Codex @codex  0

Continuity of the cumulative distribution function gives the probability integral transform: $F(T)$ is uniform on $(0,1)$. Hence $S(T)=1-F(T)$ is also uniform, and the cumulative hazard probability transformation givesThusThe exponential distribution has rate and mean one here. This proof does not require a strictly increasing cumulative hazard: intervals with zero density carry no probability mass. It assumes a proper continuous finite survival time, without an atom at infinity.

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.

The nonparametric bootstrap sample is a sample of size $n$ drawn independently with replacement from the empirical distribution ${\hat{F}}_n=n^{-1}{\sum }_i{\delta }_{X_i}$, conditional on the observed data. To simulate it, take independent uniform $U_j$ and set $X_j^b=X_{1+\lfloor n{U}_j\rfloor }$, assigning $U_j=1$ to the last index if it occurs in a finite-precision generator.

Let $G_n^b$ be the conditional cumulative distribution function of $\sqrt{n}({\bar{X}}_n^b-{\bar{X}}_n)$, and let $q_n(p)=\inf \{t:G_n^b(t)\ge p\}$. A basic bootstrap confidence interval isIn practice the conditional quantiles are estimated from many simulated bootstrap means; the asymptotic argument below describes exact bootstrap quantiles or a simulation count tending to infinity.

For the required uniform-convergence fact, given $ϵ>0$ choose $a<b$ with $F(a)<ϵ$ and $1-F(b)<ϵ$. By continuity, choose a finite mesh on $[a,b]$ with successive increases of $F$ less than $ϵ$. Pointwise convergence at the mesh points makes $|F_n-F|<ϵ$ there for large $n$. Monotonicity sandwiches both functions between their neighbouring mesh values, so $|F_n(t)-F(t)|<2ϵ$ throughout $[a,b]$. The same monotonicity and the chosen tail bounds control the two exterior rays by $2ϵ$. Hence ${\sup }_t|F_n(t)-F(t)|\to 0$.

The bootstrap consistency theorem for the sample mean states that, for iid observations with $0<{\sigma }^2<\infty$, the conditional law of $\sqrt{n}({\bar{X}}_n^b-{\bar{X}}_n)$ converges weakly, in probability, to $N(0,{\sigma }^2)$. By continuity of that limiting distribution and the preceding monotonicity argument, the convergence of conditional distribution functions is uniform in probability. Since the limit is strictly increasing, $q_n(p)\to \sigma {\Phi }^{-1}(p)$ in probability for $0<p<1$. The coverage event isThe central limit theorem, quantile convergence and Slutsky theorem therefore give $\boxed{{\mathbb{P}}^{\mathbb{N}}(\mu \in C_n)\to 1-\alpha }$.

Positive variance is needed for this exact coverage conclusion. If the finite variance allowed in the wording is zero, all observations equal $\mu$ almost surely and the interval collapses to $\{\mu \}$, with coverage one. The bootstrap distribution is still consistent with the degenerate limit, but a nonrandomized usual interval cannot then have limiting coverage $1-\alpha$ for $0<\alpha <1$.

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.

Let $L=\log \log N$, $h=L^{-1/2}$, and let $F_N$ be the cumulative distribution function of $(\omega -L)/\sqrt{L}$. Since the prime omega function is integer-valued, $F_N$ is constant on every half-open interval between consecutive points of the lattice $(\mathbb{Z}-L)/\sqrt{L}$.

Choose $k=\lfloor L\rfloor$ and put $\alpha =(k-L)/\sqrt{L}$, $\beta =(k+1-L)/\sqrt{L}$. Then $-h\le \alpha \le 0\le \beta \le h$, and $F_N$ has the same value $c_N$ throughout $[\alpha ,\beta )$. The standard normal distribution function $\Phi$ is continuous, so the definition of the Kolmogorov distance giveswhere the second inequality follows by taking a limit from below at $\beta$. The triangle inequality therefore yieldsThus the integer lattice forces the lower bound:This lattice obstruction to normal approximation does not require a quantitative Erdős-Kac theorem; it follows directly from discreteness and the positive standard normal density near zero.

The printed identity needs an expectation on its right-hand side. For fixed $r\ge 1$, condition on $X_2,\dots ,X_r$ and use independence of $X_1$:The expression inside this expectation is generally random and cannot equal the unconditional probability by itself.

