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A copula is a joint distribution function on a unit cube whose one-dimensional marginal distributions are uniform distributions. Applying each continuous marginal distribution to its random variable produces uniform marginals by the probability integral transform; their copula records the remaining dependence. Independent uniform coordinates and one common uniform coordinate give different copulas with the same marginals.

This goodness-of-fit criterion integrates the squared difference between the empirical distribution function and a specified continuous null distribution. Under a fully specified null, the probability integral transform makes its law distribution-free; its limiting law is ${\int }_0^1{B}^0(u{)}^{2}du$ for a Brownian bridge. Parameter fitting changes that calibration.

Given a birth-mass distribution and age distribution, an evolutionary state occupies the mass-age set where its starting lifetime has elapsed and its ending lifetime has not. The state number fraction is the probability measure of this set. Initial mass function tail coordinates and the probability integral transform can turn these weighted regions into simple geometric areas.

Every optimal leader bid satisfies $b(v)\le v$, because bidding zero guarantees payoff zero and winning probability is at most one. Hence $F(b(v))\le F(v)$. With identically distributed independent continuous valuations, the probability integral transform gives $\mathbb{E}F(V)=1/2$. The leader's unconditional winning probability is therefore at most $1/2$, even when some high leader types have a conditional advantage. This argument does not depend on selecting a unique optimum.

If $U$ has a uniform distribution on $(0,1)$ and $Q$ is the quantile function of a cumulative distribution function $F$, then $Q(U)$ has the desired probability distribution. Indeed, right continuity givesso $\mathbb{P}(Q(U)\le x)=\mathbb{P}(U\le F(x))=F(x)$. This short argument works for both jumps and smooth cumulative distribution functions. MIT's continuous-probability notes introduce the inverse-transformation construction. ^[ref].

For the three-value example, send $U\le 0.2$ to $0$, $0.2<U\le 0.7$ to $1$, and $U>0.7$ to $3$. These intervals have exactly the desired widths $0.2$, $0.5$, $0.3$. For the density example, output $\sqrt{U}$: for $0\le y\le 1$, $\mathbb{P}(\sqrt{U}\le y)=\mathbb{P}(U\le y^2)=y^2$. Independent uniform draws give independent random variables after this deterministic transformation. This direction of construction works even when $F$ has jumps; applying the probability integral transform in the other direction requires its own continuity hypotheses.

For independent observations from a continuous distribution function, the probability integral transform makes the Kolmogorov-Smirnov statistic distribution-free. Choose $c_{n,\alpha }$ from its uniform-sample distribution to have probability $1-\alpha$ of not exceeding that critical value. Clipping the empirical distribution plus/minus this value to the unit interval gives simultaneous coverage at every argument. This is a band for the entire distribution, rather than separate pointwise intervals.

// Target: statistical-modelling.bigb

The one-sample Kolmogorov-Smirnov test checks goodness of fit by comparing an empirical distribution function $F_n$ with a specified cumulative distribution function $F_0$. Its Kolmogorov-Smirnov statistic $D_n$ is the largest vertical gap anywhere between the two curves. Every threshold asks the same concrete question: what proportion of the observations are at or below it, and what probability does the proposed probability distribution assign there? A large discrepancy supplies evidence against the null hypothesis.

For example, four stipulated observations $(1/10,2/5,4/5,9/10)$ compared with the uniform distribution on $[0,1]$ give $D_4=3/10$. Just before $4/5$, only two observations have been counted, so the empirical distribution function approaches $1/2$, while $F_0$ approaches $4/5$. The Kolmogorov-Smirnov statistic section calculates every jump-side gap. These are exact illustrative numbers, not experimental measurements or a claim of statistical significance.

The ordinary one-sample calibration assumes independent and identically distributed real observations and a simple hypothesis $H_0:F=F_0$, with $F_0$ a continuous function fixed before inspecting these observations. All statistical parameters of $F_0$ must be specified.^[ref] Under that null hypothesis, the probability integral transform gives $U_i=F_0(X_i)$ with the uniform distribution on $(0,1)$, retaining independence. Consequently the sampling distribution of $D_n$ depends on the sample size $n$, but not on the particular continuous $F_0$. This is the useful sense in which the calibration is distribution-free.

