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Let $S,T$ be independent random variables with exponential distributions of respective positive rates $\alpha ,\beta$. Condition on $T=t$ and use the survival function of $S$:At $u=0$ this is $P(S>T)$. Dividing by that probability proves that the positive difference conditional on $S>T$ again has an exponential distribution with rate $\alpha$. The memorylessness of the exponential distribution persists after an independent competing event has already occurred. Replacing the difference by a negative sentinel when $S\le T$ adds an atom of a measure of mass $\alpha /(\alpha +\beta )$ at that sentinel without changing the positive tail.

For $r\ge 0$, integration gives the survival function and cumulative distribution functionSolving $F_R(r_p)=p$ gives the quantile function^[ref]

For the chosen positioning model $\sigma =2\mathrm{cm}$,The disk of radius one component standard deviation therefore contains about 39.35% of outcomes; the familiar one-dimensional normal interval percentage does not apply to a two-dimensional disk. The 95% and 99% radial thresholds areA circle defined by $R=r_p$ itself still has probability zero; its enclosed disk carries the stated probability.

The closed-form quantile function also gives inverse transform sampling: if $U$ has uniform distribution on $(0,1)$, then $\sigma \sqrt{-2\log U}$ has Rayleigh distribution. Using $1-U$ instead gives the same law. Endpoints are excluded so that the logarithm stays finite.

Reliability theory uses probability theory and statistical inference to describe whether a component or system continues to perform its specified function. For a nonnegative failure time $T$, its reliability at time $t$ is the survival function $R(t)=\mathbb{P}(T>t)$. If $T$ has probability density function $f$, the hazard function $h(t)=f(t)/R(t)$ describes the instantaneous failure rate conditional on survival, where $R(t)>0$.

System structure matters. For independent components with reliabilities $R_i(t)$, a series reliability system fails at the first component failure, so $R_{\mathrm{series}}(t)={\prod }_i{R}_i(t)$. A parallel reliability system works while at least one component works, so $R_{\mathrm{parallel}}(t)=1-{\prod }_i(1-R_i(t))$. Two independent components each having reliability $0.9$ therefore give $0.81$ in series and $0.99$ in parallel. These product formulas depend on independence; a common cause of failure can invalidate them.^[ref]

The residual lifetime process of a renewal process is the time remaining until the next renewal:The time since the last renewal is $A_t=t-S_{N_t}$, and $A_t+E_t$ is the length of the renewal interval containing a fixed time. Between renewal epochs, $E_t$ decreases at speed one. On reaching zero it resets to a freshly drawn interarrival time. For strictly positive iid interarrivals, this deterministic motion and independent resetting make $E_t$ a Markov process, even when the count $N_t$ alone is not Markov.

Its initial and equilibrium laws can differ. An ordinary process started at a renewal has $E_0={\xi }_1$. A stationary observation instead has the equilibrium residual-life distribution, whose survival function is the integrated interarrival tail divided by its mean. For deterministic interarrivals of length $c$, the initial residual is $c$, but the equilibrium residual is uniform on $(0,c)$. For exponential interarrivals the two laws agree, reflecting the memoryless property.

An unobserved survival time is right-censored when it is known to exceed the last event-free follow-up time. For a potential survival time $T$ and potential observation-ending time $C$, recordWhen $\delta =1$, the occurrence was observed at $Y$; when $\delta =0$, only $T>Y=C$ is known. An intact machine at the study's closing date and an event-free subject who stops attending follow-up can both yield right censoring, but the reasons for stopping matter for independent censoring. A subject right-censored at year two has contributed two years to person-time at risk and belongs to earlier risk sets. Neither recording a failure at year two nor discarding that subject preserves this information.

Under independent censoring and separate parameters for the observation mechanism, the part of the survival likelihood involving an absolutely continuous probability distribution for $T$ is $f(Y{)}^{\delta }S(Y{)}^{1-\delta }$. Thus a right-censored observation contributes the survival function $S(Y)$, rather than the event probability density function $f(Y)$. See Lu Tian's Stanford notes, Section 2.1, equations (2.1)–(2.2). The Kaplan–Meier six-subject cohort shows exactly how this distinction changes the calculation.

The right endpoint of a distribution with distribution function $F$ is $x_F=\sup \{x:F(x)<1\}$, possibly infinite. The survival function is positive below $x_F$.

