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Exam MAS-I — Survival & Reliability Models Flashcards

Survival and reliability analysis for CAS Exam MAS-I: the survival, density, hazard, and cumulative-hazard functions and the identities linking them; exponential, Weibull, and gamma lifetime models; series, parallel, and k-out-of-n system reliability for independent components; mean residual (excess) life; and empirical estimation from right-censored data via the Kaplan-Meier product-limit and Nelson-Aalen estimators - with fully worked hazards, system reliabilities, an MTTF, and a Kaplan-Meier calculation.

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  1. Survival & hazard
    Define the **survival function** $S(t)$ and give its relationship to the cdf $F(t)$ and density $f(t)$.
    For a nonnegative lifetime $T$, $S(t)=\Pr(T>t)=1-F(t)$ is the probability of surviving past time $t$. Properties: $S(0)=1$, $S$ is nonincreasing, and $S(\infty)=0$. The density is $f(t)=-\frac{d}{dt}S(t)=F'(t)$, so $S(t)=\int_t^\infty f(u)\,du$.
  2. Survival & hazard
    Define the **hazard (failure) rate** $h(t)$ and state the three equivalent forms.
    The hazard rate is the instantaneous failure intensity given survival to $t$: $h(t)=\lim_{\Delta t\to 0}\frac{\Pr(t<T\le t+\Delta t\mid T>t)}{\Delta t}$. Equivalent forms: $h(t)=\frac{f(t)}{S(t)}=-\frac{d}{dt}\ln S(t)$. It is a rate (can exceed $1$), not a probability. In reliability it is called the **force of mortality** or **failure rate**.
  3. Survival & hazard
    Define the **cumulative hazard** $H(t)$ and state how it recovers $S(t)$.
    $H(t)=\int_0^t h(u)\,du = -\ln S(t)$. Inverting, $S(t)=e^{-H(t)}$. Since $S$ decreases from $1$ to $0$, $H$ increases from $0$ to $\infty$. This is the master identity of survival analysis: any one of $S$, $f$, $h$, $H$ determines the other three.
  4. Survival & hazard
    Given $S(t)=e^{-0.02t}$ for $t\ge 0$, find $f(t)$, $h(t)$, and $H(t)$.
    Density: $f(t)=-S'(t)=0.02e^{-0.02t}$. Hazard: $h(t)=\frac{f(t)}{S(t)}=\frac{0.02e^{-0.02t}}{e^{-0.02t}}=0.02$ (constant). Cumulative hazard: $H(t)=-\ln S(t)=0.02t$. This is an exponential lifetime with rate $\lambda=0.02$; its constant hazard is the signature of the exponential.
  5. Survival & hazard
    A lifetime has hazard $h(t)=0.01+0.002t$ for $t\ge 0$. Find $S(t)$ and the probability of surviving past $t=10$.
    Cumulative hazard: $H(t)=\int_0^t(0.01+0.002u)\,du = 0.01t + 0.001t^{2}$. Survival: $S(t)=e^{-H(t)}=e^{-(0.01t+0.001t^{2})}$. At $t=10$: $H(10)=0.01(10)+0.001(100)=0.1+0.1=0.2$, so $S(10)=e^{-0.2}\approx 0.8187$. The linearly increasing hazard makes this an aging (wear-out) component.
  6. Survival & hazard
    Recover the survival function from a hazard: given $h(t)=\frac{2t}{1+t^{2}}$ for $t\ge 0$, find $S(t)$.
    $H(t)=\int_0^t \frac{2u}{1+u^{2}}\,du$. With substitution $w=1+u^{2}$, $dw=2u\,du$: $H(t)=\big[\ln(1+u^{2})\big]_0^t=\ln(1+t^{2})$. $S(t)=e^{-H(t)}=e^{-\ln(1+t^{2})}=\frac{1}{1+t^{2}}$. Check: $S(0)=1$ and $S(t)\to 0$, as required.
  7. Lifetime distributions
    State the **exponential** lifetime model: survival, hazard, mean, variance, and its defining property.