For the exponential distribution of rate one, $F_0(x)=1-e^{-x}$ for $x\ge 0$. Take $x_i=-\log (1-u_i)$ for the four stipulated $u_i$ above. The probability integral transform changes the horizontal coordinates but preserves both the order statistics and their cumulative heights, so the same Kolmogorov-Smirnov statistic $3/10$ appears on both scales. More generally, for continuous $F_0$, the jump-side formulas depend only on $F_0(X_{(i)})$. Strict increase everywhere is unnecessary: intervals on which $F_0$ is constant have probability zero under the null hypothesis. This explains the invariance without assuming a strictly positive probability density function.

The observed Kolmogorov-Smirnov statistic is a discrepancy, whereas its p-value is a tail probability under the null hypothesis:Here $D_n^{\mathrm{new}}$ is computed from a fresh sample of the same sample size. Choose a significance level $\alpha$ before the test and reject when the calibrated p-value is at most $\alpha$. The value $D_4=3/10$ alone supplies neither a p-value nor a rejection decision. Failure to reject does not establish that the null hypothesis is true. A finite-sample calibration and the asymptotic Kolmogorov-Smirnov theorem are distinct; a cutoff for $D_n$ must be rescaled before use with $\sqrt{n}{D}_n$.^[ref]

Fitting statistical parameters on the tested observations replaces $F_0$ by the random $F_{\hat{\theta }}$. The resulting transformed observations need not be independent and identically distributed uniforms, so the ordinary reference law fails. The Lilliefors test addresses a particular fitted normal distribution. More general parametric bootstrap calibration must simulate from the fitted statistical model and repeat the fitting in every replicate; holding the fitted statistical parameters fixed during those comparisons calibrates a different procedure. Such a parametric bootstrap requires its own validity justification rather than inheriting exactness from the simple-null test.

For a discrete probability distribution, $F_0(X)$ is generally not uniform. Equal observations also arise from rounding, so tied data require a calibration that accounts for the discrete probability distribution or the rounding rule. The discrepancy remains meaningful, but continuous-null p-values are not automatically valid. Statistical dependence likewise changes the reference law. The two-sample Kolmogorov-Smirnov test compares two empirical distribution functions and has a different calibration.^[ref]

Write $S(t)=\mathbb{P}(T>t)$ for the survival function; this is the function denoted $F_T$ in the survival notation. For a proper continuous event-time distribution with hazard function $h$, its cumulative hazard function is $H(t)={\int }_0^{t}h(s)ds=-\log S(t)$. The probability integral transform makes $S(T)$ uniform on $(0,1)$, including when $S$ has flat intervals: those intervals have zero event probability. ConsequentlyThus the cumulative hazard probability transformation gives $\boxed{H(T)\sim \mathrm{Exponential}(1)}$. No strict monotonicity of $H$ is needed on intervals that carry no event mass. The usual proper continuous survival law is essential; an atom of subjects who never fail would instead require separate treatment of the mass at infinity.

A constant number star formation rate over $10\mathrm{Gyr}$ produces a uniform present-age distribution. The birth-age distribution under constant star formation therefore givesOutside this age interval the cumulative distribution function is zero or one. The census here retains stellar remnants as objects, as required by the population model.

Write $m=M/M_{\odot }$. The initial mass function is proportional to $m^{-3}$ on $m\ge 0.2$. Normalizing the tail integral gives the power-law initial-mass-function tail coordinate:For lower thresholds $X=1$. Because the birth mass and formation time have independence, the probability integral transform makes $(X,Y)$ uniform on the unit square. $X$ is an upper-tail coordinate, so larger masses have smaller $X$.

A red giant of current age $t$ lies in the stellar lifetime regions in a mass-age diagramThere are no red giants below $m=\sqrt{0.8}$ in this model. For $\sqrt{0.8}<m<1$ the upper boundary is the population's age limit, while for $m\ge 1$ both lifetime boundaries are present. A white dwarf lies above $t/\mathrm{Gyr}=10/m^2$.

In uniform probability coordinates the main-sequence boundary is $Y=20X$ and the end of the giant phase is $Y=25X$. The giant region is therefore $20X\le Y<25X$, clipped to the unit square: it is the triangle with vertices $(0,0),(1/25,1),(1/20,1)$. Its area, and hence the individual giant fraction, isThe white dwarf region $Y\ge 25X$ is the triangle with vertices $(0,0),(0,1),(1/25,1)$, givingThe remaining $97.5\%$ are on the main sequence. Boundaries have zero probability and do not affect these counts. Treating every post-giant object as a white dwarf is a stipulated toy-model assumption; real high-mass stars have other stellar remnants.