A semi-log plot uses a logarithmic scale on one axis and a linear scale on the other. Which axis is logarithmic matters. Plotting $x$ against $\log y$ makes positive exponential functions appear as straight lines; plotting $\log x$ against $y$ instead makes an affine function of a logarithm appear straight. The plotting operations change positions while retaining the original tick values.^[ref]

Take the original dimensionless mathematical model $y(t)=3e^{0.4t}$ for $0\le t\le 8$. It has exponential growth, with $y(0)=3$, $y(4)\approx 14.8591$, and $y(8)\approx 73.5976$. Taking a base-ten logarithm givesThe slope in these transformed positions is $0.4/\ln 10\approx 0.173718$, although the original relative growth rate $y^{\prime }/y$ is $0.4$. Equal steps of two in $t$ multiply $y$ by $e^{0.8}\approx 2.22554$. The corresponding ordinary gains increase, so the left plot bends while the right plot is straight.

One practical use is checking a proposed exponential distribution for failure times. For rate $\lambda >0$ and $t\ge 0$, its cumulative distribution function is $F(t)=1-e^{-\lambda t}$, so the reciprocal survival function $y=1/[1-F(t)]$ satisfies ${\log }_{10}y=\lambda t/\ln 10$. Estimated probabilities can therefore be placed on a logarithmic vertical axis and compared with a straight line. Agreement supports compatibility with the chosen statistical model; a roughly straight finite sample does not establish a unique physical mechanism.^[ref]

Survival analysis studies the probability distribution of a survival time: how long from a specified starting point until a specified occurrence. The occurrence need not be death. It might be a machine's first breakdown, a customer's first renewal, or a person's recovery. The meaning of "survival" is simply that the chosen occurrence has not happened yet. Both the starting point and the outcome must be defined before interpreting a survival function.

The distinctive difficulty is incomplete observation. A machine still working when observation stops supplies a lower bound on its survival time, rather than an observed breakdown time. Right censoring lets us retain that information. The Kaplan–Meier estimator estimates a survival function using the changing risk sets, without choosing an exponential distribution or another particular survival distribution. A hazard function instead describes the current event rate among those still event-free; its cumulative hazard connects those local rates to the whole survival function.

This framework turns different follow-up lengths into usable evidence about waiting times. It supports questions such as the chance of remaining event-free for another year, or the restricted mean survival time within an observed horizon. Lu Tian's Stanford teaching materials introduce these interpretations in Survival Distributions, Hazard Functions, Cumulative Hazards, Section 1.1.

For a nonnegative survival time $T$, the survival function iswhere $F$ is the cumulative distribution function. Thus $S(t)$ is the probability that the chosen occurrence is still in the future at time $t$. We consistently use the strict inequality $T>t$: a probability mass at $t$ has already left the survival function at $t$ itself. With this convention $S$ is a right-continuous function and a nonincreasing function taking values in $[0,1]$. If $\mathbb{P}(T=0)=0$, then $S(0)=1$; if $T$ is finite with probability one, then $S(t)\to 0$ as $t\to \infty$.

For $S(a)>0$ and $b>a$, conditional probability givesFor example, $S(2)=0.8$ and $S(5)=0.6$ mean that $0.6/0.8=0.75$ of those still event-free at year two remain event-free through year five. Their intervening event probability is $0.25$, whereas the original population's probability of an occurrence in $(2,5]$ is $0.8-0.6=0.2$. Conditioning changes the denominator.

The area under the survival function up to a horizon gives the restricted mean survival time. A positive terminal plateau in a plotted Kaplan–Meier estimator does not establish that the occurrence will never happen: terminal censoring and survival-mean identifiability explains why the unobserved tail remains undetermined.

For a continuous event time $T$ with probability density function $f_{\theta }$, survival function $S_{\theta }$ and hazard function $h_{\theta }=f_{\theta }/S_{\theta }$, right censoring records $Y=\min (T,C)$ and $\delta =1_{\{T\le C\}}$. Suppose $T$ and $C$ are conditionally independent given the observed covariates, suppressed here. Write $g_{\eta }$ and $G_{\eta }$ for the censoring-time probability density function and survival function. The joint observed-data likelihood contribution isAn event contributes its event-time probability density function times the probability of remaining uncensored until then. A censored observation contributes its censoring-time probability density function times the probability of remaining event-free until then.