    For rate $\lambda>0$: $S(t)=e^{-\lambda t}$, $f(t)=\lambda e^{-\lambda t}$, $h(t)=\lambda$ (constant). Mean $E[T]=\frac{1}{\lambda}$ (the MTTF), variance $\frac{1}{\lambda^{2}}$. Defining property: **memorylessness** — $\Pr(T>s+t\mid T>s)=\Pr(T>t)$. A used component is stochastically as good as new.
  8. Lifetime distributions
    A component has an exponential lifetime with mean $500$ hours. Find the hazard rate and the probability it survives past $800$ hours.
    Mean $=\frac{1}{\lambda}=500$, so the hazard $\lambda=\frac{1}{500}=0.002$ per hour (constant). $S(800)=e^{-\lambda(800)}=e^{-0.002(800)}=e^{-1.6}\approx 0.2019$. So about a $20.2\%$ chance of lasting beyond $800$ hours.
  9. Lifetime distributions
    An exponential component has already survived $300$ hours. Given $\lambda=0.002$, find the probability it survives an **additional** $400$ hours.
    By memorylessness the past survival is irrelevant: $\Pr(T>300+400\mid T>300)=\Pr(T>400)=e^{-\lambda(400)}=e^{-0.002(400)}=e^{-0.8}\approx 0.4493$. The used component behaves like a brand-new one — a unique feature of the constant-hazard exponential.
  10. Lifetime distributions
    Define **MTTF** (mean time to failure) and give two ways to compute it from the survival function.
    MTTF $=E[T]$, the expected lifetime of a component. Direct: $E[T]=\int_0^\infty t\,f(t)\,dt$. Via survival (often easier): for $T\ge 0$, $E[T]=\int_0^\infty S(t)\,dt$. For the exponential, $\int_0^\infty e^{-\lambda t}\,dt=\frac{1}{\lambda}$, recovering MTTF $=1/\lambda$.
  11. Lifetime distributions
    A lifetime has $S(t)=\left(\frac{100}{100+t}\right)^{3}$ for $t\ge 0$ (a Pareto/Lomax form). Find the MTTF.
    Use MTTF $=\int_0^\infty S(t)\,dt = \int_0^\infty 100^{3}(100+t)^{-3}\,dt$. $=100^{3}\left[\frac{(100+t)^{-2}}{-2}\right]_0^\infty = 100^{3}\cdot\frac{(100)^{-2}}{2}=\frac{100^{3}}{2\cdot 100^{2}}=\frac{100}{2}=50$. MTTF $=50$. (General Pareto mean $=\frac{\theta}{\alpha-1}=\frac{100}{3-1}=50$.)
  12. Lifetime distributions
    State the **Weibull** lifetime model and how the shape parameter $\tau$ controls the hazard.
    Weibull survival: $S(t)=e^{-(t/\theta)^{\tau}}$, with scale $\theta>0$ and shape $\tau>0$. Hazard: $h(t)=\frac{\tau}{\theta}\left(\frac{t}{\theta}\right)^{\tau-1}\propto t^{\tau-1}$. **$\tau>1$:** increasing hazard (wear-out / aging). **$\tau<1$:** decreasing hazard (infant mortality / burn-in). **$\tau=1$:** constant hazard — reduces to the exponential.
  13. Lifetime distributions
    A Weibull lifetime has shape $\tau=2$ and scale $\theta=100$. Find the hazard at $t=50$ and the probability of survival past $t=100$.
    Hazard: $h(t)=\frac{\tau}{\theta}\left(\frac{t}{\theta}\right)^{\tau-1}=\frac{2}{100}\left(\frac{t}{100}\right)^{1}=\frac{2t}{100^{2}}=\frac{2t}{10000}$. At $t=50$: $h(50)=\frac{2(50)}{10000}=0.01$ per unit (rising in $t$ since $\tau>1$). Survival: $S(100)=e^{-(100/100)^{2}}=e^{-1}\approx 0.3679$.
  14. Lifetime distributions
    Distinguish **infant mortality**, **random failure**, and **wear-out** on the bathtub hazard curve, and which $\tau$ models each.