For the binary calculations, use a coeval binary population: components share their birth time even though their birth masses are independently drawn and their subsequent evolution is independent. Hence $(X_1,X_2,Y)$ is uniform in the unit cube. At a fixed age coordinate $Y=y$, a component is a giant when $y/25<X<y/20$, and a white dwarf when $0<X<y/25$. Define the conditional state fractionsComponents have conditional independence given $y$; averaging over the common age must be done after forming the conditional pair probabilities. The continuous-birth binary state fractions below are consequently volumes in this cube, not products of the marginal single-star fractions.

For $n\ge 2$, the probability integral transform makes $U_i=F(X_i)$ iid uniform on $[0,1]$. Continuity of $F$ is enough; strict increase is unnecessary. Their extreme order statistics have joint densitychoose the observations giving the two extremes and put the remaining $n-2$ between them. Transform to $r=v-u$ and integrate $u$ from zero to $1-r$. The uniformized sample range therefore hasEquivalently $R\sim \mathrm{Beta}(n-1,2)$. For $n=1$, the range is identically zero instead.

The exact failure probability isThus the exact integer sample size is the least $n\ge 2$ making this quantity at most $\delta$. For $\epsilon \downarrow 0$ with $n\epsilon \to \vartheta >0$, it tends to $(1+\vartheta )e^{-\vartheta }$. This function decreases strictly from one to zero on positive $\vartheta$, because its derivative is $-\vartheta e^{-\vartheta }$. For $0<\delta <1$ there is therefore exactly one positive solution ofThis gives the extreme-order-statistic tolerance interval sample-size ruleWhen $\delta$ is small, $\vartheta =\log (1/\delta )+\log (1+\vartheta )$, so a useful first refinement is $\vartheta \approx \log (1/\delta )+\log \log (1/\delta )$. This is an approximation, not an exact integer guarantee; use the preceding exact failure probability to check rounding and very stringent tail requirements. The interval here encloses population content, rather than estimating a fixed distribution parameter.

The empirical distribution function is ${\hat{F}}_n(x)=n^{-1}{\sum }_i{1}_{X_i\le x}$. For continuous F, the probability integral transform gives independent uniform variables $U_i=F(X_i)$. Strict increase of F is unnecessary. More explicitly, generate the observations by $X_i=F^{-1}(U_i)$ with the generalized quantile inverse. Then $1_{X_i\le x}=1_{U_i\le F(x)}$, so ${\hat{F}}_n(x)=G_n(F(x))$, where $G_n$ is the uniform empirical distribution.

Continuity and the limits of F ensure that its range contains $(0,1)$. The endpoint discrepancies for $G_n$ are zero. Hencewhich depends only on n. Equivalently it is ${\max }_{1\le i\le n}\{i/n-U_{(i)},U_{(i)}-(i-1)/n\}$, a function of uniform order statistics. This proves the distribution-free property of the Kolmogorov-Smirnov statistic even when F has flat portions.

Choose $c_{n,\alpha }$ from this common distribution so that $\mathbb{P}(D_n\le c_{n,\alpha })=1-\alpha$. The Kolmogorov-Smirnov confidence band isThe event that all these inequalities hold is exactly the event used to choose the critical value, giving simultaneous coverage $1-\alpha$. The Glivenko-Cantelli theorem states that for an independent identically distributed sample from any distribution function, continuous or not, ${\sup }_x|{\hat{F}}_n(x)-F(x)|\to 0$ almost surely.

For an integrable r-argument U-statistic kernel, replace it by $h_s(x_1,\dots ,x_r)=(r!{)}^{-1}{\sum }_{\pi }h(x_{\pi (1)},\dots ,x_{\pi (r)})$. Every permutation has the same expectation by exchangeability, so this symmetrization of a U-statistic kernel preserves the target functional. For a symmetric U-statistic kernel h and $n\ge r$, the U-statistic isEach summand has expectation theta, hence the statistic is unbiased.

For the population variance, use $h(x,y)=(x-y{)}^2/2$. Independence gives $\mathbb{E}(X_1-X_2{)}^2=2\mathrm{Var}(X_1)$ when the second moment is finite. The corresponding statistic simplifies using ${\sum }_{i<j}(X_i-X_j{)}^2=n{\sum }_i(X_i-\overline{X}{)}^2$:Thus variance as a U-statistic is precisely the usual unbiased sample variance, for $n\ge 2$.

