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A real functional $\Phi$ is Hadamard differentiable at $s_0$ if there exists a continuous linear functional $L={\dot{\Phi }}_{s_0}:S\to \mathbb{R}$ such that, whenever $t_j\ne 0$, $t_j\to 0$ and $h_j\to h$ in the normed vector space,If $\Phi$ is defined only on a subset, require $s_0+t_j{h}_j$ to lie in its domain. The variation of the directions $h_j$ is essential: convergence along each fixed direction alone is a weaker differentiability condition. Tangential Hadamard differentiability restricts the limit directions to a specified subspace; here the derivative is defined on all of $S$.

We prove the functional delta method using compact sets, so do not need an unjustified application of the Skorokhod representation theorem on an arbitrary nonseparable normed vector space. As usual for function-space convergence in distribution, the limiting random element is tight, and the real statistics are measurable, or convergence is understood through outer expectations. These are automatic under the usual separable complete random-element formulation; the empirical-process formulation used below has a tight separably supported limit.

For $t\ne 0$ putHadamard differentiability says $R_{t_j}(h_j)\to 0$ for every convergent sequence of directions, since continuity gives $L(h_j)\to L(h)$. It implies a Hadamard remainder near a compact set: for every compact $K\subset S$ and $ϵ>0$ there are $a>0$ and $\delta >0$ such thatTo prove it, suppose it failed. Choose $t_j\to 0$ and $h_j$ at distance less than $1/j$ from $K$ with $|R_{t_j}(h_j)|\ge ϵ$. Choose $k_j\in K$ with $\Vert h_j-k_j\Vert <2/j$. A subsequence of $k_j$ converges by compactness, so the corresponding $h_j$ converge too. The defining derivative limit contradicts the lower bound on $R_{t_j}(h_j)$.

Now let $Z_n=r_n(X_n-s_0)\stackrel{d}{\to }X$ and $t_n=1/r_n$. Given $\eta >0$, tightness of a probability measure provides a compact $K$ with $\mathbb{P}(X\in K)>1-\eta$. For the open $\delta$-neighbourhood $K^{\delta }$, the Portmanteau theorem givesOn $K^{\delta }$ the preceding uniform remainder is less than $ϵ$ for all sufficiently large $n$. Consequently ${\mathrm{limsup}}_n\mathbb{P}(|R_{t_n}(Z_n)|\ge ϵ)\le \eta$. Since $\eta$ is arbitrary, this remainder tends to zero in probability. We have proved the actual expansionContinuity of $L$ and the continuous mapping theorem give $L(Z_n)\stackrel{d}{\to }L(X)$. The vanishing remainder and Slutsky theorem therefore yieldThis derivation uses the derivative definition, rather than assuming a uniform Fréchet remainder over every bounded set. If outer expectations are needed, the same compact-neighbourhood argument uses outer probabilities and gives the corresponding empirical-process functional delta method.

For the statistical application, use the sample's empirical distribution function and the plug-in estimatorHere the printed $L^{\infty }$ denotes bounded pointwise functions with the supremum norm, namely ${\ell }^{\infty }(\mathbb{R})$, rather than equivalence classes modulo Lebesgue null sets. Assume Hadamard differentiability at the true $F$ and the usual measurability of the scalar estimator. The Donsker theorem for empirical distribution functions and the proved functional delta method giveNo continuity of $F$ is needed for this application. Since the derivative is continuous and linear and the bridge is a centred Gaussian process, $Z_F$ is a centred Gaussian real random variable, possibly identically zero.

Finally, a finite real weak limit ensures the required boundedness in probability. Choose $M$ so large that $\mathbb{P}(|Z_F|\ge M)<ϵ/2$. Applying the closed-set part of the Portmanteau theorem to $\{z:|z|\ge M\}$ givesHence there is $n_0$ such that, for every $n\ge n_0$,In particular the upper-tail condition in the question holds. The plug-in estimator has estimation error $O_{\mathbb{P}}(n^{-1/2})$, even when its limiting variance is zero.

