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For the standard Cauchy distribution tail, use the indicatorThen $\int \theta (x)f(x)dx$ is the desired probability, and the ordinary importance sampling estimator is $n^{-1}{\sum }_{i}f(x_i)1_{\{x_i\ge k\}}/g(x_i)$. The proposal must cover the nonzero integrand, namely the upper tail. A proposal restricted to $[0,k]$ cannot directly estimate this indicator integral; part (v) instead uses a complementary bounded-interval identity.

Under the reference convention ${\alpha }_1=0$, the hypothesis becomes absence of the free group contrast ${\alpha }_2$. Introduce an inclusion indicator $J$: set ${\alpha }_2=0$ when $J=0$, and use its stated proper normal prior when $J=1$. Assign positive prior probabilities $q_0,q_1$ to those models, and retain $\mu ,{\beta }_2$ in both.

A birth draws $u$ from a proposal probability density function $g(u)$ and appends it as ${\alpha }_2^{\prime }=u$; the reverse death deletes it. The Jacobian is one. At fixed $\theta ,{\beta }_2$, the likelihood ratio isThus the birth acceptance ratio before capping iswhere $p_{\alpha }$ is the normalized $N(0,{\sigma }_1^2)$ prior. Again, taking $g=p_{\alpha }$ cancels that factor. Use the reciprocal death rule and ordinary within-model parameter updates.

After warm-up, the Markov chain Monte Carlo occupation fraction estimates the desired probability:This requires the chain to explore both models and satisfy its usual ergodic convergence conditions. Under the original continuous normal prior alone, the equality event has posterior probability zero; the explicit model indicator supplies the necessary prior atom. If intercept selection from part (v) is also performed, use two indicators and count all sampled states with $J=0$, regardless of the intercept indicator.

Introduce a model indicator $M\in \{0,1\}$. In $M=0$, set $\mu =0$, hence $\theta =1$, and retain the two group effects. In $M=1$, include $\mu$ with the prior induced by the gamma baseline. Choose positive prior model probabilities $p_0,p_1$ summing to one; the supplied within-model parameter priors alone do not specify these probabilities.

Let $r=(a_2,b_2)$, use the same proper prior $p(r)$ in both models, and denote their likelihoods by $L_0(r)$ and $L_1(\mu ,r)$. The unnormalized joint model targets areFor a birth move, draw an auxiliary $u$ from a positive probability density function $q(u)$ and map $(r,u)$ to $({\mu }^{\prime }=u,r^{\prime }=r)$. The reverse death move simply deletes $\mu$. This is a dimension-matching bijection with absolute Jacobian one. If the birth and death selection probabilities are $b_0$ and $d_1$, respectively, the birth and death moves for Bayesian variable selection rule givesThe reverse acceptance is the reciprocal ratio, capped at one, evaluated at the current $\mu$. Choosing $q=p_{\mu }$ cancels the added-parameter prior; a data-informed proposal may mix better but must retain its proposal factor. Within each model, also update its continuous parameters to ensure exploration.

For an explicit likelihood ratio, with $E$ and $T$ from part (ii),This supplies a complete reversible-jump Markov chain Monte Carlo move. Its acceptance probabilities enforce detailed balance between the two dimensions. The lower-dimensional state is a separate point-mass model, not a zero-probability equality test inside the continuous full-model prior.

Let $Z$ be the unobserved part of the fourth count associated with the component of probability $\theta /3$. Its complementary subcount has probability $7/12$. Conditional on the observed fourth count and a parameter value,The complete-data log-likelihood now has the simple formThe constant term can depend on the completed counts, but not on $\theta$. The split therefore converts the troublesome $\log (7+4\theta )$ term into a binomial-like maximization problem. Since $Z$ is unobserved, it should be averaged conditionally in an EM algorithm, not guessed or treated as observed.