    The bathtub curve plots $h(t)$ in three phases: **Infant mortality:** decreasing hazard (early defects burn in) — Weibull $\tau<1$. **Useful life:** roughly constant hazard (random shocks) — exponential, $\tau=1$. **Wear-out:** increasing hazard (aging/fatigue) — Weibull $\tau>1$. Reliability engineering blends these phases to describe a component's whole life.
  15. Reliability & systems
    Define the **reliability function** $R(t)$ and how it relates to the survival function and hazard.
    In reliability theory $R(t)=\Pr(T>t)=S(t)$ — the probability a component is still functioning at time $t$. The two terms are interchangeable. Thus $R(t)=e^{-H(t)}$, the failure rate is $h(t)=-\frac{R'(t)}{R(t)}$, and MTTF $=\int_0^\infty R(t)\,dt$.
  16. Reliability & systems
    State the **series system** reliability formula for $n$ independent components and explain when it fails.
    A series system works **only if every** component works, so it fails when **any** one fails. With independent components of reliability $R_i$, the system reliability is $R_S=\prod_{i=1}^{n} R_i = R_1 R_2 \cdots R_n$. Because each $R_i\le 1$, $R_S$ is never greater than the weakest component — series links degrade reliability.
  17. Series & parallel
    State the **parallel system** reliability formula for $n$ independent components and explain when it fails.
    A parallel (redundant) system fails **only if all** components fail, so it works if **any** one works. The system **un**reliability is the product of component unreliabilities: $R_P = 1 - \prod_{i=1}^{n}(1-R_i)$. Because each $(1-R_i)\le 1$, $R_P$ is never less than the strongest component — redundancy improves reliability.
  18. Series & parallel
    A series system has three independent components with reliabilities $0.95$, $0.90$, and $0.99$. Find the system reliability.
    Series: multiply the component reliabilities. $R_S = 0.95 \times 0.90 \times 0.99$. $0.95\times 0.90 = 0.855$; $0.855\times 0.99 = 0.84645$. $R_S \approx 0.8465$. Note the system is **less** reliable than its weakest part ($0.90$) — the hallmark of a series arrangement.
  19. Series & parallel
    Two independent components, each with reliability $0.80$, are placed in **parallel**. Find the system reliability.
    Parallel: the system fails only if both fail. Each failure probability is $1-0.80=0.20$. $R_P = 1 - (1-0.80)(1-0.80) = 1 - (0.20)(0.20) = 1 - 0.04 = 0.96$. The redundant pair ($0.96$) is far more reliable than a single unit ($0.80$).
  20. Series & parallel
    A system has two components in **series**, each of which is itself a **parallel** pair of units with unit reliability $0.90$. Find the overall reliability.
    First each parallel pair: $R_{\text{pair}}=1-(1-0.90)^{2}=1-0.01=0.99$. Then the two pairs are in series: $R_S = 0.99 \times 0.99 = 0.9801$. Overall reliability $\approx 0.9801$. Redundancy at the unit level lifts each block to $0.99$ before the series multiplication.
  21. Reliability & systems
    Each component of a $3$-component **series** system is exponential with the same rate $\lambda=0.01$. Find the system survival function and its MTTF.
    Series survival: $R_S(t)=\prod_i e^{-\lambda t}=e^{-3\lambda t}=e^{-0.03t}$. The system itself is exponential with combined rate $3\lambda=0.03$ (hazards of independent series components **add**). MTTF $=\frac{1}{3\lambda}=\frac{1}{0.03}\approx 33.33$, one-third the single-component MTTF of $100$.
  22. Reliability & systems
    Why do the **hazard rates add** for independent components in series, and what does this imply for the system MTTF?
    Series survival is $R_S(t)=\prod_i e^{-H_i(t)}=e^{-\sum_i H_i(t)}$, so $H_S(t)=\sum_i H_i(t)$ and therefore $h_S(t)=\sum_i h_i(t)$ — the hazards add. For identical exponential parts each with rate $\lambda$, $h_S=n\lambda$ and the system MTTF $=\frac{1}{n\lambda}$, shrinking as components are added in series.
  23. Reliability & systems
    Define a **$k$-out-of-$n$** system and give the reliability formula for identical independent components.