Choose a proposal probability density function $g$ and a finite envelope constant $M$ such that $f(y)\le Mg(y)$ wherever the target has positive density. A rejection sampling trial draws $Y\sim g$ and, independently, $U\sim \mathrm{Uniform}(0,1)$; accept $Y$ whenotherwise repeat with fresh independent draws. For any measurable set $D$,In particular $\mathbb{P}(\mathrm{accept})=1/M$, so the conditional accepted density is $f$. More explicitly, summing over the possible first accepted trial givesThus the algorithm terminates almost surely and its output has exactly the requested law.

The envelope constant satisfies $M\ge 1$ and is the expected number of proposal trials for one output, by the geometric distribution of the waiting time. This is the quantity often called the rejection rate $M$. The actual probability of rejecting a particular proposal is $1-1/M$, and the expected number of rejections before one acceptance is $M-1$; these conventions should not be conflated.

To generate a standard normal distribution, first generate its absolute value, whose half-normal distribution density isWith the given exponential distribution proposal, the density ratio isIts maximum is at $y=\lambda$, hence the normal rejection sampling with an exponential envelope constant isDraw $Y\sim \mathrm{Exp}(\lambda )$ and a fresh uniform $U$; accept when $U\le e^{-(Y-\lambda {)}^2/2}$. An accepted $Y$ has density $h$. Independently choose a sign, positive or negative with probability $1/2$, and output $Z$ with that sign and magnitude $Y$. On either half-line its density is $h(|z|)/2=(2\pi {)}^{-1/2}{e}^{-z^2/2}$, proving the normal output law.

To optimize the rejection rate, differentiateIts derivative is $\lambda -1/\lambda$, negative before one and positive after one; equivalently its second derivative is $1+1/{\lambda }^2>0$. ThusThe rejection probability is consequently about $0.23983$.

To use only uniforms, apply inverse transform sampling: for an independent $U_1\in (0,1)$,has survival probability $\mathbb{P}(Y>y)=e^{-\lambda y}$ for $y\ge 0$. Use a second independent uniform for acceptance, and, after acceptance, a third independent uniform to choose the sign. Repeat with fresh uniforms after any rejection. No exponential generator is then needed.

An unordered bootstrap sample is determined by the multiplicities $N_i\ge 0$ of the $n$ distinct observations, with ${\sum }_i{N}_i=n$. Conversely each such vector specifies exactly one sample up to rearrangement. Place $n$ identical marks into $n$ boxes using $n-1$ separators. The stars and bars argument therefore givesThese bootstrap count vectors are not equally likely: the probability of a vector is $n!/(n^n{\prod }_i{N}_i!)$, counting its possible ordered sequences. The ordered index sequences themselves number $n^n$.

The first simulation uses independent standard Cauchy distributions and returnsThe tail probability and estimator moments areThe indicator count has a binomial distribution with parameters $n,\theta$; the infinite variance of a Cauchy observation is irrelevant because this estimator averages bounded indicators.

For the next code block, each row is an independent sample of size $n$ from the empirical distribution of the original $x_i$. Its entry in the vector $v$ is the corresponding bootstrap indicator mean ${\hat{\theta }}_b^{\ast }$. The last expression is the variance of these $B=199$ replicate estimates, with denominator $B-1$:Condition on the original data and write a star on bootstrap expectations. Every resampled indicator has a Bernoulli distribution with success probability $\hat{\theta }$, and the $n$ draws within a row are independent. Thus the conditional bootstrap variance of a sample mean givesIt estimates the sampling variance of the tail-probability estimator, not the variance of the raw Cauchy data. Its square root is an estimated standard error. Finite $B$ leaves bootstrap simulation noise. Averaging over the original data, its expectation is $(n-1)\theta (1-\theta )/n^2$, so it is not exactly unbiased for the true variance $\theta (1-\theta )/n$ at finite $n$.