Define $N_i={\sum }_{k=1}^N{1}_{\{X_k=a_i\}}$. Grouping equal claim sizes gives the pathwise identityConditional on $N=n$, the $n$ independent claims receive category labels with probabilities $p_i$. There are $n!/(n_1!\cdots n_m!)$ assignments producing prescribed category counts, and every assignment has probability ${\prod }_i{p}_i^{n_i}$. Thus the conditional probability distribution is a multinomial distribution:when the nonnegative counts sum to $n$, and is zero otherwise.

For $0\le z_i\le 1$, condition first on $N$ and use the probability generating function of a Poisson distribution:This factorization is the joint probability generating function of independent Poisson distribution variables. HenceA category with $p_i=0$ has zero count almost surely. The categorical thinning of a Poisson claim count produces independence after averaging over the random total, even though the counts are constrained to sum to $n$ conditional on that total.

Use the preceding construction with type-$i$ claims of deterministic size $a_i$. For $\Lambda ={\sum }_i{\lambda }_i>0$, the answer isThe characteristic function verifies this directly:Hence $T$ has the stated compound Poisson distribution. Distinctness of the $a_i$ makes the atomic probabilities unambiguous.

A zero coefficient is allowed. It corresponds to zero-sized claims in this representation and contributes nothing to $T$. If a representation using only positive claims is preferred, remove that category: its count parameter becomes ${\Lambda }_{+}={\sum }_{i:a_i>0}{\lambda }_i$, with probabilities ${\lambda }_i/{\Lambda }_{+}$ on the positive values. The zero claim sizes in a compound Poisson representation explain why the count parameter need not be unique. If $\Lambda =0$, or if all positive-size categories have zero rate, the aggregate is identically zero.

Let $M_X(t)=\mathbb{E}{e}^{tX}$. Conditioning on $N\sim \mathrm{Pois}(\lambda )$ givesWhere the moment-generating function is finite around zero, differentiating the cumulant-generating function givesIn particular $\mathbb{E}S=\lambda \mathbb{E}X$ and $\mathrm{Var}S=\lambda \mathbb{E}{X}^2$. These compound Poisson cumulants use raw claim moments. The first two formulas also follow by conditioning: $\mathbb{E}[S\mid N]=N\mathbb{E}X$ and $\mathrm{Var}(S\mid N)=N\mathrm{Var}X$, soThis is the law of total variance. They therefore remain valid with finite first two moments even when no positive exponential moment exists.

Under per-claim excess of loss reinsurance, put $Y=\min (X,M)$ and let $W=(X-M{)}_{+}$ be the positive part of the excess. The insurer and reinsurer totals are sums of these payments over the same Poisson distribution count. Write $\overline{F}(x)=\mathbb{P}(X>x)$. Using the tail integral formula for moments on the tail probabilities of $Y$ and $W$ givesFor example, $\mathbb{E}{Y}^2={\int }_0^{\infty }\mathbb{P}(Y^2>t)dt=2{\int }_0^{M}x\overline{F}(x)dx$ after $t=x^2$, and shifting by $M$ gives the reinsurer formula. The moment formulas are finite whenever the corresponding claim moments are finite.

For the specified Pareto distribution, with shape three and minimum $d>0$,If $0\le M\le d$, every claim exceeds the reinsurance retention. ThereforeIf $M\ge d$, the tail integrals instead giveConsequently the requested sum of variances isOn the first branch, $V^{\prime }(M)=\lambda (4M-3d)$; it decreases until $M=3d/4$ and then increases. On the second branch $V^{\prime }(M)=\lambda d^3/M^2>0$. For $\lambda >0$ the minimum is unique; if $\lambda =0$, all variances are zero and every reinsurance retention minimizes them. ThusThe curve starts at $3\lambda d^2$, reaches this minimum, passes through $V(d)=2\lambda d^2$ with matching slope $\lambda d$ on both branches, and then increases concavely toward the horizontal asymptote $3\lambda d^2$. The variance-minimizing retention for shape-three Pareto claims is shown below.