    A $k$-out-of-$n$ system works if **at least $k$** of its $n$ components work. (Series is the $n$-out-of-$n$ case; parallel is $1$-out-of-$n$.) For i.i.d. components each with reliability $p$, the number working is Binomial$(n,p)$, so $R = \sum_{j=k}^{n}\binom{n}{j} p^{j}(1-p)^{n-j}$.
  24. Reliability & systems
    A $2$-out-of-$3$ system has identical independent components each with reliability $0.90$. Find the system reliability.
    Works if at least $2$ of $3$ work; let $X\sim\text{Bin}(3,0.9)$, need $\Pr(X\ge 2)$. $\Pr(X=2)=\binom{3}{2}(0.9)^{2}(0.1)=3(0.81)(0.1)=0.243$. $\Pr(X=3)=(0.9)^{3}=0.729$. $R=0.243+0.729=0.972$. The $2$-of-$3$ majority system ($0.972$) beats a single unit ($0.90$) without full triple redundancy.
  25. Series & parallel
    A bridge-free system has component A in **series** with a **parallel** block of B and C. Reliabilities: $R_A=0.98$, $R_B=0.85$, $R_C=0.85$. Find the system reliability.
    Parallel block B$\parallel$C: $R_{BC}=1-(1-0.85)(1-0.85)=1-(0.15)^{2}=1-0.0225=0.9775$. Then A in series with that block: $R_S = R_A \times R_{BC}=0.98\times 0.9775$. $=0.95795$. $R_S \approx 0.9580$.
  26. Series & parallel
    Two independent components have exponential lifetimes with rates $\lambda_1=0.01$ and $\lambda_2=0.02$, arranged in **parallel**. Find the probability the system survives past $t=50$.
    Component survivals at $t=50$: $R_1=e^{-0.01(50)}=e^{-0.5}\approx 0.6065$, $R_2=e^{-0.02(50)}=e^{-1}\approx 0.3679$. Parallel: $R_P = 1-(1-R_1)(1-R_2)=1-(0.3935)(0.6321)$. $(0.3935)(0.6321)\approx 0.2487$. $R_P \approx 1-0.2487 = 0.7513$.
  27. Reliability & systems
    Why is a **parallel** system's lifetime distribution generally NOT exponential even when its components are exponential?
    For a parallel pair the system lifetime is the **maximum** of the component lifetimes, $T_S=\max(T_1,T_2)$, with $R_P(t)=1-(1-e^{-\lambda_1 t})(1-e^{-\lambda_2 t})$. Differentiating shows the hazard $h_S(t)$ is **not constant** — it rises from $0$ as the system ages. The memoryless property is lost; redundancy creates an increasing failure rate even from constant-hazard parts.
  28. Mean residual life
    Define the **mean residual life** (mean excess life) $e(t)$ and give its integral formula.
    $e(t)=E[T-t\mid T>t]$ is the expected remaining lifetime given survival to $t$. Formula: $e(t)=\frac{1}{S(t)}\int_t^\infty S(u)\,du$. In loss models this is the **mean excess loss** $e_X(d)=E[X-d\mid X>d]$ above a deductible $d$. At $t=0$ it equals the MTTF: $e(0)=E[T]$.
  29. Mean residual life
    What is the mean residual life $e(t)$ of an **exponential** lifetime, and why?
    For an exponential with rate $\lambda$, $e(t)=\frac{1}{S(t)}\int_t^\infty e^{-\lambda u}\,du = \frac{1}{e^{-\lambda t}}\cdot\frac{e^{-\lambda t}}{\lambda}=\frac{1}{\lambda}$. The mean residual life is **constant** at $\frac{1}{\lambda}=$ MTTF for every $t$, a direct consequence of memorylessness: a survivor's expected remaining life never changes.
  30. Mean residual life
    A lifetime is uniform on $[0,100]$. Find the mean residual life $e(40)$.
    For Uniform$[0,100]$, $S(t)=\frac{100-t}{100}$ for $0\le t\le 100$. $\int_t^{100} S(u)\,du = \int_t^{100}\frac{100-u}{100}\,du = \frac{(100-t)^{2}}{200}$. $e(t)=\frac{1}{S(t)}\cdot\frac{(100-t)^{2}}{200}=\frac{100}{100-t}\cdot\frac{(100-t)^{2}}{200}=\frac{100-t}{2}$. $e(40)=\frac{100-40}{2}=30$. (Expected remaining life of a uniform survivor is half the time left.)