In the importance-sampling block, the transformation $Y=2/(1-U)$ gives, for $y\ge 2$,The variable $Y$ is a proposal draw supported entirely in the integration region. The deterministic value computed from it is the importance sampling weightConsequentlyThe code uses ordinary importance sampling of a Cauchy tail, not self-normalization; $Y$ itself is neither an observation from the target Cauchy law nor the estimator's summand. The weight compensates for sampling from its proposal law.

The final block bootstraps these same proposal observations $Y_i$ and recomputes the weight average in each row. Since $W$ is a deterministic function of $Y$, this is equivalent to resampling the observed weights $W_i$. If ${\tilde{\theta }}_b^{\ast }$ denotes a replicate, the final output is its sample variance over the $B$ replicates. Its conditional expectation isIts expectation over the original proposal sample is $(n-1)\mathrm{Var}(W)/n^2$.

The variance reduction can be calculated, rather than inferred just from the appearance of the weights. Write $A=\arctan (1/2)$. Thenbecause an antiderivative of the integrand without its prefactor is $\frac{1}{2}(\arctan y-y/(1+y^2))$. HenceBy comparison, $\theta (1-\theta )\approx 0.125803$. For $n=100$ the true estimator variances are approximately $9.55253\times 10^{-7}$ and $1.25803\times 10^{-3}$ respectively. The importance estimator has about 1/1,317 of the indicator estimator's variance, because all proposals land in the tail and the weights remain in the narrow interval $[2/(5\pi ),1/(2\pi ))$.

Thus the last bootstrap output should usually be much smaller than the first, and its expected value is smaller by the same variance ratio. This is not a deterministic ordering of two finite-run outputs: if all the original Cauchy draws happen to be at most two, the first bootstrap variance is zero, while unequal importance weights can still give a positive second variance.

The Metropolis–Hastings algorithm proposes $y\sim q(\cdot \mid x)$ and moves from the current state $x$ with acceptance probabilityOn rejection it retains $x$. The target need only be known up to its normalizing constant, but it must define a proper probability distribution. The accepted off-diagonal probability flow iswhich is symmetric in $x,y$ and proves detailed balance. This permits considerable flexibility in proposals, including joint moves when a posterior distribution has strongly correlated parameters, but requires proposal tuning and can waste computation on rejections.

A Gibbs sampler instead updates a coordinate or block from its exact full conditional distribution, holding the other coordinates fixed. This is a special Metropolis–Hastings algorithm update with acceptance probability one: the joint density factors into the unchanged marginal density and the proposed conditional density, which cancel in the acceptance ratio. The coordinate update preserves the target; composing such updates therefore preserves it too, although an entire deterministic sweep need not itself be reversible.

Use Gibbs sampling when the conditional laws are easy to simulate and suitably chosen blocks mix well. Scalar updates can mix poorly under strong posterior distribution correlation, whereas a blocked Gibbs sampler can alleviate that dependence. Prefer a flexible Metropolis proposal when conditional draws are difficult or a well-tuned joint proposal is more effective. Variations include random or systematic coordinate scans, random-walk Metropolis algorithms, independence proposals, and using Metropolis updates for difficult blocks inside a Gibbs scheme. Both methods generate dependent draws; usable long-run estimates require an appropriate irreducible, aperiodic chain and assessment of mixing, not merely a formal update rule.

For the change-point calculation, put $v={\sigma }^2$ andMultiplication of the Gaussian likelihood function by the specified prior distribution gives the formal posterior distribution kernel, with respect to $d\theta d\varphi dv$ and counting measure on $m$,

There is an empty component under an improper prior at $m=n$. The sum $S_n$ is independent of $\varphi$, whose prior distribution is flat on all of $\mathbb{R}$. For every fixed finite $\theta$ and $v>0$, the kernel is a strictly positive constant in $\varphi$, henceThe configuration has positive prior probability $1/n$. Integrating over, for example, a bounded interval of $\theta$ and $1\le v\le 2$ already makes the joint normalization infinite. Thus the printed prior distribution does not produce a posterior probability distribution. This failure is present for every data set, not just for a special arrangement of observations.