The two aggregate payouts are dependent. Conditioning on the common count gives covariance $\lambda \mathrm{Cov}(Y,W)+\lambda \mathbb{E}Y\mathbb{E}W=\lambda \mathbb{E}[YW]=\lambda M\mathbb{E}W$, so $V(M)+2\mathrm{Cov}(S_I,S_R)=3\lambda d^2$ for every reinsurance retention, as required by $S_I+S_R=S$. The covariance of two components of a compound Poisson sum accounts for this distinction between a sum of variances and the variance of a sum.

There is a moment domain qualification for this heavy-tailed example. For every $t>0$, ${\int }_d^{\infty }{e}^{tx}3d^3{x}^{-4}dx=\infty$, and moments of order $j\ge 3$ also diverge. Thus its positive moment-generating function is not finite and higher finite cumulants cannot be obtained by differentiating around zero. Its mean and variance calculations above are valid by conditioning or by the first two right derivatives at zero of the Laplace transform of a nonnegative random variable.

In a regular model let ${\ell }_N(\theta )$ be the log-likelihood, $U_N(\theta )=\nabla {\ell }_N(\theta )$ its score, $I_N(\theta )$ its total Fisher information, and $\hat{\theta }$ the unrestricted maximum-likelihood estimator. For a point null $\theta ={\theta }_0$ in a p-dimensional parameter space, the standard statistics areThese define the Wald test, score test and likelihood-ratio test. With an interior true parameter, increasing information, suitable differentiability and asymptotic normality, all have asymptotic chi-squared distribution with p degrees of freedom under the null. Observed information can replace expected information when it is asymptotically equivalent.

For q smooth independent restrictions $g(\theta )=0$, replace the Wald statistic by $g(\hat{\theta }{)}^T[G{I}_N(\hat{\theta }{)}^{-1}{G}^T{]}^{-1}g(\hat{\theta })$, where G is the derivative of g at the estimate; compare with chi-squared q. The likelihood-ratio statistic uses the difference between unrestricted and null-restricted maxima. The score statistic is computed at the restricted maximum and uses efficient information for the tested coordinates: with nuisance coordinates lambda this is $I_{\psi \psi }-I_{\psi \lambda }{I}_{\lambda \lambda }^{-1}{I}_{\lambda \psi }$. Inverting the acceptance inequalities for point nulls produces approximate coverage-$1-\alpha$ confidence regions; for example the Wald region consists of theta satisfying $(\hat{\theta }-\theta {)}^T{I}_N(\hat{\theta })(\hat{\theta }-\theta )\le {\chi }_{p,1-\alpha }^2$. Analogous score and likelihood-ratio inequalities give their confidence regions.

For the Poisson family, put $Y_{-1}=1$ to include the initial observation, and defineMultiplying the initial and conditional Poisson probabilities gives, on the supported observation sequences,The parameter-free term is ${\sum }_{i=0}^n[Y_i\log Y_{i-1}-\log (Y_i!)]$, interpreting $0\log 0$ as zero. A positive count after a zero preceding count is outside the support.

For $S_n>0$, strict concavity gives unconstrained maximum $\tilde{\theta }=S_n/T_n$, and the upper endpoint restriction givesThere is an endpoint qualification in the printed question. With probability $e^{-\theta }$, $Y_0=0$ and all later counts are zero. Then $S_n=0$, $T_n=1$ and the likelihood is $e^{-\theta }$, which has no maximum on the stated space $0<\theta \le 1$. The formula produces zero only as an extended estimate on the closure $[0,1]$. This is a likelihood supremum at an excluded boundary, and zero cannot be called an actual maximum in the original space.

