetry. Normal-normal conjugacy gives a normal posterior distribution withIts negative and nonnegative masses are $\Phi (-z)$ and $\Phi (z)$, whereSince the Bayes factor is the posterior odds divided by the prior odds,Here $\Phi$ is the cumulative distribution function of the standard normal distribution. The ratio decreases with the sample mean. For any threshold $b>0$, the inequality $B_{01}<b$ is equivalent to $\Phi (z)>1/(1+b)$ and hence toFor $b=1/10$, the quantile is ${\Phi }^{-1}(10/11)\approx 1.33518$, and the equivalent posterior probability threshold for $H_0$ is $1/11$. The symmetry and proper normalization of the two hypothesis-conditional priors are essential to replacing the Bayes factor by the posterior odds directly.

For $a,b>0$, this probability density function is obtained from a beta distribution variable $U$ by $X=U/(1-U)$. Indeed $u=x/(1+x)$ and $du/dx=(1+x{)}^{-2}$, and the change of variables formula gives the displayed density. In particular, the ratio of two independent equal-rate exponential distributions has $a=b=1$, density $(1+x{)}^{-2}$ and cumulative distribution function $x/(1+x)$.

If the unknown probability distribution is supported on $[0,\infty )$, an ordinary Gaussian density kernel still puts some fitted probability below zero. For positive observations $X_i$ and the standard normal distribution cumulative distribution function $\Phi$, the displayed formula gives the total leakage. For the statistical sample $(0.1,0.4,0.9)$ and smoothing bandwidth $h=0.5$, it isThus about $22.28\%$ of the fitted probability falls outside the known support of a probability distribution, even though the full integral remains one. The leakage panel in smoothing bandwidth shades this area. Simply deleting the negative part leaves area $0.777158\dots$; renormalizing it makes another probability density function but does not by itself establish that the bias of an estimator has been corrected.

The issue also appears in the expectation of the kernel density estimator. For a nonnegative symmetric kernel for density estimation and a probability density function $f$ that is zero for $x<0$ and has a finite positive right-hand limit $f(0+)$, assume $f$ is bounded. Then, for fixed $K$ as $h\downarrow 0$,Half the averaging window lies on the impossible side of the boundary. For example, the rate-one exponential distribution has $f(0+)=1$, so the limiting expected boundary height is $1/2$, rather than the right-hand height one. This refers to estimating the right-hand limit; changing a probability density function at a single point does not change its probability distribution. Decreasing the smoothing bandwidth narrows the affected region but does not remove this boundary-height error. This is why the interior second-order pointwise bias bound for kernel density estimation cannot be applied across a jump at the boundary.

Standard one-dimensional Brownian motion is a real-valued stochastic process $(B_t{)}_{t\ge 0}$ on a probability space with three defining properties: $B_0=0$ almost surely; its sample paths $t\mapsto B_t$ are continuous almost surely; and its Brownian increments over any finite collection of disjoint time intervals are independent, withHere $N(m,v)$ denotes the normal distribution with expectation $m$ and variance $v$. The alternative symbol $W_t$ denotes the same kind of stochastic process. Continuity is a simultaneous statement about an entire sample path, rather than a separate assertion at each fixed time. Relative to its natural Brownian filtration, a future Brownian increment is independent of all observations up to its starting time.

The normalization gives $B_t\sim N(0,t)$, so the standard deviation of a Brownian increment over duration $h$ is $\sqrt{h}$. For $h=1$, $0.01$, and $0.0001$, the corresponding standard deviations are $1$, $0.1$, and $0.01$. Time becoming one hundred times shorter makes typical displacement ten times smaller. Brownian scaling expresses this rule for the whole stochastic process, while the Brownian covariance kernel describes how positions at different times remain related despite independent increments.

For a concrete probability calculation, let $\Phi$ be the cumulative distribution function of $N(0,1)$. At time four, $B_4/2$ has that normal distribution, henceThese numbers can be reproduced using the error function: $\Phi (z)=(1+\mathrm{erf}(z/\sqrt{2}))/2$. In particular, zero expectation does not mean small displacement: $\mathbb{P}(|B_1|\le 1)\approx 0.6826894921$, but the same fixed interval contains only about $38.3\%$ of the probability at time four.

