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For $Q(u)=\inf \{x:F(x)\ge u\}$, $0<u<1$, the quantile function is a monotone function and is left-continuous. The distribution function is right-continuous, giving $Q(u)\le x$ exactly when $u\le F(x)$. If ${\sup }_{v<u}Q(v)<Q(u)$, a point between them has $F(x)<u$ and contradicts this equivalence for some $v\in (F(x),u)$. A jump just to the right of $u$ is possible when $F$ is flat on an interval.

For an atomless measure $\mu$ on $\mathbb{R}$ with cumulative distribution function $F$, the monotone transport to a target with quantile function $G^{-1}$ is $T=G^{-1}\circ F$, defined $\mu$-almost everywhere. It minimizes well-defined costs $d(x-y)$ for convex continuous $d$. With atoms in the source, the common-quantile transport plan $(F^{-1},G^{-1}{)}_{\#}\mathcal{U}(0,1)$ remains available but need not be induced by a map.

Put $Q(u)=\inf \{x\in \mathbb{R}:F(x)\ge u\}$ for $0<u<1$. This is the quantile function, rather than an ordinary inverse requiring strict increase. The limits of the distribution function at infinity make $Q(u)$ finite, and the fact that $F$ is a right-continuous function ensures $F(Q(u))\ge u$. ConsequentlyIncreasing $u$ shrinks the set in the infimum, so the quantile function is a monotone function. To prove the left continuity of the quantile function, fix $u$ and let $L={\sup }_{v<u}Q(v)\le Q(u)$. If $L<Q(u)$, choose $L<y<Q(u)$. Then $F(y)<u$, so some $v\in (F(y),u)$ satisfies $Q(v)>y$, a contradiction. Thus $Q(v)\uparrow Q(u)$ as $v\uparrow u$. At a flat stretch of the distribution function, the quantile function may jump immediately to the right, consistent with this left-continuous convention.

For the uniform order statistic, count the observations at most $u$. That count has the binomial distribution with parameters $n,u$, givingDifferentiating makes adjacent terms telescope: use $r\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ and $(n-r)\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$. The probability density function is thereforeIn particular, $U_{(j)}$ has the Beta distribution with parameters $j,n-j+1$.

Write $m=Q(1/2)$ for the specified median. Continuity of the distribution function gives $F(m)=1/2$ and no mass at $m$, even if the distribution function has a flat stretch there. Thus $B={\sum }_i{1}_{\{X_i<m\}}$ has the binomial distribution with parameters $n,1/2$. Except on a null event, the order-statistic confidence interval for a median covers $m$ exactly when $j\le B\le n-j$. Symmetry of the binomial distribution makes its two failure probabilities equal. Alternatively, the probability integral transform givesHence the coverage is exactly $1-\alpha$, withThe asymmetric open/closed endpoint convention does not change the coverage because the continuous probability distribution assigns no mass to the median.

The generalized quantile function isThe tail limits of a cumulative distribution function make the defining set nonempty and its infimum finite. Monotonicity and right continuity imply $F^{-1}(u)\le x$ exactly when $u\le F(x)$. To see the direction involving the infimum, let points in the defining set decrease towards it; right continuity gives $F(F^{-1}(u))\ge u$, so every larger $x$ also qualifies. Consequently, for a uniform distribution variable $U$,The probability-zero endpoints $U=0,1$ can be assigned arbitrary outputs. This proves inverse transform sampling even for distributions with atoms or flat portions; continuity or strict monotonicity of $F$ is not required.

Write ${\lambda }_n$ for Lebesgue measure on ${\mathbb{R}}^n$. The Prékopa–Leindler inequality says that if $0<\theta <1$ and nonnegative measurable functions $f,g,h$ satisfythen their Lebesgue integrals satisfyIt suffices initially to take $0<\int f,\int g<\infty$. Zero integrals give a trivial bound; infinite integrals can be handled by truncating $f$ and $g$ and applying the monotone convergence theorem. As usual, a zero factor makes the asserted lower bound zero.

Here is a one-dimensional proof using quantile functions. Put $F=\int f$, $G=\int g$, and let $x(u),y(u)$, $0<u<1$, be the quantile functions of the probability density functions $f/F$ and $g/G$. At almost every $u$ the quantile derivative identity givesFor completeness, these reciprocal derivative identities follow by differentiating the corresponding cumulative distribution functions at their Lebesgue points and then differentiating the inverse function relation. Sampling a quantile function at a uniform argument samples the density itself, so the exceptional set, including locations where the density vanishes or is infinite, has probability zero. Gaps in a density's support can give jumps in its quantile function; they do not invalidate the argument below.