If the censoring model has separate parameters $\eta$ with no constraints linking them to $\theta$, its factor can be omitted when maximizing over $\theta$. The event-model part for independent subjects is thenwhere $H_{\theta }$ is the cumulative hazard function. Independent censoring alone does not permit dropping a censoring term that itself contains the event-model parameters. The familiar event-versus-censoring formula is thus a factorization under an ignorable observation mechanism, rather than an unconditional rule for every missing event time.^[ref]

A survival time is the elapsed time to an event of interest, which may be death, failure, recovery or completion of a task. Its survival function and hazard function describe the event-time law. Right censoring records only that the event time exceeds the observed follow-up time.

A switching-current distribution of a Josephson junction records the bias electric current at the first detected transition to a sustained nonzero electric potential difference. Interpret it using a stipulated monotone ramp $dI/dt=r>0$, a reproducible initial trapped state, and an instantaneous escape hazard function $\Gamma (I)$. Neglect retrapping and readout delay. The survival function $S$ obeys $dS/dt=-\Gamma S$, so changing variable from time to electric current givesHere $p$ is a probability density function per unit electric current, not a rate per unit time. Division by the surviving population is essential: only trials still trapped can escape. A typical switching current is consequently not identical to the noiseless superconducting critical current.

For an independent illustrative check, choose constant ${\Gamma }_0=1000{\mathrm{s}}^{-1}$ and $r=1\mu \mathrm{A}{\mathrm{s}}^{-1}$. ThenHalf the trials have switched after $\ln 2/{\Gamma }_0=0.693147\mathrm{ms}$, corresponding to $0.693147\mathrm{nA}$ of ramp. Doubling $r$ doubles the mean increment without changing the underlying escape rate. Real junction rates vary sharply with bias; this example isolates the statistical conversion.

Repeating the preparation, measuring the ramp, resolving the voltage onset, and allowing cooling after each dissipative event are prerequisites for this interpretation. Missed events, finite detection delay, changing temperature and retrapping alter the measured distribution. Even a correctly reconstructed rate needs the independent thermal and spectral checks before identifying its escape mechanism.

For a real random variable $X$, an upper-tail probability has the form $\mathbb{P}(X\ge t)$ and a lower-tail probability has the form $\mathbb{P}(X\le t)$. Each is the probability of an event lying beyond a chosen threshold in one direction. A tail probability need not be small: the threshold determines how much of the probability distribution it includes. A concentration inequality bounds such probabilities, often for deviations from an expected value.

Endpoint conventions matter. Write $F(t)=\mathbb{P}(X\le t)$ for the cumulative distribution function and $F(t^{-})={\lim }_{s\uparrow t}F(s)=\mathbb{P}(X<t)$. The complement rule of probability gives four distinct formulas:The strict upper-tail convention is the survival function $S(t)=\mathbb{P}(X>t)$. Its value differs from the inclusive upper tail probability by exactly the probability mass at the threshold:A singleton with positive probability mass is an atom of a measure. Thus $1-F(t)$ excludes an atom of a measure at $t$, whereas $1-F(t^{-})$ includes it. These endpoint identities follow directly from the cumulative distribution function properties in Kyle Siegrist, Distribution and Quantile Functions: endpoint probabilities and right tails.

As the threshold increases, both upper tail probabilities are nonincreasing functions. The strict survival function is a right-continuous function; the inclusive upper tail probability takes the upper endpoint value at a downward jump, equal to the left limit of the survival function. Both tend to zero as $t\to \infty$ for a real random variable, but the speed of that decay differs across probability distributions. The cumulative distribution function properties and their proof from continuity of probability measures are given in MIT 6.436J, Lecture 4: random variables and cumulative distribution function properties.

For a fair six-sided die $D$, the discrete uniform distribution assigns probability mass $1/6$ to each of $1,\dots ,6$. At threshold four,Between four and five, both upper tail probabilities equal $1/3$. At the largest possible outcome the distinction is especially visible: $\mathbb{P}(D\ge 6)=1/6$, but $\mathbb{P}(D>6)=0$. The diagram shows exact probabilities for this mathematical model; its dots specify which endpoints belong to each step function.