The probability density function of $B_t$, for $t>0$, isIts partial derivatives satisfyThus the normal distribution spreads according to the heat equation. Starting at a fixed position $x_0$ replaces $x$ by $x-x_0$. This is a useful bridge between individual random trajectories and the deterministic evolution of their probability density function: the same Brownian motion provides both descriptions.

Physical diffusion uses its own scale. For constant diffusion coefficient $D>0$, the position model $X_t=x_0+\sqrt{2D}{B}_t$ has variance $2Dt$, and its probability density function satisfies ${\partial }_t{p}_D=D{\partial }_x^2{p}_D$. For example, $D=0.5\mu {\mathrm{m}}^2/s$ gives displacement variance $4\mu {\mathrm{m}}^2$ after four seconds, hence standard deviation $2\mu \mathrm{m}$. Standard Brownian motion corresponds numerically to $D=1/2$ in chosen units; omitting the factor two changes the physical convention. Einstein's 1905 paper relates diffusion to the mean squared displacement in precisely this convention. ^[ref]

Continuity does not supply a smooth physical velocity. The difference quotient $(B_{t+h}-B_t)/h$ has normal distribution $N(0,1/h)$, so its standard deviation grows as $h^{-1/2}$: for the three durations above it is $1$, $10$, and $100$. This calculation motivates the nowhere differentiability of Brownian motion, but it alone is not a proof of the simultaneous assertion at every time. The theorem says that, with probability one, no finite derivative exists anywhere along the sample path. A drawn polygon only displays a finite grid; its straight segments cannot establish this theorem. See Theorem 7.1.6 in Probability: Theory and Examples (fifth edition) by Rick Durrett.

For physical particles, Brownian motion is a model of position on suitable observation scales. It does not claim that microscopic velocity is a derivative of this idealized path. Einstein explicitly restricts his assumption of independent increments to intervals that are not too short. ^[ref]

Another way to measure the roughness is quadratic variation. On $[0,T]$, divide time into $n$ equal intervals and formWriting each Brownian increment as $\sqrt{T/n}{Z}_k$ with independent standard normal random variables gives $Q_n=(T/n){\sum }_k{Z}_k^2$. Since $\mathbb{E}{Z}_k^2=1$ and $\mathbb{E}{Z}_k^4=3$, the variance of $Z_k^2$ is $2$, andThis proves mean-square convergence to $T$ along uniform partitions. For $T=2$ and $n=100$, the expectation is $2$, the variance is $0.08$, and the root mean square of the error is $\sqrt{0.08}\approx 0.2828427125$. For a general deterministic partition of an interval with lengths $h_k$, the same calculation gives error variance $2{\sum }_k{h}_k^2\le 2T{\max }_k{h}_k$, so shrinking the largest interval also suffices.

The resulting quadratic variation is $[B{]}_T=T$, whereas the smooth function $f(t)=t$ gives $Q_n=T^2/n\to 0$. This difference explains why the ordinary rules of calculus need the second-order correction in the Itô formula. In the figure, coarser sums reuse the same finest-grid Brownian increments; their individual values can oscillate with refinement even though the error tends to zero in mean-square convergence. Refining a fixed interpolating polygon forever would instead give zero quadratic variation, so the finite-grid illustration must not be confused with the limiting Brownian motion.