Let $X_i=Y_i-{\theta }_0$. Under the symmetric continuous null hypothesis, positive and negative signs have probability $1/2$ and zeros have probability zero. For the sign test, take $N_{+}={\sum }_i{1}_{\{X_i>0\}}$ and reject for small values: $N_{+}\sim \mathrm{Bin}(n,1/2)$ under the null. Choose $k$ with ${\mathbb{P}}_0(N_{+}\le k)\le \alpha$; randomization at the next value attains exact size if desired. The lower tail is appropriate for $\theta <{\theta }_0$.

For the Wilcoxon signed-rank test, rank $|X_i|$ increasingly and put $W_{+}={\sum }_i{R}_i{1}_{\{X_i>0\}}$. Continuous observations have no absolute-value ties almost surely. Conditional on the absolute values, the signs are independent fair signs because symmetry makes the sign independent of magnitude. Hence $W_{+}$ has the distribution of ${\sum }_{r=1}^{n}r{B}_r$ for independent Bernoulli random variables $B_r$ with parameter $1/2$. Again reject in the lower tail, using this exact subset-sum distribution or a normal approximation. Independence gives the requested null moments:For a symmetric kernel $h$ of fixed order $m$, a one-sample U-statistic is $U_n={\left(\genfrac{}{}{0ex}{}{n}{m}\right)}^{-1}{\sum }_{i_1<\cdots <i_m}h(X_{i_1},\dots ,X_{i_m})$. State the U-statistic central limit theorem: if $\mathbb{E}{h}^2<\infty$, $\eta =\mathbb{E}h$, and ${\zeta }_1=\mathrm{Var}(\mathbb{E}[h\mid X_1])>0$, then $\sqrt{n}(U_n-\eta )\Rightarrow N(0,m^2{\zeta }_1)$. The nondegeneracy condition is important.

For $h(x,y)=1_{\{x+y>0\}}$, its order-two U-statistic satisfies the exact Wilcoxon signed-rank statistic identityTo prove it, in an unordered pair the sign of the sum is the sign of whichever observation has larger absolute value. Counting the pairs for which a positive observation is larger contributes its rank minus one; its own positive-sign indicator supplies the remaining one.

Write $F_0$ for the distribution function of $X$ under the null. Symmetry gives $\eta =1/2$ andThe probability integral transform makes $F_0(X_1)$ uniform on $(0,1)$, so ${\zeta }_1=1/12$. Therefore $\sqrt{n}(U_n-1/2)\Rightarrow N(0,1/3)$. The centred $N_{+}$ is $O_p(\sqrt{n})$ and is negligible on the scale $n^{3/2}$ of $W_{+}$. Combining the exact decomposition and its mean givesThe lower-tail approximate level-$\alpha$ rejection rule compares this standardized statistic with ${\Phi }^{-1}(\alpha )$; a half-unit continuity correction can refine discrete-tail calculations.

For a fully specified continuous null distribution $F_0$, compare the empirical distribution function $F_n$ with $F_0$. The probability integral transform makes $F_0(Y_i)$ iid uniform under the null, giving distribution-free calibration. Three common discrepancies areThese define the Kolmogorov-Smirnov test, Cramér–von Mises criterion and Anderson–Darling test, respectively. Reject for large values. The supremum detects the largest discrepancy, the unweighted integral combines discrepancies throughout the distribution, and the reciprocal weight emphasizes the tails.

The Donsker theorem for empirical distribution functions gives $\sqrt{n}(F_n-F_0)\Rightarrow B^0\circ F_0$ for a Brownian bridge $B^0$, with covariance $\min (u,v)-uv$. Consequently the standard continuous-null limits are ${\sup }_{0\le u\le 1}|B^0(u)|$, ${\int }_0^1{B}^0(u{)}^{2}du$, and ${\int }_0^1{B}^0(u{)}^2/[u(1-u)]du$ for $\sqrt{n}{D}_n,C_n,A_n$, respectively; the weighted last limit also requires control near the endpoints. Exact finite-sample or asymptotic null quantiles provide critical values.

If parameters in $F_{\theta }$ are estimated from the same observations, these fully specified-null critical values generally fail. A first-order expansion isso the estimated parameter changes the limiting process. One must use the fitted-model null law, adjusted asymptotic theory, or a parametric bootstrap that re-estimates the parameters in each replicate.