  31. Mean residual life
    How does the **sign of the hazard's trend** relate to whether mean residual life increases or decreases (DFR vs IFR)?
    **Increasing hazard (IFR, e.g. Weibull $\tau>1$):** an aging component's expected remaining life **decreases** with $t$. **Decreasing hazard (DFR, e.g. Weibull $\tau<1$):** mean residual life **increases** with $t$ — survivors that pass the burn-in phase are sturdier. **Constant hazard (exponential):** $e(t)$ is constant. Thus $e'(t)$ and $h'(t)$ generally have opposite signs.
  32. Mean residual life
    A lifetime has $S(t)=e^{-0.05t}$. A maintenance contract pays for the **expected remaining life** of a unit that has survived to $t=20$. What value does it use?
    This is exponential with $\lambda=0.05$, so by memorylessness $e(20)=\frac{1}{\lambda}=\frac{1}{0.05}=20$. The contract values the expected remaining life at $20$ time units — exactly the original MTTF, regardless of the $20$ units already survived.
  33. Mean residual life
    Relate the mean residual life $e(d)$ to the **expected value with a deductible** in loss models.
    Splitting the unconditional mean of losses above $d$: $E[(X-d)_+] = E[X-d\mid X>d]\cdot \Pr(X>d) = e(d)\,S(d)$. So the expected payment per loss with deductible $d$ equals the mean excess loss times the survival probability. Equivalently $e(d)=\frac{E[(X-d)_+]}{S(d)}$ — the per-payment severity above the deductible.
  34. KM & Nelson-Aalen
    What problem do the **Kaplan-Meier** and **Nelson-Aalen** estimators solve, and what is **right censoring**?
    They estimate $S(t)$ (or $H(t)$) **nonparametrically** from lifetime data without assuming a distribution. **Right censoring:** for some subjects we only know the lifetime exceeds their last observed time (study ended, lost to follow-up) — the exact failure time is unknown but bounded below. KM and NA correctly use this partial information through the at-risk set, rather than discarding censored observations.
  35. KM & Nelson-Aalen
    State the **Kaplan-Meier** (product-limit) estimator and define $r_i$ and $s_i$.
    At each distinct **death time** $t_i$, let $r_i$ = number **at risk** (alive and uncensored) just before $t_i$ and $s_i$ = number of **deaths** at $t_i$. $\hat S(t)=\prod_{t_i\le t}\left(1-\frac{s_i}{r_i}\right)$. The estimate is a right-continuous step function, dropping only at observed death times; censored times reduce later $r_i$ but cause no drop.
  36. KM & Nelson-Aalen
    State the **Nelson-Aalen** estimator of the cumulative hazard and how it gives an alternative estimate of $S(t)$.
    $\hat H(t)=\sum_{t_i\le t}\frac{s_i}{r_i}$ — a sum (not a product) of the per-time hazard contributions. An alternative survival estimate is $\tilde S(t)=e^{-\hat H(t)}$. Nelson-Aalen is preferred for estimating $H$ directly and tends to be more stable with small samples; it generally gives slightly **higher** survival estimates than KM.
  37. KM & Nelson-Aalen
    Ten units are on test. Failures occur at $t=2$ ($1$ unit), $t=5$ ($2$ units), and $t=8$ ($1$ unit); no censoring. Compute the Kaplan-Meier estimate $\hat S(8)$.
    Build the risk sets: $t=2$: $r_1=10$, $s_1=1 \Rightarrow 1-\frac{1}{10}=0.9$. $t=5$: $r_2=9$, $s_2=2 \Rightarrow 1-\frac{2}{9}=\frac{7}{9}\approx 0.7778$. $t=8$: $r_3=7$, $s_3=1 \Rightarrow 1-\frac{1}{7}=\frac{6}{7}\approx 0.8571$. $\hat S(8)=0.9\times 0.7778\times 0.8571 \approx 0.6000$.
