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The survivor function of a nonnegative event time isHere the paper uses $F$ for the survival function, rather than for the cumulative distribution function; the latter is $1-F(t)$ for a continuous distribution. The survivor function is nonincreasing, starts at 1 when there is no atom at zero, and tends to zero for an almost surely finite event time.

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.

Interpret the invertibility assumption on $G$ as continuity and strict increase on the relevant range, so that $\nu$ is an atomless measure. Suppose a non-decreasing transport map $T$ with $T_{\#}\mu =\nu$ exists. Then $\mu$ is also an atomless measure: if $\mu (\{x\})>0$, the pushforward measure would give $\nu (\{T(x)\})\ge \mu (\{x\})>0$. Thus $F$ is continuous.

At any point $x$ where the non-decreasing representative is defined, the definition of a monotone function givesUsing the pushforward measure identity and continuity of the two cumulative distribution functions, we obtainConsequentlyThe monotone rearrangement in part (c) is optimal for the convex difference cost, so $T$ has the same cost and solves the Monge optimal transport problem. The meaningful uniqueness is up to a $\mu$-null set; arbitrary values away from the source do not affect transport or cost. Existence of the non-decreasing transport map supplies the source condition missing in part (c).

A simple choice is an independent Laplace distribution at each pixel, giving a Laplace prior for sparse images:Its negative log density penalizes the L1 norm. Relative to a normal distribution, it combines a sharp peak at zero with heavier tails, allowing most pixels to lie near the background level while a few have appreciable intensity. Quantitatively, $\mathbb{P}(|u_j|>\tau )=e^{-\lambda \tau }$, so the expected number of appreciable pixels is $d{e}^{-\lambda \tau }$. Choosing $\lambda \tau$ large makes this small. This is a simple sparsity model: it does not enforce that nearby active pixels form connected objects, nor does its continuous density give exact zero pixels positive probability. Additional spatial modelling would be needed to insist on contiguous shapes.

To sample, draw independent $U_j\sim \mathcal{U}(0,1)$ and use inverse transform sampling:The endpoint events have probability zero and can be assigned arbitrary finite values. To justify the rule, the cumulative distribution function of a centered Laplace distribution isThe displayed rule is $F^{-1}(U_j)$, so $\mathbb{P}(u_j\le x)=\mathbb{P}(U_j\le F(x))=F(x)$. Independence of random variables then gives the stated product prior distribution.

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.

Since $Y_1=X_1^2$ and a standard normal variable has second and fourth moments $1$ and $3$,Thus $T_n$ has the chi-squared distribution with $n$ degrees of freedom and

Convergence in distribution $W_n\Rightarrow W$ means that the cumulative distribution functions satisfy $F_{W_n}(x)\to F_W(x)$ at every continuity point of $F_W$. The Lévy continuity theorem states that this is equivalent to pointwise convergence of characteristic functions to the characteristic function of $W$, provided the limiting function is continuous at zero.

For fixed $t$, independence and a Gaussian integral giveUsing $\log (1-z)=-z-z^2/2+O(z^3)$,Therefore ${\phi }_{V_n}(t)\to e^{-t^2/2}$, the characteristic function of the standard normal distribution. Lévy's theorem now givesThis derives the limit directly, without assuming the central limit theorem.

The scaled maximumhas cumulative distribution function $\mathbb{P}(U\le u)=u^n$ for $0\le u\le 1$. ThusThe mean squared error of the maximum likelihood estimator is therefore

Each $X_i$ from the uniform distribution on $[0,2\theta ]$ has mean $\theta$ and variance ${\theta }^2/3$. The sample mean is unbiased and has variance ${\theta }^2/(3n)$, hence

Let ${\epsilon }_i$ be independent standard logistic random variables and define the latent variableSet $Y_i=1$ exactly when $Z_i>0$. If $F$ is the logistic cumulative distribution function, its symmetry and the identity $F^{-1}(p)=\mathrm{logit}(p)$ givewhich is precisely the fitted logistic regression.

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 $N$ be the homogeneous Poisson point process in ${\mathbb{R}}^3$ with intensity measure $\lambda dx$, and map each point $x$ to its radius $g(x)=|x|$. For an interval $(a,b]\subseteq {\mathbb{R}}_{+}$, the pushforward measure of the intensity isThe Mapping theorem for Poisson point processes therefore shows that the radii form a Non-homogeneous Poisson point process with intensity function

If $R$ is the distance to the closest star, the event $\{R>r\}$ says that the ball of radius $r$ contains no points. The zero-count probability of a Poisson distribution givesDifferentiating its cumulative distribution function yields the probability density function

Conditionally on $Y_1,\dots ,Y_m$, the cumulative distribution function of $X^{\ast }$ is the weighted empirical distributionFor fixed $t$, the strong law of large numbers gives, almost surely,whileThe ratio therefore converges almost surely to $F(t)$. This is precisely the claimed convergence in distribution of the self-normalized importance sampling resample.

Apply part (b) simultaneously at every rational $t$; the intersection of these probability-one events still has probability one. On that event, monotonicity and continuity extend the convergence from the rationals to every real $t$.

More directly, fix $\epsilon >0$. Since $F$ is a continuous cumulative distribution function, choose rational pointsso that the two tails have $F(t_0)<\epsilon$ and $1-F(t_k)<\epsilon$, while every increment $F(t_i)-F(t_{i-1})<\epsilon$. Part (b) makes $|F_m^{\ast }(t_i)-F(t_i)|<\epsilon$ at all these finitely many points for all sufficiently large $m$. If $t_{i-1}\le t\le t_i$, monotonicity givesso $|F_m^{\ast }(t)-F(t)|<2\epsilon$; the same estimate follows in the tails. This proves the uniform convergence of distribution functions to a continuous limit:almost surely.

Differentiate the cumulative distribution function for $y>0$, using ${\Phi }^{\prime }(z)=(2\pi {)}^{-1/2}{e}^{-z^2/2}$ and $d\sqrt{2y}/dy=1/\sqrt{2y}$. The probability density function isIts value at zero can be assigned arbitrarily without changing any probability. Although this probability density function diverges as $y\downarrow 0$, $F_Y(0)=0$ and there is no point mass at zero. Equivalently, $Y$ has a gamma distribution with shape $1/2$ and rate one.

Since $Y=Z^2/2$ is nonnegative, its cumulative distribution function is zero for $y<0$. For $y\ge 0$,The last equality uses the symmetry of the standard normal distribution. Thus

Differentiating the cumulative distribution function in part (ii) gives the probability density function $n{t}^{n-1}$ on $(0,1)$. HenceThus an unbiased estimator based on the sample maximum is

For $0\le t\le 1$, independence and the uniform distribution giveTherefore the cumulative distribution function isIt contains no $\theta$, so $M/\theta$ is a pivotal quantity.

For a random variable with a probability density function, a singleton has zero probability:Equivalently, its cumulative distribution function is continuous, so it has no jump at $x$.

The cumulative distribution function isThe total integral of the nonnegative probability density function is one.

The cumulative distribution function must tend to one at infinity, so $\boxed{k=1}$. Differentiation gives the probability density function $f_X(x)=e^{-x}$ for $x>0$ and zero for $x<0$; there is no atom at zero. This is an exponential distribution of rate one. Its first two moments are ${\int }_0^{\infty }x{e}^{-x}dx=1$ and ${\int }_0^{\infty }{x}^2{e}^{-x}dx=2$, so

For $t_0\ge 0$, integration of the exponential distribution gives its cumulative distribution function:For $t_0<0$ the event is empty and its probability is zero.