The formal full conditionals expose exactly where the requested sampler breaks. For $m<n$, completing the two mean squares givesThe variance kernel is that of $\mathrm{IG}(n/2,S_m/2)$ when $S_m>0$, and the discrete conditional probabilities are proportional to $e^{-S_m/(2v)}$. At $m=n$ the conditional kernel for $\varphi$ is constant and cannot be normalized; it is not a normal distribution with an infinite variance. Therefore no valid Gibbs sampler, and no value of $P(m=n\mid x)$, exists for the model as printed. A chain that simply substitutes an arbitrary distribution for this missing conditional would be sampling a different model.

Here is a fully specified proper-prior repair retaining the no-change configuration. Choose fixed hyperparametersindependently, with $t_{\theta },t_{\varphi },a_0,b_0>0$. Its Gaussian change-point posterior with proper priors isIt is normalizable even for degenerate data: the likelihood function is at most $(2\pi {)}^{-n/2}{v}^{-n/2}$, and its integral against the variance prior distribution is bounded byThe normal mean priors integrate to one, and the normalizing constant is positive. This proves posterior propriety for the repaired model.

Completing the conditional squares gives the Gaussian change-point Gibbs updates. DefineThe four updates areAt $m=n$, $M_{\varphi }={\mu }_{\varphi },V_{\varphi }=t_{\varphi }$, giving its proper prior distribution unchanged. Starting from any finite means, positive variance and allowed $m$, cycle through these four updates using the latest values. The sum-of-squares terms can be evaluated from cumulative data sums; normalize the discrete log weights after subtracting their maximum to avoid numerical underflow. The proper positive conditionals allow movement among all split values and throughout the continuous parameter support.

Under this specified repair, $P(m=n\mid x)$ is the posterior probability of no change within the observed sequence: all $n$ data points belong to the first mean regime. For $N$ retained draws after initialization has ceased to affect the estimates, its Markov chain Monte Carlo estimator isAlternatively average the displayed conditional probability of $m=n$ at the sampled continuous parameters, giving an estimate by Rao-Blackwellization. Autocorrelation must be accounted for in its Monte Carlo uncertainty. These are probabilities and estimators for the explicitly repaired model; the original unnormalizable kernel supplies neither.

The odds ratio compares event odds, not event probabilities. Let group 1 denote the elective group and group 0 the reference group. Their fitted event probabilities are ${\hat{p}}_1=9/1515\simeq 0.00594$ and ${\hat{p}}_0=1215/75057\simeq 0.01619$. ThusThe risk ratio is similarly about $0.367$. The unadjusted observations show a lower adverse-event frequency in the elective group, rather than an increase. This is an association, not evidence that choosing the procedure prevents adverse outcomes. The direction is opposite to the reported adjusted odds ratio above one; the two analyses compare different populations or different conditional risks.

Zero observed events do not mean that the underlying event probability cannot be estimated. For independent observations having a common Bernoulli distribution, the binomial likelihood with no events is $(1-p{)}^{1515}$, maximized at $\hat{p}=0$. A zero-event binomial upper confidence bound shows the remaining uncertainty: the one-sided 95% upper bound solves $(1-p_U{)}^{1515}=0.05$, givingThus the point estimate is zero, but a nonzero mortality risk is compatible with the observations. This bound is an illustration under independent equal-risk sampling; clustered data or unequal risks require their own uncertainty calculation.

The empirical unadjusted odds ratio is also defined:Its log odds ratio is $-\infty$, so the ordinary normal approximation for a log-odds confidence interval is unavailable. Exact likelihood-based limits can still describe uncertainty; a specified continuity correction would instead yield a finite, method-dependent point estimate.

In an unpenalized logistic regression with a separate coefficient for elective exposure, every exposed death outcome is zero. Decreasing that coefficient toward $-\infty$ increases the exposed observations' likelihood contributions and leaves the reference observations unchanged. This is quasi-complete separation: there is no finite ordinary maximum-likelihood estimate of the exposure coefficient, even after adding background predictors. It is reasonable to say that the usual finite adjusted coefficient could not be fitted; it is too strong to say that no risk estimate or statistical information is possible. A specified penalized likelihood or Bayesian inference with proper coefficient priors can provide a finite adjusted estimate, whose dependence on the regularization must be reported.