The observed negative second derivative is $S_n/{\theta }^2$. Conditional expectation gives $\mathbb{E}{Y}_0=\theta$ and $\mathbb{E}{Y}_i=\theta \mathbb{E}{Y}_{i-1}={\theta }^{i+1}$. Therefore for $0<\theta <1$,This is the bounded information in a subcritical Poisson branching process. Observing more generations of one family does not provide unbounded information. Indeed $\mathbb{P}(Y_n>0)\le \mathbb{E}{Y}_n={\theta }^{n+1}\to 0$, and zero is absorbing, so extinction occurs at a finite generation almost surely. The total observed progeny $S_{\infty }$ is finite; eventually $S_n=S_{\infty }$, $T_n=1+S_{\infty }$, and the extended estimate equals $S_{\infty }/(1+S_{\infty })$, a discrete nondegenerate random variable rather than a consistent increasingly precise estimate.

Here is a direct proof of the failure of a chi-squared Wald limit which does not rely solely on the information bound. Under any interior null ${\theta }_0$, the event $Y_0=1$, $Y_1=0$ has probability ${\theta }_0{e}^{-2{\theta }_0}>0$ and forces every later count to be zero. On it the genuine constrained estimate is $1/2$ for every $n\ge 1$, so the usual expected-information Wald statistic isThus it has a persistent atom tending to $4(1/2-{\theta }_0{)}^2$. The fact that atoms obstruct a continuous distributional limit rules out a chi-squared limit, including when ${\theta }_0$ equals one half and that atom is at zero. Any definition adopted for the separate all-zero sample event leaves this positive-count counterexample intact. Hence the Wald statistic does not have the usual asymptotic chi-squared null law for this increasing observation horizon.

There is a necessary distinction between a failure bound averaged over the unknown state, a worst-case bound, and a separate bound for each candidate state. The last interpretation is false: a forger can discard the inputs and always prepare $|{\psi }_0{⟩}^{\otimes N}$. If the actual state is $|{\psi }_0⟩$, every authenticity test passes with probability one. The meaningful uniform statement concerns worst-case failure, or average failure under any fixed prior giving both states positive probability. The PDF does not specify that prior; the argument below gives an explicit bound for either interpretation.

Put $c=|⟨{\psi }_0|{\psi }_1⟩|\in (0,1)$. Any unconditional strategy producing $N$ output qubits, including ancillas, adaptive measurement in quantum measurements and correlated outputs, is a quantum channel $\mathcal{E}$. Its inputs and ideal outputs are the density operatorsThe projective measurements on the different banknotes commute, and their all-pass effect is exactly $P_i$. Therefore the probability of at least one rejection, conditional on input $i$, isThis remains true for entangled outputs; no independence of their individual test outcomes is assumed.

Use trace distance $D(\rho ,\sigma )=\frac{1}{2}\Vert \rho -\sigma {\Vert }_1$. The pure-state trace distance and fidelity identity givesFor completeness, two pure-state projectors differ by a matrix with nonzero eigenvalues $\pm \sqrt{1-|⟨u|v⟩{|}^2}$, which proves this identity. Also the pure-target upper bound on trace distance gives $D({\rho }_i,P_i)\le \sqrt{e_i}$. To see the mixed-state step, write ${\rho }_i={\sum }_j{q}_j|u_j⟩⟨u_j|$ and use convexity of the trace norm followed by the Cauchy-Schwarz inequality:Finally, trace-distance contraction under quantum channels gives $D({\rho }_0,{\rho }_1)\le D({\sigma }_0,{\sigma }_1)$. This follows from $D(\rho ,\sigma )={\max }_{0\le E\le I}\mathrm{Tr}[E(\rho -\sigma )]$: the adjoint channel is positive and unital, so its image of any effect $E$ is still an effect.

The triangle inequality now yieldsThus the cloning failure bound from trace-distance contraction isand in particular $\max (e_0,e_1)\ge {\Delta }^2/4$. For positive priors ${\pi }_0+{\pi }_1=1$, another Cauchy-Schwarz inequality givesso the average rejection probability satisfiesFor equal priors, one may take $p_0={\Delta }^2/8$; the same choice is a strict lower bound on worst-case rejection. This proves a quantitative version of the no-cloning theorem for the bank's all-pass test. It concerns an unconditional attempt: heralded success cannot be postselected while omitting failed attempts from the accounting.