The continuity of the cumulative distribution function gives the probability integral transform: $F(X_i)$ is uniform on $[0,1]$. Strict monotonicity or a probability density for $F$ is not required. Since $F$ is nondecreasing,By independence, $P(U\le u)=u^n$ for $0\le u\le 1$, so $U$ has probability density function $n{u}^{n-1}$. Average the conditional probability in part (b) using this distribution:Thus $\boxed{P(\tau =k)=n/[(k+n-1)(k+n)]}$. The expression telescopes to total mass one, so the waiting time is finite almost surely.

Let $F_n(x)=n^{-1}{\sum }_i{1}_{\{X_i\le x\}}$. The Kolmogorov-Smirnov statistic is $D_n={\sup }_x|F_n(x)-F(x)|$, often reported as $\sqrt{n}{D}_n$. Under the simple null, the probability integral transform makes $U_i=F(X_i)$ independent uniform variables on $[0,1]$. Strict monotonicity and continuity identify the empirical distribution under this change of coordinates, soIts null law is therefore independent of $F$.

For the composite null, the joint density is ${\theta }^{-n}{\prod }_i{1}_{\{0\le x_i\le \theta \}}$. Its parameter dependence is only through $M={\max }_i{X}_i$, proving sufficiency by the Fisher-Neyman factorization theorem; the larger statistic $(J,M)$ is sufficient too. Ties have probability zero. For fixed $j$, the density of $(J=j,M=\xi )$ is ${\xi }^{n-1}/{\theta }^n$, $0<\xi <\theta$, because the other coordinates range independently over $[0,\xi ]$. Dividing the joint density by this marginal gives conditional densityThus the remaining observations are conditionally independent $U[0,\xi ]$.

For $n\ge 2$, discard the maximum and apply the Kolmogorov-Smirnov test to the $n-1$ ratios $X_i/M$, $i\ne J$, against $U[0,1]$. Use the ordinary null critical values for sample size $n-1$. Its conditional, hence unconditional, level is independent of $\theta$. Keeping the fitted maximum and using the ordinary $n$-sample critical values would not have that justification.

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.

Use the usual continuous-distribution convention $F(0)=0$, $F(1)=1$, and write $f=F^{\prime }$. After observing $b$, the follower can win by matching it, because ties favor the follower. Its best response is to match when $v_2>b$ and choose zero when $v_2<b$; the indifferent equality has probability zero. Thus the leader wins with probability $F(b)$ and a leader of type $v$ maximizesBids above $1$ are dominated by bidding $1$. This is the leader optimization in a sequential private-value all-pay contest. Since $F$ is a concave function, $h_v$ is concave. An interior optimum satisfies $vf(b)=1$, with the usual endpoint conditions when that equation has no interior solution.

For precision, select the smallest maximizer when the leader is indifferent. This defines a Stackelberg equilibrium and supplies the printed strict conditional comparison at the threshold type. Let $q=F^{-1}(1/2)$. The density $f(q)$ is positive: if it vanished there, its nonnegative nonincreasing continuation would force $F$ to remain $1/2$ up to $1$, which is impossible. At the median,If this is positive, every maximizer is strictly greater than $q$, so the leader wins with probability greater than $1/2$. If it is negative, every maximizer is strictly smaller than $q$. If it is zero, $q$ is a maximizer, and the smallest maximizer is at most $q$. Therefore the median-density threshold for the leader in an all-pay contest isStrict concavity of $F$ would make the optimum unique, removing the selection convention. With mere concavity, the printed assertion needs that convention at equality. For example, $F(b)=b$ and $v=1$ make all bids optimal, and choosing $b=3/4$ makes the leader more likely to win even though the strict threshold is not exceeded.

The same issue can occur at an interior type, rather than just an endpoint. A continuously differentiable concave distribution isIts density decreases from $2$ to $5/4$, is constant on $[1/5,2/5]$, and then decreases to $1/24$; integration gives $F(1)=1$. Its median is $31/100$. At $v=4/5=1/f(q)$, every bid in $[1/5,2/5]$ maximizes $h_v$, and choosing $2/5$ gives winning probability $49/80>1/2$. This confirms the genuine best-response selection at a flat leader objective issue. The smallest-maximizer convention avoids it; the threshold type has zero ex ante probability.