Using the reference event rate as the background-risk benchmark givesThis is the expected count if the elective group had the reference group's baseline risk and the procedure added no effect. It is substantially above the observed count, illustrating the importance of the background-risk comparison.

Similarity of background factors alone does not fix the elective group's event risk if the procedure itself changes it. If one additionally retained the reported common adjusted odds ratio $R=2.7$ and used $p_0=1215/75057$ as the common baseline risk, then conversion from odds to probability would giveThese are different assumptions: about 25 is the baseline/no-effect benchmark; about 64 includes an odds multiplier of 2.7.

Use a concurrent comparison: test the same specimens for both cocaine and mephedrone, rather than treating the historical cocaine percentage as an exactly known 2010 parameter. Assume representative sampling of privates, effectively independent specimens, stable underlying rates during the sampling period, and sufficiently accurate assays. I use a two-sided 5% test of equal positive-test probabilities, with 80% statistical power at $p_C=0.010$ and $p_M=0.005$. This is a design to distinguish the rates at a twofold alternative; a significance test does not prove that their ratio is exactly two.

The paired design needs the joint positive-test distribution. For specimen $i$, write $C_i,M_i\in \{0,1\}$ and $D_i=C_i-M_i$. Put $q_{11}=P(C_i=M_i=1)$ andThen $\mathbb{E}{D}_i=\delta$ and $\mathrm{Var}(D_i)=q-{\delta }^2$. Under equal marginal rates, the two discordant outcomes have equal probabilities; conditional on the total number of discordant specimens, their split is a fair binomial distribution. This gives the McNemar test. Its large-sample version rejects for a sufficiently large absolute difference between the discordant counts, divided by their estimated standard deviation.

For planning, suppose simultaneous positives are negligible, so $q\simeq 0.015$. Under the alternative the difference in positive counts has mean $n\delta$ and standard deviation $\sqrt{n(q-{\delta }^2)}$; the approximate upper null rejection boundary is $z_{0.975}\sqrt{nq}$. Requiring 80% upper-tail rejection probability gives the paired binary sample size calculationAdd the mephedrone assay to about 5,000 existing tests, roughly ten weeks at 500 per week. An exact conditional-power calculation under the negligible-overlap model gives about $77.6\%$ at 4,707 specimens and $80.2\%$ at 5,000, so the rounding also compensates for the conservative discrete test. If positivity for the two drugs were independent within a specimen, $q_{11}=0.00005$, giving a very similar requirement of about 4,676. For the stated marginal rates, $q\le 0.015$, so negligible overlap is a conservative variance choice within this approximation. Substantial co-use reduces the discordant proportion and changes the necessary size. Repeated tests on the same person, unit-level clustered data or inaccurate assays can instead reduce effective information and require more specimens.

A pre-specified one-sided 5% comparison would use $1.645$ in place of $1.96$ and require about 3,708 specimens before allowing for discreteness under the negligible-overlap assumption. These are different testing conventions. Treating the historical 1% as fixed would give a one-sample calculation, but would ignore uncertainty or changes in the actual 2010 cocaine rate.

Set $\alpha =\log \{{\beta }^{(1)}/{\beta }^{(0)}\}$ and absorb the common multiplier into $h_0(t)=m(t){\beta }^{(0)}$. The model becomesIt is a Cox proportional-hazards model; the unknown baseline does not need a parametric form. The equality hypothesis is precisely $\alpha =0$. The absolute multipliers and $m$ have a scale ambiguity, but their hazard ratio $e^{\alpha }$ is identifiable when both groups supply information.

