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The quantile function converts a cumulative probability level into a value. For a cumulative distribution function $F$ and $0<p<1$, use the generalized inverseIt selects the first threshold reaching the requested level. The infimum handles jumps and flat portions, so this construction does not require an ordinary inverse function. Other endpoint conventions are possible at $p=0$ or $p=1$; the open probability interval avoids those issues. NIST introduces the inverse-direction question as the percent point function. ^[ref].

For the finite example under probability distribution,Thus $Q_X(0.1)=0$, $Q_X(0.5)=1$, $Q_X(0.9)=3$. The value at $p=0.2$ is still zero, because $F_X(0)=0.2$ already reaches that level. For $F_Y(y)=y^2$ on $[0,1]$, the quantile function is $Q_Y(p)=\sqrt{p}$, giving $Q_Y(0.5)=1/\sqrt{2}\approx 0.70710678$.

A median is a central threshold; $Q(1/2)$ supplies one, though a median need not be unique. The same quantile function becomes a sampling algorithm through inverse transform sampling. Jumps in $F$ become flat bands in $Q$, precisely the mechanism needed to assign positive probability to one outcome.

For $r\ge 0$, integration gives the survival function and cumulative distribution functionSolving $F_R(r_p)=p$ gives the quantile function^[ref]

For the chosen positioning model $\sigma =2\mathrm{cm}$,The disk of radius one component standard deviation therefore contains about 39.35% of outcomes; the familiar one-dimensional normal interval percentage does not apply to a two-dimensional disk. The 95% and 99% radial thresholds areA circle defined by $R=r_p$ itself still has probability zero; its enclosed disk carries the stated probability.

The closed-form quantile function also gives inverse transform sampling: if $U$ has uniform distribution on $(0,1)$, then $\sigma \sqrt{-2\log U}$ has Rayleigh distribution. Using $1-U$ instead gives the same law. Endpoints are excluded so that the logarithm stays finite.

Let the numerical support of a probability distribution be $x_1<\cdots <x_n$. Its cumulative distribution function counts how many values have been passed:It is zero below $x_1$, equals $j/n$ on $x_j\le t<x_{j+1}$ for $1\le j<n$, and is one at and above $x_n$. Every jump is $1/n$, even when the horizontal gaps differ.

For inverse transform sampling, take an ideal continuous uniform distribution for $U$ on $[0,1)$. Partition that interval into $n$ intervals of width $1/n$ and outputEach interval has probability $1/n$, proving the desired discrete uniform distribution. Using the floor function, the index is $j=1+\lfloor nU\rfloor$; for consecutive integer values the output is $X=a+\lfloor nU\rfloor$. With the usual quantile function convention $Q(u)=\inf \{t:F_X(t)\ge u\}$, one instead writes $Q(u)=x_{\lceil nu\rceil }$ for $0<u\le 1$, using the ceiling function. The endpoint conventions differ only at a finite set of points with probability zero under the ideal continuous uniform distribution.

For $S=\{1,2,4,8\}$, the four intervals $[0,1/4)$, $[1/4,1/2)$, $[1/2,3/4)$, and $[3/4,1)$ select $1,2,4,8$, respectively. Thus $U=3/10$ selects $2$, while $U=9/10$ selects $8$. The upper half of the interval selects values at least four, recovering probability $1/2$. Equal widths refer to the sampling probability coordinate, not to distances between output values. This is an exact mathematical construction assuming an ideal continuous uniform distribution; a finite computer random-number source needs its own justification of equal output chances.

To draw $N(\mu ,1)$ restricted to a positive value, let $a=\Phi (-\mu )$ and return $\mu +{\Phi }^{-1}(a+(1-a)U)$ for uniform $U\in (0,1)$. For restriction to a negative value use $\mu +{\Phi }^{-1}(aU)$. These are inverse transform sampling constructions. Repeatedly drawing an untruncated normal until its sign agrees is also exact, but inefficient for rare signs.

A triangular probability density function rises linearly from zero at $a$ to its mode $c$, then falls linearly to zero at $b$. Its peak is $2/(b-a)$. Its inverse transform sampling rule is $a+\sqrt{u(b-a)(c-a)}$ for $u\le (c-a)/(b-a)$, and $b-\sqrt{(1-u)(b-a)(b-c)}$ otherwise.

A uniform random variable has a uniform distribution. Independent such variables can supply independent inverse transform sampling; reusing a single variable for two transforms generally creates dependence.