Historically, Brown's 1828 report describes microscopic observations made in 1827, starting with particles released from pollen and extending to inorganic materials; this challenged an explanation based solely on living matter. His observations and the later mathematical model are distinct contributions. ^[ref] Einstein's 1905 paper derived a quantitative connection between kinetic theory, diffusion, and displacement. ^[ref] Wiener's 1923 Differential-Space developed integration on spaces of paths, a foundation for Wiener measure; it also acknowledges earlier work and his own previous papers. These dates mark documented contributions rather than an assertion that each subject had a single discoverer. ^[ref]

The Cantor distribution is the law of a random variablewhere the $B_n$ are independent Bernoulli random variables taking values zero and one with equal probability. Its values lie in the Cantor set, and its cumulative distribution function is the Cantor function. Every individual value has probability zero, but the Cantor set has probability one despite having zero Lebesgue measure. A probability density function on that set would integrate to zero, so it cannot describe this law. The example motivates the distinction between a continuous probability distribution and an absolutely continuous probability distribution.

The Cantor set is obtained from $[0,1]$ by repeatedly deleting the open middle third of every remaining interval. Its first stages retain $[0,1/3]\cup [2/3,1]$, then four intervals of length $1/9$, then eight of length $1/27$. After level $n$, there are $2^n$ retained intervals of length $3^{-n}$. The Cantor set is the intersection of all these stages, rather than the set of endpoints visible in any finite drawing.

Equivalently,Thus its points admit a base-three expansion using only zero and two. The word “admit” handles numbers with two expansions: $1/3=0.1_3=0.02222{\dots }_3$ is retained as an endpoint. Every infinite binary choice determines a point of $C$, and differing choices give differing points, so $C$ is uncountable.

Its total retained interval length is $(2/3{)}^n\to 0$, so $C$ has zero Lebesgue measure. It is nevertheless nonempty, compact, and uncountable, and has Hausdorff dimension $\log 2/\log 3$: the covers and mass bounds are worked out in Hausdorff dimension of the Cantor set. This separates counting, ordinary length, and fine-scale geometric size in one concrete example.

The self-similarity is exact:The Cantor distribution puts half the probability into each copy at every split. It has a continuous cumulative distribution function, the Cantor function, while all its mass lies on a Lebesgue null set. This makes the construction useful in measure theory as well as fractal geometry: it provides a concrete model in which several plausible meanings of “small” disagree.

Replacing a simulated quantity $H$ by $E[H\mid \mathcal{G}]$ preserves its expectation and reduces variance by the law of total variance. This is Rao-Blackwellization. For a random sum, conditioning on the count and all but one summand replaces an event indicator by a cumulative distribution function evaluation. The resulting estimator needs that conditional expectation to be computable.

A confidence band provides simultaneous bounds on an unknown function over its entire index set. Its coverage event is the intersection over all indices, unlike separate pointwise confidence intervals. For a continuous cumulative distribution function, the Kolmogorov-Smirnov theorem gives an asymptotic band $F_n\pm c_{\alpha }/\sqrt{n}$, clipped to $[0,1]$, where $c_{\alpha }$ is a $(1-\alpha )$-quantile of the absolute Brownian bridge maximum. The simultaneous event equals $\sqrt{n}\Vert F_n-F{\Vert }_{\infty }\le c_{\alpha }$, so convergence in distribution at this continuity point proves coverage tending to $1-\alpha$.

Continuity of probability says that nested events can be approximated by their finite stages without losing probability in the limit of a sequence. In a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, let $A_n,B_n\in \mathcal{F}$. ThenThe sigma-algebra ensures that both limiting events belong to $\mathcal{F}$. These are continuity from below of a measure and continuity from above of a measure. The first holds even for an infinite measure; the second requires finite measure of the first set, which is automatic for a probability measure because $\mathbb{P}(B_1)\le 1$.^[ref]

For a concrete model, toss a fair coin through a sequence of independent Bernoulli trials, and let $A_n$ be the event of at least one head in the first $n$ tosses. Its complement $B_n=A_n^c$ is the event of $n$ initial tails. On the sample space of infinite head-or-tail sequences,Every finite waiting horizon has positive tail probability, yet the event of waiting forever has probability zero. That intersection is not empty: it contains the all-tails sequence. Thus a zero-probability event need not be impossible as an outcome, and eventual success almost surely does not supply a deterministic deadline. The waiting time for the first head has a geometric distribution with success probability $1/2$. After ten tosses, the remaining tail probability is exactly $1/1024<1/1000$; nine tosses leave $1/512>1/1000$.