For an untied event at $a_k$, let $R_k$ be the risk set just before the event, and let $i_k$ be the subject who fails. The conditional event probability is proportional to that subject's hazard, so the common baseline cancels:Multiply these factors to form the Cox partial likelihood. One can maximize it and compare its maximum with its value at $\alpha =0$ using a partial likelihood-ratio test, or use the corresponding score test.

Explicitly, with group risk-set sizes $r_{k0},r_{k1}$ and observed event indicator $d_{k1}$ for group 1, the score at zero and its information areUnder the null the event label has exactly this Bernoulli distribution conditional on its current risk set. Hence $U^2/V$ is the usual one-degree-of-freedom log-rank test statistic, with an asymptotic chi-squared distribution when $V>0$ and enough informative events are observed. Censoring can differ between groups, provided it is independent of event time within the conditioning model. If there are recorded ties, use the appropriate tied-risk-set version: for $d_k$ events the null variance contribution is $d_k(r_k-d_k)r_{k0}{r}_{k1}/[r_k^2(r_k-1)]$.

Average the covariate over its possible event subjects using the conditional probabilities from part (a). Define the risk-weighted sumswith $z_i^0=1$. ThenThis is the mean covariate of the next event subject, conditional on the current risk set and an event. It is generally not the unweighted average over the original sample. For $\beta =0$ it reduces to the unweighted current risk-set mean.

Label an orthonormal basis of the $n$-dimensional Hilbert space by $j=0,\dots ,n-1$, with indices modulo $n$. LetThese are the shift and phase Heisenberg-Weyl operators. Alice holds the input $S$ and one half $A$ of the shared maximally entangled stateShe measures $SA$ in the generalized Bell basisIts inner products areThere are $n^2$ orthonormal vectors in an $n^2$-dimensional space, so this is a complete projective measurement.

For an unknown normalized state $|u⟩={\sum }_j{u}_j|j⟩$, its contraction with measurement outcome $(p,q)$ gives Bob's unnormalized stateEach outcome has probability $1/n^2$, independently of the input. Alice communicates the two $n$-valued labels, and Bob applies the ordered inverseThe order matters because shift and phase operators do not generally commute. Normalizing the branch recovers the unknown state exactly, and all outcomes are corrected, so success is deterministic. The quantum resource is one maximally entangled pair of local dimension $n$. The classical resource is an $n^2$-valued message, with information content $2{\log }_{2}n$ bits; $\lceil {\log }_2(n^2)\rceil$ ordinary bits suffice in fixed-length binary encoding.

This is faithful qudit teleportation even for an input entangled with a reference $R$. Expand a joint pure state as ${\sum }_j|j{⟩}_S|r_j{⟩}_R$, without requiring the reference vectors to be orthogonal. The same contraction and correction act on each system basis vector, producing $n^{-1}{\sum }_j|j{⟩}_B|r_j{⟩}_R$. Thus the whole joint state, not just a reduced state, is preserved. Linearity extends the conclusion to mixed inputs. This is teleportation as an identity channel on a reference.

The event that $H_1$ is hit but $H_2$ is not has probability $1/3-1/12=1/4$. Dividing by the probability $1/3$ that $H_1$ is covered, using the definition of conditional probability, givesThis conditioning concerns angular coverage of a sector; the two house-hit events are not independent.

The independent uniform distributions give a constant joint probability density $1$ on the unit square. For $0<x<1$, integrate over $x<u<v^2$:For the random quadratic with uniform coefficients, real roots require its discriminant to be nonnegative, equivalently $U\le V^2$. The boundary has probability zero, and the area under this parabola isOn that event the roots are $-V\pm \sqrt{V^2-U}$, both nonpositive. Bounding both absolute values by one therefore amounts to bounding the more negative root:Since $0\le V\le 1$, squaring the inequality is legitimate and gives $U\ge 2V-1$. The admissible unit-square region is $\max (0,2v-1)\le u\le v^2$. Its probability isDividing by the real-root probability gives the requested conditional probability:

The cumulative distribution function is $F(x)=P(X\le x)$. For the differentiable distribution in the question, its probability density function is $f(x)=F^{\prime }(x)$, so probabilities over intervals are obtained by integrating $f$. The survival probability is $P(X>x)=1-F(x)$. Differentiating immediately givesThe strict inequality is consistent with the definition using $X\le x$. This relation connects the probability density function with the decline of the survival probability.