  38. KM & Nelson-Aalen
    Eight units are tested. Deaths at $t=3,7,9$; a unit is **censored** at $t=5$. Risk counts: $r=8$ at $t=3$, then $r=6$ at $t=7$, $r=5$ at $t=9$ (one death each). Compute $\hat S(9)$.
    Censoring at $t=5$ removes a unit from the risk set without a survival drop. $t=3$: $1-\frac{1}{8}=0.8750$. $t=7$: $1-\frac{1}{6}\approx 0.8333$ (the death at $t=3$ and the censoring at $t=5$ both leave $6$ at risk just before $t=7$). $t=9$: $1-\frac{1}{5}=0.8000$. $\hat S(9)=0.8750\times 0.8333\times 0.8000 \approx 0.5833$.
  39. KM & Nelson-Aalen
    Using the same data — $10$ units, deaths $1$ at $t=2$, $2$ at $t=5$, $1$ at $t=8$, no censoring — compute the Nelson-Aalen $\hat H(8)$ and the implied $\tilde S(8)=e^{-\hat H(8)}$.
    Sum the hazard contributions $\frac{s_i}{r_i}$: $\hat H(8)=\frac{1}{10}+\frac{2}{9}+\frac{1}{7}=0.1000+0.2222+0.1429=0.4651$. $\tilde S(8)=e^{-0.4651}\approx 0.6281$. This exceeds the Kaplan-Meier $\hat S(8)\approx 0.6000$, illustrating that Nelson-Aalen typically gives a slightly higher survival estimate.
  40. KM & Nelson-Aalen
    How does a **censored** observation at time $c$ enter the Kaplan-Meier calculation, and why does it not cause $\hat S$ to drop?
    A unit censored at $c$ stays counted in the at-risk set $r_i$ for all death times $t_i\le c$, then is removed before any later death time. Because no death is recorded at $c$, there is no factor $\left(1-\frac{s_i}{r_i}\right)$ for it — $\hat S$ does not step down there. Its only effect is to shrink the later risk sets, which makes subsequent death-time drops slightly larger. This is how KM extracts information from incomplete (right-censored) lifetimes.
  41. Lifetime distributions
    Define the **maximum likelihood estimator** of an exponential rate from complete lifetime data, and give it for a sample mean of $250$.
    For i.i.d. exponential$(\lambda)$ data $t_1,\dots,t_n$, the log-likelihood is $\ell(\lambda)=n\ln\lambda - \lambda\sum t_i$. Setting $\ell'(\lambda)=\frac{n}{\lambda}-\sum t_i=0$ gives $\hat\lambda=\frac{n}{\sum t_i}=\frac{1}{\bar t}$. With $\bar t=250$: $\hat\lambda=\frac{1}{250}=0.004$, so the estimated MTTF is $\frac{1}{\hat\lambda}=250$.
  42. Lifetime distributions
    A test runs $20$ identical exponential units; total observed run-time across all units is $4{,}000$ hours and $8$ failures occur (Type I censoring). Estimate the failure rate $\hat\lambda$ and MTTF.
    For exponential data with censoring, the MLE of the rate is $\hat\lambda=\frac{\text{number of failures}}{\text{total time on test}}=\frac{8}{4000}=0.002$ per hour. MTTF $=\frac{1}{\hat\lambda}=\frac{1}{0.002}=500$ hours. Only the failure count and aggregate exposure are needed — a standard reliability life-test result.
  43. Reliability & systems
    Compare **series** and **parallel** systems of identical components on reliability and on system MTTF (assume constant-hazard parts).
    **Series ($n$ identical, reliability $p$):** $R_S=p^{n}\le p$; reliability **falls** as parts are added. For exponential parts the MTTF $=\frac{1}{n\lambda}$ — adding parts shortens system life. **Parallel ($n$ identical):** $R_P=1-(1-p)^{n}\ge p$; reliability **rises** toward $1$. For exponential parts the MTTF $=\frac{1}{\lambda}\sum_{j=1}^{n}\frac{1}{j}$ — adding redundancy lengthens system life. Series trades reliability for nothing; parallel buys reliability at the cost of redundant hardware.