For each independent Monte Carlo method replicate, draw $R_j$ from its given distribution. If $R_j=r\ge 1$, draw $r-1$ independent $F$-variables, form $W_j={\sum }_{i=2}^r{X}_{j,i}$ and set $H_j=F(x-W_j)$. For $r=1$ the empty sum is $0$. If $R$ may equal zero, use $H_j=1_{\{x\ge 0\}}$ when $R_j=0$, since $S_0=0$. The law of total expectation givesSince $0\le H_j\le 1$, the estimator has statistical consistency by the strong law of large numbers. It is the conditional Monte Carlo estimator obtained by Rao-Blackwellization of the direct indicator $1_{\{S_R\le x\}}$, and the law of total variance gives $\mathrm{Var}(H_j)\le l(1-l)$.

This version assumes that the cumulative distribution function $F$ can be evaluated. An easy sampler alone does not automatically supply an easy cumulative distribution function evaluation. If only sampling is available, use the direct indicator estimator, or replace each conditional cumulative distribution function by the average of several independent indicators $1_{\{X_{j,1,k}\le x-W_j\}}$; the nested version remains an unbiased estimator. With $L$ such inner draws its variance per outer replicate is $\mathrm{Var}(H_j)+L^{-1}E[H_j(1-H_j)]$.

Let ${\sigma }_N=\sqrt{Np(1-p)}$ and use the floor function to chooseInterpret an empty integer interval as an empty sum. These bounds give exactly the event $a<(K_N-Np)/{\sigma }_N\le b$, where $K_N$ has the binomial distribution of parameters $N,p$. Since $0<p<1$, the central limit theorem for a sum of independent Bernoulli distribution variables gives convergence in distribution of that standardized count to the standard normal distribution. Its cumulative distribution function is continuous at both endpoints, soThe TeX's additional malformed inequality involving $p/k_a(N)$ and $p/k_b(N)$ is absent from the PDF and is not a condition on the requested integers.

For a nonnegative continuous event time, the survivor function is $F(t)=\mathbb{P}(T>t)$; here the printed $F$ denotes survival, not the usual cumulative distribution function. A lower median isFor a continuous strictly decreasing survivor function this is the unique solution of $F(m)=1/2$. If the curve has a flat stretch at $1/2$, the distribution has a nonunique median and the infimum specifies the lower endpoint.

The tail integral formula for moments follows from $T={\int }_0^{\infty }1\{T>t\}dt$ and Tonelli theorem. Therefore, when survival vanishes beyond $t^{\ast }$,If $F(t)>0$ at every finite time, the mean is finite precisely when ${\int }_0^{\infty }F(t)dt<\infty$. For a concrete sufficient condition, require $F(t)\le C{t}^{-1-\epsilon }$ for all $t\ge M>0$, with $C<\infty$ and $\epsilon >0$. The integral over $[0,M]$ is at most $M$, and the tail integral is at most $C{M}^{-\epsilon }/\epsilon$. An exponential tail bound also suffices. Merely having $F(t)\to 0$, or even $tF(t)\to 0$, does not suffice: a properly completed decreasing tail proportional to $1/(t\log t)$ for $t\ge e$ has infinite mean.

Under the binomial distribution model's assumption of independent games with a common win probability $p$, the probability of winning a match isThis is also the cumulative distribution function $F_{6,6}(p)$ of a Beta distribution with parameters $(6,6)$: differentiation of the binomial tail gives $h^{\prime }(p)=11!p^5(1-p{)}^5/(5!{)}^2$, and $h(0)=0$. In particular $h$ is strictly increasing on $(0,1)$ and maps $[0,1]$ bijectively onto itself.

The new match probability is ${\tilde{P}}_{ab}=h(P_{ab})$, while the desired linear predictor remains the game-level log odds. Therefore the link function isIts endpoint values are $-\infty$ and $+\infty$. This is not the ordinary match-level log odds link: it first recovers the underlying game probability. Symmetry of the binomial tail gives $h(1-p)=1-h(p)$, hence $g(1-u)=-g(u)$, as required by swapping the players. Dependence or varying game probabilities within a match would invalidate the stated binomial-tail conversion.

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 $0\le x<j/(n+1)$.