For $n\ge 1$, substitute $s=e^{t/n}$ in the probability generating function. The moment-generating function of $X_n/n$ isThe domain follows from convergence of the generating series; at or above that positive threshold the expectation diverges, even though a rational continuation can be written.

Conditioning on survival, the probability mass function from part (ii) becomes a positive-integer geometric distribution:Its conditional moment-generating function is $e^{t/n}/[n+1-n{e}^{t/n}]$, tending to $1/(1-t)$ for fixed $t<1$. We can prove the required limit directly, without invoking a continuity theorem. Set $r_n=n/(n+1)$. For every $x\ge 0$, the strict upper tail is exactlySince $\log r_n=-1/n+O(n^{-2})$ and $\lfloor nx\rfloor /n\to x$,At $x=0$ both tails equal one. This is the exponential limit of surviving geometric branching. Conditioning is essential: the unconditional variable has probability $n/(n+1)$ of being zero, whereas the rare surviving population has size of order $n$.

The geometric family law on the nonnegative integers has probability generating functionIts expected value is $G^{\prime }(1)=1$, so this is a geometric-offspring critical branching process. Starting from $G_0(s)=s$, verify the rational iterate by induction. If $G_n(s)=[n-(n-1)s]/[n+1-ns]$, thenwhich has the same form with $n$ replaced by $n+1$. Thus, for $n\ge 1$,The stated disk is the actual power-series convergence domain, not merely a place where the rational continuation can be evaluated. Its simple pole is at $s=1+1/n$.

The extinction probability at generation $n$ is the constant coefficient:For later use, expanding the remaining geometric series givesIn particular $P(X_n>0)=1/(n+1)$. At $n=0$ the initial population is one deterministically; its PGF is $s$, and the displayed radius involving $1/n$ is used only for $n\ge 1$.

For independent random variables, $\{\min (X,Y)>u\}=\{X>u,Y>u\}$ and $\{\max (X,Y)\le v\}=\{X\le v,Y\le v\}$. Factor their probabilities to obtain the distribution functions of independent minima and maxima:These formulas also allow atoms; no probability density function assumption is needed.

For independent unit-rate exponential distributions, $P(U>u)=e^{-2u}$ for $u\ge 0$, so $U\sim \mathrm{Exp}(2)$. To establish the independent minimum and gap of two exponential variables, set $D=V-U$. On either ordering branch, the change of variables is $(X,Y)=(u,u+d)$ or $(u+d,u)$ with unit absolute Jacobian determinant. Both contribute $e^{-2u-d}$; ties have probability zero. Thus the joint probability density isIt factors into normalized exponential probability density functions, proving $U\sim \mathrm{Exp}(2)$, $D\sim \mathrm{Exp}(1)$ and their independence.

The maximum is $V=U+D$. A half-scaled unit-rate exponential has rate two, so the independent pair $(D,U)$ has the same joint law as $(X,Y/2)$. ConsequentlyUsing the expected values, variances and independence of the summands,The spacing factorization explains why an order statistic can be represented as a sum of independent waiting times.

For events with $P(B)>0$, the conditional probability is $P(A\mid B)=P(A\cap B)/P(B)$. Applying the definition in both orders gives Bayes' theorem:Conditionally on $N=n$, independent fair tosses give a binomial distribution for the head count, so $P(H=1\mid N=n)=n{2}^{-n}$. The joint probability mass function is consequentlyUse the law of total probability and the differentiated geometric series, ${\sum }_{n\ge 1}n{r}^n=r/(1-r{)}^2$, to obtain $P(H=1)=4/9$. The posterior coin count after exactly one head is thereforeThese probabilities sum to one by the same series identity. The conditioning weights possible coin counts by their likelihood of producing exactly one head.