Let $q=|B|$ and choose $h$ uniformly from the finite family $H$. All bounds below must hold for every pair of distinct inputs $x,y\in A$.

A universal hash family, also called universal2, satisfies $P[h(x)=h(y)]\le 1/q$. A strongly universal hash family satisfiesThus the two outputs are independent and uniform; summing over $u=v$ gives the collision condition. An almost strongly universal hash family permits a larger conditional guessing bound. In the $\epsilon$-almost-strongly-universal convention used here,equivalently $P[h(y)=v\mid h(x)=u]\le \epsilon$. Necessarily $\epsilon \ge 1/q$, with $\epsilon =1/q$ giving strong universality. Stating this normalization matters because the name alone does not specify the allowed forgery probability.

Here is an explicit polynomial almost strongly universal hashing construction and a one-use message authentication protocol. Use a publicly specified finite field ${\mathbb{F}}_q$, where $q=2^t$. Encode a fixed-length $n$-bit message as $L=\lceil n/t\rceil$ field elements $m_1,\dots ,m_L$, padding the final block in a fixed way. Both parties know the length, so this encoding is injective. They share independent uniform secret field elements $a,b$, requiring $2t$ shared secret bits. DefineFor every message, the output is uniform because of $b$. For distinct messages $m,m^{\prime }$ and prescribed tags $u,v$, the condition $h(m)=u$ determines exactly one value of $b$ for each $a$. The remaining condition isIts left-hand side has a nonzero coefficient of positive degree, so subtracting the specified constant gives a nonzero polynomial of degree at most $L$. By the root bound for a polynomial, there are at most $L$ possible field elements $a$. Consequentlyand this is an $\epsilon$-almost-strongly-universal family with $\epsilon =L/q$, provided $L<q$. The powers start at one deliberately: a message coefficient at power zero could give a known constant tag shift, allowing a deterministic substitution.

Alice sends the message and its tag $u=h_{a,b}(m)$ over the public channel. Bob accepts only if his independently computed tag agrees. An attacker who has seen no tag guesses one with probability $1/q$. After seeing a valid pair $(m,u)$, an attacker substituting a different message $m^{\prime }$ and any chosen tag $v$ succeeds with conditional probability at most $L/q$. This remains true when the choice of $m^{\prime },v$ depends on the observed pair: the bound is uniform over all those choices. More explicitly, the observed tag leaves $a$ uniform because each possible $a$ has exactly one compatible secret mask $b$.

For a desired forgery bound $\eta$, choose $t=\lceil {\log }_2(n/\eta )\rceil$. Since $L\le n$ and $q\ge n/\eta$, the substitution bound is at most $\eta$. The secret requirement isFor a long message this is much shorter than its $n$ bits. This is Wegman–Carter authentication: a keyed hash plus a fresh secret one-time pad on the tag. It authenticates the message rather than concealing its contents. Use the key once for the guarantee just established; multiple authenticated messages require fresh masks and a reuse security analysis. Authenticate lengths and sequence identifiers too if variable lengths or replay protection are required.

Compute the honest one-qubit density operators in the computational basis. Writing $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|u⟩=(5|0⟩+|1⟩)/\sqrt{26}$, the two quantum state ensembles giveThe equality includes the off-diagonal coherence; comparing only the basis outcome probabilities would not suffice.

The protocol is perfectly hiding against Bob when Alice is honest. Independence of the honest preparations makes his full received density operator ${\rho }_0^{\otimes N}={\rho }_1^{\otimes N}$. For any POVM $\{E_z\}$, including collective measurements and measurements using ancillas, the Born rule gives the same outcome probabilities $\mathrm{Tr}(E_z{\rho }_0^{\otimes N})=\mathrm{Tr}(E_z{\rho }_1^{\otimes N})$. Thus no measurement supplies information about the committed bit.