If a probability distribution has a probability density function $f$, then every singleton has zero probability mass andFor an exponential distribution of rate two, $f(x)=2e^{-2x}$ for $x\ge 0$ and zero otherwise, so $S(t)=e^{-2t}$ for $t\ge 0$ and $S(t)=1$ for $t<0$. Hence $\mathbb{P}(X\ge 1)=e^{-2}\approx 0.135335$. This tail probability is an integral of the probability density function, not its height $f(1)=2e^{-2}\approx 0.270671$. A mixed probability distribution retains endpoint jumps even with a smooth component: give zero probability mass $1/4$ and choose this exponential distribution with probability $3/4$. Then $\mathbb{P}(X\ge 0)=1$, $\mathbb{P}(X>0)=3/4$, and $S(t)=(3/4)e^{-2t}$ for $t\ge 0$.

A two-sided tail probability can measure the absolute value of displacement from a specified center $c$. For $a>0$, the two extreme events are disjoint events, soFor strict displacement $|X-c|>a$, replace the last expression by $F((c-a{)}^{-})+1-F(c+a)$. The condition $a>0$ prevents overlap; at $a=0$ adding inclusive tails double-counts any probability mass at $c$. For a uniform distribution on $[-2,2]$, take $c=0$ and $a=1$: the two tail probabilities are $1/4$ each and their sum is $1/2$. Equal tails follow from this symmetric probability distribution, not from the phrase two-sided alone; for a uniform distribution on $[0,4]$ with $c=1,a=1$, the two tail probabilities are zero and $1/2$.

The tail-sum formula for expectation turns a whole curve of tail probabilities into an expected value. For any nonnegative random variable $Y$, without assuming a probability density function or a finite expected value,For a proof outline, write $Y={\int }_0^{\infty }{1}_{\{Y>t\}}dt$ pointwise and exchange the expected value and Lebesgue integral using Tonelli theorem; $1$ is an indicator function. The inclusive version has the same Lebesgue integral: positive probability masses occur at only a countable set of thresholds, which has Lebesgue measure zero. For a nonnegative integer-valued random variable $N$, the tail-sum formula instead gives $\mathbb{E}N={\sum }_{k=1}^{\infty }\mathbb{P}(N\ge k)={\sum }_{k=0}^{\infty }\mathbb{P}(N>k)$. The integer hypothesis is essential for these sums. The integral identity, its indicator function proof, and the integer version appear in Kyle Siegrist, Additional Properties of Expected Value: integration and summation of right tails.

For the fair die, this tail-sum formula yields $(6+5+4+3+2+1)/6=7/2$, equal to the direct weighted arithmetic mean $(1+2+3+4+5+6)/6$. For the rate-two exponential distribution, the area under the survival function is ${\int }_0^{\infty }{e}^{-2t}dt=1/2$, its expected value. The first panel below shades an integral of the probability density function for a single threshold; the second shades an integral of the survival function across all nonnegative thresholds. These are analytical illustrations, with the displayed curves truncated at three; their exact total areas use an infinite integration range.

Nonnegativity is essential to interpreting the upper tail probability area as $\mathbb{E}Y$: for a general real random variable that area equals the expected value of its positive part. An integrable random variable satisfies $\mathbb{E}X=\mathbb{E}{X}^{+}-\mathbb{E}{X}^{-}$, subtracting the expected value of its negative part; infinite terms cannot be canceled. Even when $Y$ is finite almost surely, its expected value can be infinite. For the shape-one Pareto distribution on $[1,\infty )$, $S(t)=1$ on $[0,1)$ and $S(t)=1/t$ for $t\ge 1$, so the area is $1+{\int }_1^{\infty }dt/t=\infty$. The finite versus infinite expected value distinction is illustrated in Kyle Siegrist, Definitions and Basic Properties: exponential and Pareto expected values.

Why are tail probabilities useful? They answer threshold questions directly, and a concentration inequality can bound them without knowing the full probability distribution. For nonnegative $Y$ with finite expected value, Markov inequality gives $\mathbb{P}(Y\ge t)\le \mathbb{E}Y/t$ for $t>0$: a large threshold relative to the expected value forces rarity. The quantile function reverses the question by choosing a threshold for a prescribed cumulative level. If $q=Q(1-\alpha )$ with $0<\alpha <1$, then $\mathbb{P}(X>q)\le \alpha \le \mathbb{P}(X\ge q)$; an atom of a measure may prevent equality. For the die, $Q(0.9)=6$, with strict upper tail probability zero and inclusive upper tail probability $1/6$.