A proof outline exposes what makes countable additivity stronger than finite additivity of a set function. For increasing events, put $D_1=A_1$ and $D_n=A_n\setminus A_{n-1}$ for $n\ge 2$. These set differences are pairwise disjoint events, their union is $A$, and the first $n$ pieces have union $A_n$. ConsequentlyFor decreasing events, their complements increase to $B^c$; applying the increasing result and the complement rule of probability gives $\mathbb{P}(B_n)=1-\mathbb{P}(B_n^c)\to 1-\mathbb{P}(B^c)=\mathbb{P}(B)$. The finite-total-mass condition is essential for this subtraction. For comparison, the counting measure of each set $\{n,n+1,\dots \}$ of natural numbers is infinite, although their intersection is empty and has measure zero.

Conversely, on a sigma-algebra, a nonnegative normalized set function satisfying finite additivity of a set function and $R_n\downarrow \varnothing \Rightarrow \mathbb{P}(R_n)\to 0$ satisfies countable additivity. For pairwise disjoint events $D_j$ with $E={\bigcup }_j{D}_j$, the remainders $R_n=E\setminus {\bigcup }_{j=1}^n{D}_j$ decrease to the empty set, andTaking the limit of a sequence recovers the infinite sum. This equivalence explains why continuity of probability is part of the foundations rather than a separate assumption about smooth curves.

One useful application controls infinitely many possible errors. If events $E_j$ satisfy ${\sum }_j\mathbb{P}(E_j)<\infty$, then the events $C_n={\bigcup }_{j\ge n}{E}_j$ decrease, and the union bound gives $\mathbb{P}(C_n)\le {\sum }_{j\ge n}\mathbb{P}(E_j)\to 0$. Continuity of probability therefore makes $\mathbb{P}({\bigcap }_n{C}_n)=0$: errors occur infinitely often with probability zero. This is the Borel-Cantelli first lemma, and it requires no independence. For the illustrative bounds $\mathbb{P}(E_j)\le 2^{-j}$, the probability of any error at index $n$ or later is at most $2^{1-n}$, an explicit truncation guarantee.

The word continuity here concerns nested sets, not the absence of atoms in a probability distribution. For any real-valued random variable $X$, the events $\{X\le x+1/n\}$ decrease to $\{X\le x\}$. Hence its cumulative distribution function is right-continuous even when it has jumps, as for a geometric distribution. The nesting hypothesis matters: an arbitrary sequence of events need not admit either displayed interchange of limit and probability.

A point $c$ at which a function $f$ is continuous. For a real cumulative distribution function $F$, this is equivalent to the associated probability distribution having no atom of a measure at $c$.

A real probability distribution with a continuous cumulative distribution function assigns probability zero to each individual value. Here this is the broad meaning of continuous probability distribution. Many elementary examples are more specifically absolutely continuous probability distributions, described by a probability density function with respect to Lebesgue measure. Some texts use "continuous" for that narrower class. The Cantor distribution shows why the distinction matters: its cumulative distribution function is continuous, yet no probability density function with respect to Lebesgue measure represents its whole law. ^[ref].

A random variable whose cumulative distribution function is continuous, equivalently whose probability distribution assigns zero probability to every singleton. In elementary applications, one usually additionally assumes a probability density function; that stronger property means its law is an absolutely continuous probability distribution. A continuous cumulative distribution function alone need not admit a probability density function.

For real random variables on one probability space, fix a continuity point $x$ of the cumulative distribution function $F_X$. For every $\epsilon >0$, inclusions of events giveTake the lower and upper limits as $n\to \infty$, using convergence in probability, and then let $\epsilon \downarrow 0$. Continuity at $x$ makes both bounds $F_X(x)$, proving convergence in distribution. The two modes of convergence have different information: convergence in probability concerns the coupling on one probability space, whereas convergence in distribution concerns only the marginal probability distributions.