The unconditional comparison is valid for every optimal selection. Zero effort guarantees the leader payoff zero, so an optimal bid satisfiesConsequently $F(b(v))\le F(v)$. If $V$ has the continuous distribution $F$, the probability integral transform makes $F(V)$ uniform on $[0,1]$. Taking expectations proves the ex ante follower advantage in a sequential all-pay contest:For strictly increasing atom-free $F$, optimal bids in fact satisfy $b(v)<v$ for almost every $v\in (0,1)$, so the first inequality is strict. Conditional advantage for unusually high leader types is therefore compatible with an unconditional follower advantage.

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.

Let $T=10\mathrm{Gyr}$ and let $t$ denote present stellar age, rather than time measured from the galaxy's birth. With a fixed initial mass function, constant star formation rate also gives a constant number of births per unit time, as in the birth-age distribution under constant star formation. Counting the supplied stellar evolution model's remnants as part of the population, ages have a uniform distribution on $[0,T]$. Integrating the constant birth rate therefore givesOutside this interval the cumulative fraction is zero or one as appropriate.

Write $m=M/M_{\odot }$ for the birth-mass parameter and $M_{\min }=0.2M_{\odot }$. The normalization cancels when integrating the initial mass function:Thus the mass-tail fraction isIt is one at and below the lower cutoff. The normalization constant has no effect on any number fraction.

For evolutionary-state selection by stellar lifetimes, a red giant has completed its main sequence lifetime but not its giant lifetime. Its stellar lifetime regions in a mass-age diagram are bounded byThe lower boundary meets $t=T$ at $m=\sqrt{0.8}$ and the upper boundary meets it at $m=1$. A white dwarf lies above the upper lifetime curve, within the same age interval.

The probability integral transform, applied to the decreasing mass-tail coordinate, makes $X$ uniform on $(0,1)$: $\mathrm{Pr}(X<x)=x$. The independent birth age gives $Y=t/T$ uniform on $(0,1)$, so number fractions are areas in a unit square. Since $m^{-2}=25X$, the red giant region becomes $20X<Y<25X$, a triangle with vertices $(0,0)$, $(1/25,1)$ and $(1/20,1)$. Its area is $\frac{1}{2}(1/20-1/25)=1/200$. The white dwarf region $Y>25X$ is a triangle of area $\frac{1}{2}(1/25)=1/50$. Hence the present individual-star fractions areThe rest are on the main sequence under the given toy lifetime model.

For the binary stars, draw two independent mass-tail coordinates $X_1,X_2$, but only one shared age coordinate $Y$: components born together are coeval. The coeval binary population therefore occupies a uniform unit cube $(X_1,X_2,Y)$. At a fixed age $Y=y$, the state probabilities for either component areTheir evolutionary states have conditional independence given $y$. They generally do not have unconditional independence, because sharing an age correlates their states. The binary fractions below are slice areas averaged over $0<y<1$.

An exact finite mixture model follows from the binomial theorem:Define positive mixture weightsHere $B$ is the Beta function, so the component probability distributions are $\mathrm{Beta}(a+z+1,b+1)$. All coefficients are computable from the observed integer counts.

Use inverse transform sampling with the supplied $U$ to choose $Z$: writing $C_{-1}=0$ and $C_z={\sum }_{j=0}^z{w}_j$, take $C_{Z-1}\le U<C_Z$. Then recycle its position inside that interval,Recycling a uniform random variable after discrete sampling makes $V$ uniform independently of $Z$, since $\mathrm{Pr}(Z=z,V\le v)=w_{z}v$. It is also independent of all supplied random variables with gamma distributions. Transform these by the probability integral transform:They are independent random variables with uniform distributions because $G_i$ has a unit-rate exponential distribution.

Conditional on $Z=z$, put $k=a+z+b+1$ and $r=a+z+1$. Form the $k$ values $V,W_1,\dots ,W_{k-1}$ and return their $r$th order statistic. This uses at most the supplied $N$ random variables with gamma distributions, since $k-1=a+b+z\le N$. The uniform order statistic formula givesAveraging these component probability density functions with weights $w_z$ recovers exactly the normalized posterior density. Unused inputs can be discarded. When $N=0$, $Z=0$, $k=r=1$, and the output is $U$, as required. Interval endpoints and ties have probability zero; any consistent convention there is harmless. Exactness here concerns the mathematical algorithm; finite-precision arithmetic can be stabilized with logarithmic mixture weights.