It is not binding against Alice. Here is an explicit ensemble-steering attack on quantum bit commitment. For each position, Alice prepares the normalized two-qubit pure stateretains $A$ and sends $B$. Its partial trace on $A$ is the honest common density operator. She sends the $B$ halves of $N$ independent copies while retaining all the $A$ halves, and chooses her bit only at unveiling.

To open one, she makes a projective measurement of each retained qubit in the computational basis. Outcome zero has probability $13/20$ and prepares Bob's state $|u⟩$; outcome one has probability $7/20$ and prepares $|1⟩$. She announces that list. It has exactly the honest bit-one distribution, and every declared-state test passes.

To open zero, she instead measures each retained qubit using a three-outcome POVM with effects $E_j=|v_j⟩⟨v_j|$, whereAll effects are positive. Their diagonal sums are $4/5+1/5=1$ and $4/70+16/70+5/7=1$, and the off-diagonal sum is zero, so $E_0+E_{+}+E_1=I$. This establishes that the proposed POVM is physically valid.

To verify the conditional states directly, the coefficient matrix of $|\Omega ⟩$, with Bob's index as the row and Alice's as the column, isA real rank-one effect $|v_j⟩⟨v_j|$ on Alice's half gives the unnormalized Bob state vector $C{v}_j$. For the three effects above,Their squared norms are therefore $1/2,1/4,1/4$, exactly the honest bit-zero probabilities, and Bob's conditional pure states are the corresponding declared states. Independent measurements on the retained halves produce the honest distribution of the entire list, so even checks on the declared frequencies do not defeat the attack. Every declared-state projector test again succeeds.

Alice can consequently unveil either bit with acceptance probability one, without having chosen it when sending the qubits. This is the Hughston–Jozsa–Wootters theorem mechanism implemented by an explicit measurement, rather than merely an appeal to the general impossibility of perfectly hiding and binding quantum bit commitment.

Use the intended model of independent random variables $A,B$ with exponential distributions of rates $\alpha ,\beta >0$. For $M=\min (A,B)$, independence gives the survival functionThusThe variance is $1/(\alpha +\beta {)}^2$, obtained from the standard first two exponential moments, $EM=1/(\alpha +\beta )$ and $E{M}^2=2/(\alpha +\beta {)}^2$.

For these competing exponential clocks, ties have probability zero. Integrating over Alice's firing time givesLet $H$ denote the event that the next shot hits, and use the stated success probability as $P(H\mid A<B)=1/2$. Bill's corresponding conditional success probability is one. The law of total probability givesThe Bayes' theorem therefore yields

The printed assumptions specify the two marginal exponential distributions but do not explicitly specify their independence. The answers above require that extra assumption; marginals alone do not determine the race. For a concrete counterexample, take one $Z\sim \mathrm{Exp}(1)$ and set $A=Z/\alpha$, $B=Z/\beta$. These have the required marginals, but $M\sim \mathrm{Exp}(\max (\alpha ,\beta ))$; if $\alpha >\beta$, Alice always fires first and $P(A<B\mid H)=1$. Thus the independence qualification changes actual answers, rather than merely their derivation.

All the following distributions are conditional on the point lying in the unit disk. Initially the joint probability density of $(S,T)$ is $1/4$ on the square. The disk has area $\pi$, so the conditioning event has probability $\pi /4$. Dividing by this probability gives the constant conditional density $1/\pi$ on the disk, proving its uniform distribution there.

The polar coordinates $S=r\cos \theta ,T=r\sin \theta$ have Jacobian determinant $r$. ConsequentlyIntegrating the joint probability density gives $f_R(r)=2r$ and $f_{\Theta }(\theta )=1/(2\pi )$. Since the joint density equals the product of these marginal densities on a product support, $R$ and $\Theta$ are independent. In particular $U=R^2$ has the uniform distribution on $(0,1)$.