A tail probability describes rarity under a specified probability distribution; it is not automatically a conditional probability after extra information. For $u\ge t$ with $\mathbb{P}(X\ge t)>0$,because the threshold events are nested. Thus the die has $\mathbb{P}(D\ge 5)=1/3$ but $\mathbb{P}(D\ge 5\mid D\ge 4)=2/3$. Conditioning changes the sample space and renormalizes its probability measure; the same extreme outcomes can become common in that restricted population.

The tail-sum formula expresses an expected value as the area under a survival function. For every nonnegative random variable $Y$,This Lebesgue integral and the expected value may both be $+\infty$; a probability density function is unnecessary. For a random variable $N$ taking values in the nonnegative integers, the corresponding sum isThe events $\{N>k\}$ and $\{N\ge k+1\}$ coincide because $N$ takes integer values. Thus the strict sum starts at zero, while the inclusive sum starts at one.^[ref]

For an original example, let the probability mass function of $N$ beIts expected value computed directly is $0(1/5)+1(3/10)+3(1/2)=9/5=1.8$. The strict tail probabilities at thresholds $0,1,2,3$ are $4/5,1/2,1/2,0$: their sum is again $1.8$. There is no probability mass at two, yet the event $\{N>2\}$ still has probability $1/2$. The survival function has heights $4/5,1/2,1/2$ on the successive intervals $[0,1),[1,2),[2,3)$; each interval has length one, so its integral is the same sum.

The indexing matters numerically. Starting the inclusive sum at zero adds $\mathbb{P}(N\ge 0)=1$, giving $2.8$ for this example. Starting the strict sum at one omits $\mathbb{P}(N>0)=4/5$, giving $1$. For a general nonnegative random variable, replacing $>$ with $\ge$ inside the integral is harmless: the difference is $\mathbb{P}(Y=t)$, nonzero at only a countable set of thresholds. For each positive integer $m$, at most $m$ thresholds can have probability mass at least $1/m$; their countable union contains all thresholds with positive probability mass and has Lebesgue measure zero. This reasoning concerns an integral, not equality of the two survival function conventions at every point.

For a continuous original example, take the uniform distribution on the interval $[0,4]$. Its survival function is $S(t)=1-t/4$ for $0\le t\le 4$ and zero for $t>4$. The area is a triangle of base four and height one:Directly integrating against the probability density function gives the same expected value, ${\int }_0^{4}y(1/4)dy=2$. The integral of the probability density function alone is one; it is the survival function whose area is the expected value.

The proof counts layers of each outcome. For a realized value $y\ge 0$, the indicator function $1_{\{y>t\}}$ equals one over an interval of length $y$, soTonelli theorem permits the interchange because the indicator function is nonnegative; no finite expected value is assumed. Similarly, an outcome $N=n$ contributes one indicator random variable for each of the $n$ events $\{N\ge k\}$ that occur:Again Tonelli theorem applies to the nonnegative sum, or equivalently the monotone convergence theorem applies to its partial sums. This exchanges counting layers within each outcome with averaging across outcomes. The existing integer-valued hypothesis in the tail-sum formula explains what changes when the values are not integers; the tail integral formula for moments weights the layers to obtain higher moments.

For an integrable random variable $X$ that can have either sign, apply the nonnegative formula separately to its positive part $X_{+}=\max (X,0)$ and negative part $X_{-}=\max (-X,0)$:Integrability makes both integrals finite. More generally the difference defines an extended expected value when at least one is finite; if both are infinite, subtraction does not define an expected value.^[ref]

Tail probabilities are often easier to measure or bound than a whole probability density function. For a device's survival time, the survival function records the probability of still operating at each threshold; its integral gives the mean operating time. A finite observation window naturally gives the expected value of a capped nonnegative random variable. For a mathematical bound, $\mathbb{P}(Y>t)\le e^{-t/2}$ for all $t\ge 0$ implies $\mathbb{E}[Y]\le {\int }_0^{\infty }{e}^{-t/2}dt=2$. The tail-sum formula turns control of a whole family of events into control of an average.