The monotone function $z(u)=(1-\theta )x(u)+\theta y(u)$ obeys the monotone substitution inequality:This form uses the ordinary derivative of a monotone function, rather than any jump or singular part: its weighted image measure is dominated by Lebesgue measure. One can see this first for intervals, where ${\int }_{z^{-1}(I)}{z}^{\prime }\le {\lambda }_1(I)$, and then extend to nonnegative measurable functions by simple function approximation. Thus no assumption of strictly positive smooth probability density functions is hidden in the proof.

Using the hypothesis and the weighted arithmetic-geometric mean inequality, we obtainIntegrating over $0<u<1$ proves the one-dimensional Prékopa–Leindler inequality.

For higher dimensions, use mathematical induction and Tonelli theorem. Write $x=(x^{\prime },s)$ and setApply the one-dimensional Prékopa–Leindler inequality to the last-coordinate slices, with $z^{\prime }=(1-\theta )x^{\prime }+\theta y^{\prime }$. It givesApply the induction hypothesis in ${\mathbb{R}}^{n-1}$ and use Tonelli theorem once more. For finite total integrals, infinite slice integrals occur only on null sets and may be replaced by zero there; this only weakens the displayed premise. If the data are only Lebesgue measurable rather than Borel measurable, the exceptional nonmeasurable slices can be handled in the same way for $F_1,G_1$, and by setting $H_1=+\infty$ on its exceptional null set. The total integrals are unchanged, and the slice premise holds everywhere it is needed. The result therefore holds for general nonnegative measurable functions.

The intended monotone rearrangement isHere $G^{-1}$ is the quantile function of $\nu$, agreeing with the ordinary inverse when $G$ is continuous and strictly increasing. With an atomless measure $\mu$, its cumulative distribution function $F$ is continuous, and the probability integral transform makes $F(X)$ uniform on $(0,1)$ for $X\sim \mu$. Thus $G^{-1}(F(X))\sim \nu$. The one-dimensional monotone rearrangement theorem says this transport map minimizes the cost $d(x-y)$ for convex continuous $d$, whenever the cost integrals are well defined. Values at exceptional endpoints may be chosen arbitrarily.

The printed assumptions omit an essential source condition. Invertibility of $G$ alone does not ensure an admissible transport map. For example, $\mu ={\delta }_0$ and $\nu$ a standard normal distribution satisfy the stated condition on $G$, but $T_{\#}{\delta }_0$ is always a Dirac measure. There is no solution to the Monge optimal transport problem in this example. The boxed answer therefore requires the additional assumption that $\mu$ is an atomless measure, or an equivalent condition making the displayed map admissible. For arbitrary sources the always admissible monotone transport plan is $(F^{-1},G^{-1}{)}_{\#}\mathcal{U}(0,1)$, which need not be induced by a map.

The quantile function $X_{\mu }=F_{\mu }^{-1}$ is nondecreasing, hence a Borel measurable function. More explicitly, for every $x\in \mathbb{R}$,Thus $X_{\mu }$ is a random variable, and because Lebesgue measure is being used on $(0,1)$,Its cumulative distribution function is therefore $F_{\mu }$, soThis is inverse transform sampling.

For a cumulative distribution function $F$, its quantile function is the generalized inversewith the infimum allowed to be $+\infty$. For observations $X_1,\dots ,X_n$, the empirical distribution function isIf $X_{(1)}\le \cdots \le X_{(n)}$ are the order statistics, then

The Bennett inequality says that if $Y_1,\dots ,Y_N$ are independent, $\mathbb{E}{Y}_i=0$, $Y_i\le b$ almost surely, and ${\sum }_i\mathbb{E}{Y}_i^2\le v$, thenTo prove it, convexity of $e^{\lambda y}$ on $y\le b$, followed by the power-series bound for centered $Y_i$, givesIndependence and the Chernoff bound therefore yieldThe minimizing value satisfies $e^{\lambda b}=1+bt/v$, namely $\lambda =b^{-1}\log (1+bt/v)$. Substitution gives the stated exponent.

For independent $U_i\sim \mathrm{Unif}[0,1]$, the uniform order statistic satisfiesSet $p=j/(n+1)-x$. The event $U_{(j)}\le p$ means that at least $j$ sample points lie in $[0,p]$. If this occurs, some $j$-element subset consists entirely of such points. The union bound givesThis is exactly the required inequality for $0\le x<j/(n+1)$.

Let $Q=F^{-1}$ be the quantile function. For $p\in (0,1)$,The reverse implication is immediate from the infimum defining $Q$. For the forward implication, if $Q(p)<t$ there is a point $s\le t$ with $F(s)\ge p$. If $Q(p)=t$, choose $s_m<t+1/m$ with $F(s_m)\ge p$; monotonicity and right continuity giveThe endpoint values $p=0,1$ have zero probability under a continuous uniform variable.

ConsequentlyThus the inverse transform sampling identity is