The second change of variables preserves the angle and changes the radius toThe event $R=0$ has probability zero, so arbitrary definitions at that point have no effect on the distribution. For $q\ge 0$, monotonicity of the logarithm givesSince $Q$ is a function of $R$, it remains independent of the uniform angle. Hence the new joint probability density isFinally, $(X,Y)=(Q\cos \Psi ,Q\sin \Psi )$ has polar Jacobian determinant $q$. Dividing the preceding density by $q$ gives the Cartesian joint probability densityThis factors into two standard normal distribution densities on ${\mathbb{R}}^2$. Thus $X$ and $Y$ are independent, each with mean zero and variance one. This derivation is the radial form of the Box-Muller transform.

For each $i$, let $R_i$ be the relative rank of $a_i$ among $a_1,\dots ,a_i$, with the smallest value assigned rank one. Then $Y_i=1_{\{R_i=1\}}$. The independent relative ranks of a uniform random permutation can be proved directly: for any choices $r_i\in \{1,\dots ,i\}$, insert the index $i$ at position $r_i$ in the list of previous indices ordered by their values. After all insertions, assign the values $1,\dots ,n$ along that list. This is a unique permutation with those relative ranks, and conversely every permutation yields one such insertion sequence.

There are ${\prod }_{i=1}^{n}i=n!$ possible insertion sequences, all equally likely because the uniform random permutation is uniform. ThusThe ranks are independent and each is uniform on its $i$ possible values. It follows that the independent record indicators haveThe first indicator is the constant one, which is also independent of the others. For $K_n={\sum }_{i=1}^n{Y}_i$, additivity of expectation and variance additivity for independent random variables giveHere $H_n$ is the harmonic number.

Let $T$ be the year index of the second record. For $2\le i\le n$, independence givesThe empty product for $i=2$ is one. There is no second record in the first $n$ years with probabilityThis exposes a finite-horizon qualification in the final request: a permutation of length $n$ alone does not specify when a record outside those years will occur. Under the usual continuation by an infinite sequence of independent, identically distributed observations with a continuous distribution, the same relative-rank argument holds for every finite prefix. Then $P(T=i)=1/[i(i-1)]$ for every $i\ge 2$, and $P(T>m)=1/m$. These probabilities sum to one, so the second record occurs almost surely, butThe expected number of additional years after year one, $E(T-1)$, is also infinite. This is the first-sample case of the waiting time for a new record after a fixed sample.

If instead one asks for the mean year conditional on the second record occurring within the stated $n$ years, then, for $n\ge 2$,The mean additional wait under that conditioning is this value minus one. These are distinct questions: assigning $T=\infty$ whenever no second record occurs in the finite horizon also gives an infinite unconditional mean, whereas a censored observation $\min (T,n)$ has finite expectation $1+H_{n-1}$ by the tail-sum formula for expectation.

Normalize the probability density function using $u=t^c$:The corresponding Weibull distribution has survival functionSince the distribution is continuous, the choices $>$ and $\ge$ have the same probabilities. The conditional probability is consequentlyFor $s,t>0$ and $c>1$, the supplied inequality, applied to $s/t$, gives $(s+t{)}^c>t^c+s^c$. Taking exponentials yieldsThis expresses the increasing hazard function $h_c(t)=c{t}^{c-1}$: surviving to a later age shortens the remaining lifetime in this conditional distribution. At $c=1$ equality holds, as expected from the memoryless property of the exponential distribution.

The expectation of a complex random quantity is defined through its real and imaginary parts. Both $\cos X$ and $\sin X$ are bounded, so all expectations below exist. Expand the addition formulae and apply the expectation of a product of independent random variables identity:Combining real and imaginary parts provesFor the simple symmetric random walk, write $W_n={\sum }_{j=1}^n{\xi }_j$, with independent increments satisfying $P({\xi }_j=1)=P({\xi }_j=-1)=1/2$. The usual casino model uses these independent, or conditionally fair, increments. Their characteristic function is $(e^{iu}+e^{-iu})/2=\cos u$, so repeated factorization givesThis is real for real $u$. Equivalently, the distribution of $W_n$ is symmetric, making $E\sin (u{W}_n)=0$.