For serial stages with survival functions $S_i(a)=\exp [-{\int }_0^a{\Gamma }_i(s)ds]$, set $H_i(r)={\int }_0^{\infty }{\Gamma }_i(a)S_i(a)e^{-ra}da$. If completion of stage two produces $m$ new stage-one individuals, a common exponential mode requires $m{H}_1(r)H_2(r)=1$. Indeed its age profiles are ${\varphi }_i(a)={\varphi }_i(0)e^{-ra}{S}_i(a)$, while the incoming fluxes give ${\varphi }_2(0)=H_1{\varphi }_1(0)$ and ${\varphi }_1(0)=m{H}_2{\varphi }_2(0)$. Cancelling nonzero amplitudes proves the equation. Under ordinary positive residence-time assumptions a positive root exists exactly when $m{H}_1(0)H_2(0)>1$. The normalized profiles are $e^{-ra}{S}_i(a)/\int e^{-rs}{S}_i(s)ds$.

For an absolutely continuous distribution function with positive probability density function near its right endpoint of a distribution, write $\eta =f/(1-F)$ for its hazard function. The limits $t\eta (t)\to \alpha$ at an infinite endpoint or $(x_F-t)\eta (t)\to \alpha$ at a finite endpoint imply attraction to the Fréchet distribution or negative Weibull distribution, respectively. A continuously differentiable reciprocal hazard function with $(1/\eta {)}^{\prime }\to 0$ implies attraction to the Gumbel distribution, provided the survival function tends to zero at the endpoint.

A waiting-time distribution is the probability distribution of a nonnegative random duration $T$ until a specified event. It can be discrete, absolutely continuous, or mixed. Its survival function is $S(t)=\mathbb{P}(T>t)$ and its cumulative distribution function is $F(t)=1-S(t)$.

If the finite-time law is absolutely continuous, the probability density function is $f(t)=-S^{\prime }(t)$ almost everywhere. Mere continuity of the cumulative distribution function is insufficient for this density representation. At times with positive survival, the hazard function is $h(t)=f(t)/S(t)$; it measures the instantaneous event rate conditional on having waited that long.

Constant hazard $h(t)=\lambda >0$ with $S(0)=1$ gives $S(t)=e^{-\lambda t}$, the exponential distribution. Waiting a discrete number of independent Bernoulli trials instead gives a geometric distribution. A fixed deterministic wait has a point mass at its prescribed duration rather than an ordinary density.

Sequential independent waiting stages add, so their probability distributions combine by convolution. The two-step catalytic waiting time gives a concrete chemical-kinetics example. Waiting-time distributions also describe first-passage times and time-to-event observations in survival analysis.

The two-parameter Weibull distribution is a positive lifetime model with shape $k>0$ and scale $\lambda >0$. Its Probability density function and Survival function areIts Cumulative distribution function is $1-S(y)$. The Hazard function is $h(y)=k{\lambda }^{-1}(y/\lambda {)}^{k-1}$: it decreases for $k<1$, is constant for $k=1$, and increases for $k>1$. This permits decreasing or increasing failure rates within one family.^[ref]

The mean is $\lambda \Gamma (1+1/k)$ and the variance is ${\lambda }^2[\Gamma (1+2/k)-\Gamma (1+1/k{)}^2]$. The substitution $v=(y/\lambda {)}^k$ reduces the power-moment integrals to gamma integrals. At $k=1$ the law is exponential with rate $1/\lambda$; at $k=2$ its survival is $e^{-(y/\lambda {)}^2}$.

Inverse transform sampling is explicit: for $U$ uniform on $(0,1)$, take $Y=\lambda (-\log U{)}^{1/k}$. Conditional survival after an already attained age $a$ is $S(a+u)/S(a)$, which depends on $a$ unless $k=1$. This is why an Unknown entry age in a Weibull survival segment affects the likelihood. The scale parameter here is $\lambda$; inverse-scale parameterizations instead use its reciprocal.

Let $X$ have survival function $\mathbb{P}(X\ge x)=x^{-s}$ for $x\ge 1$, with $s>1$. Its expected value is $s/(s-1)$. The tail-sum formula for expectation givesThe absolutely continuous probability distribution gives zero mass to the integers, so $\lceil X\rceil =\lfloor X\rfloor +1$ almost surely and $\mathbb{E}[\lceil X\rceil ]=\zeta (s)+1$. Taking expectations of $\lfloor X\rfloor \le X\le \lceil X\rceil$ gives $\zeta (s)\le s/(s-1)\le \zeta (s)+1$, which rearranges to the displayed bounds on the Riemann zeta function. Finiteness is justified by $\lfloor X\rfloor \le X$ and $\lceil X\rceil \le X+1$.