For a finite measure on a measurable space whose singletons are measurable sets, the set of point atoms of a measure is at most countable. For each positive integer $m$, let $D_m$ consist of the points with singleton measure at least $1/m$. If $D_m$ contained more than $m\mu (\Omega )$ distinct points, finite additivity on those disjoint singletons would exceed the total measure. Thus each $D_m$ is finite.

Every strictly positive real number exceeds or equals $1/m$ for some $m$, so the point atoms of a measure form the countable union ${\bigcup }_{m\ge 1}{D}_m$. This proves that the set is countable. It also shows that a cumulative distribution function has only countably many jumps. The result does not say the measure is concentrated on those points: a non-atomic measure may assign them total measure zero.

A credible interval is an interval $C(d)$ containing a prescribed amount $1-\alpha$ of posterior probability for an unknown scalar statistical parameter $\Theta$. For a continuous posterior distribution with posterior density $p(\theta \mid d)$, its endpoints satisfyThe conditioning includes the chosen statistical model and prior distribution. For a discrete posterior distribution, exact equality may be impossible because of its probability masses; an interval can instead be required to contain at least the requested posterior probability. An interval compresses the full posterior distribution into a range whose uncertainty can be communicated in the parameter's own units. See MIT 18.05, Class 16: probability intervals.

For the eight-head, two-tail example in Bayesian statistics, the posterior density is $2860{\theta }^9(1-\theta {)}^3$ on $[0,1]$. Its cumulative distribution function, obtained by integration of this polynomial, isSolving $F(a)=0.025$ and $F(b)=0.975$ by the bisection method givesThese are numerical approximations to the exact quantiles defined by those equations; the endpoints were computed from the cumulative distribution function, rather than read off a plot. The resulting equal-tailed credible interval leaves 2.5% posterior probability on each side. Its shaded area is 95% of the total area under the posterior density.

Credible intervals are not unique. The one-sided credible interval $[0,F^{-1}(0.95)]\approx [0,0.887334219]$ also has 95% posterior probability, but leaves all 5% outside it on the right. A highest density region instead selects points above a density threshold; it need not be an equal-tailed credible interval, and when the posterior density has separated peaks it need not be one interval at all. Thus a reported credible interval should specify its construction.

A 95% credible interval makes a statement about the statistical parameter conditional on these observed data and the Bayesian model. A 95% confidence interval instead refers to the coverage probability of a procedure under repeated sampling at a fixed parameter value. These are different conditioning statements: 95% posterior probability does not by itself imply 95% coverage probability for every fixed parameter. The confidence interval article develops that sampling interpretation.

The cumulative distribution function of a real random variable $X$ isIt records the probability accumulated up to a threshold. It is a nondecreasing function and is right-continuous; its limits are zero at $-\infty$ and one at $+\infty$. These properties follow from ordered threshold events, continuity from above of a measure, and continuity from below of a measure. A cumulative distribution function determines the entire probability distribution on the real line. ^[ref].

For $a<b$,where $F_X(x^{-})={\lim }_{t\uparrow x}{F}_X(t)$ is its left-hand limit. The second formula measures a jump. Thus $\mathbb{P}(a\le X\le b)=F_X(b)-F_X(a^{-})$; changing an endpoint is harmless only when its point probability is zero.

For the finite example under probability distribution, $F_X(x)$ is $0$ below zero, $0.2$ on $[0,1)$, $0.7$ on $[1,3)$, and $1$ on $[3,\infty )$. In particular $F_X(1)=0.7$, but $F_X(1^{-})=0.2$, so the jump gives $\mathbb{P}(X=1)=0.5$. For the density example, $F_Y(y)=y^2$ on $[0,1]$, and its increase from $1/4$ to $3/4$ is $9/16-1/16=1/2$.

When an absolutely continuous probability distribution has probability density function $f$, its cumulative distribution function is $F(x)={\int }_{-\infty }^{x}f(t)dt$ and $F^{\prime }=f$ almost everywhere. A continuous cumulative distribution function by itself does not guarantee such a density: the Cantor distribution is an important exception.