Exam FAM — Parametric & Empirical Estimation Practice Flashcards

**Answer: (B).** The empirical variance divides by $n$, not $n-1$: $\widehat{\operatorname{Var}}(X)=\dfrac{1}{n}\sum x_i^2 - \bar x^2$. Mean: $\bar x = \dfrac{3+5+5+8+9}{5}=\dfrac{30}{5}=6$. $\sum x_i^2 = 9+25+25+64+81 = 204$, so $\dfrac{1}{n}\sum x_i^2 = \dfrac{204}{5}=40.8$. $\widehat{\operatorname{Var}}(X)=40.8 - 6^2 = 40.8 - 36 = 4.80$. The sum of squared deviations is $\sum(x_i-\bar x)^2 = 9+1+1+4+9 = 24$. Dividing by $n-1=4$ instead of $n=5$ gives the unbiased sample variance $\frac{24}{4}=6.00$ — distractor (D), the classic $n$-vs-$n-1$ error.

**Answer: (A).** The empirical survival function is $S_n(x)=\dfrac{\#\{x_i > x\}}{n}$ — strictly greater than $x$. Only the single value $15$ exceeds $11$, so $S_7(11)=\dfrac{1}{7}$. Counting values $\ge 11$ (the three $11$'s plus the $15$, i.e. $4$ of them) gives $\dfrac{4}{7}$ — distractor (D), the error of using $\ge$ instead of $>$. Note $F_7(11)=\dfrac{6}{7}$, the complement, is distractor (E).

**Answer: (B).** Mean: $\bar x = \dfrac{120}{6}=20$. $\dfrac{1}{n}\sum x_i^2 = \dfrac{3{,}000}{6}=500$. Empirical variance $= 500 - 20^2 = 500 - 400 = 100$, so the standard deviation is $\sqrt{100}=10$. Reporting the variance $100$ itself is distractor (E); forgetting to square the mean and taking $\sqrt{500}\approx 22.36$ is distractor (D).

**Answer: (B).** Track the risk set $r_i$ (lives just before each death) and deaths $s_i$: $t=3$: $r=10$, $s=1$ → $1-\frac{1}{10}=0.9$. $t=6$: $r=9$, $s=2$ → $1-\frac{2}{9}=\frac{7}{9}\approx 0.77778$. The censored life at $t=8$ leaves the risk set without a drop. Just before $t=10$ we have $10-1-2-1(\text{censored})=6$ lives. $t=10$: $r=6$, $s=1$ → $1-\frac{1}{6}=\frac{5}{6}\approx 0.83333$. $\hat S(10)=0.9\times 0.77778\times 0.83333\approx 0.58333$. Ignoring the censoring (so $r=7$ at $t=10$) gives the factor $\frac{6}{7}$ and $\hat S(10)\approx 0.6000$ — distractor (C).

**Answer: (C).** Nelson–Aalen sums the hazard increments $\dfrac{s_i}{r_i}$ over death times: $\hat H(10)=\dfrac{1}{10}+\dfrac{2}{9}+\dfrac{1}{6}=0.10000+0.22222+0.16667=0.48889$. The implied survival is $\hat S(10)=e^{-0.48889}\approx 0.61331$ (distractor E reports that survival, not the hazard). Using $\frac{1}{7}$ at the last step (ignoring the censoring's effect on the denominator) gives $0.10+0.2222+0.1429=0.4651$, near distractor (B).

**Answer: (C).** The Nelson–Aalen survival estimate is $\hat S(t)=e^{-\hat H(t)}$. $\hat S(10)=e^{-0.48889}$. Since $e^{-0.48889}\approx 0.61331$, $\hat S(10)\approx 0.6133$. This exceeds the Kaplan–Meier $\hat S(10)\approx 0.5833$ (distractor B), as Nelson–Aalen always does because $e^{-x}\ge 1-x$. Computing $1-\hat H = 1-0.48889 = 0.5111$ instead of $e^{-\hat H}$ gives distractor (A).

**Answer: (A).** With no censoring, the risk set just decreases by the deaths: $t=2$: $r=8$, $s=1$ → $1-\frac{1}{8}=0.875$. $t=4$: $r=7$, $s=2$ → $1-\frac{2}{7}=\frac{5}{7}\approx 0.714286$. $t=7$: just before, $8-1-2=5$ lives remain, so $r=5$, $s=1$ → $1-\frac{1}{5}=0.8$. $\hat S(7)=0.875\times 0.714286\times 0.8 = 0.625\times 0.8 = 0.5$. Stopping after the $t=4$ factor ($0.875\times 0.714286\approx 0.625$) gives distractor (D), the off-by-one error of omitting the last death; the simple empirical survivor $\frac{4}{8}=0.5$ happens to coincide here.

**Answer: (A).** Track the risk set just before each death (censored lives leave without a drop): $t=2$: $r=12$, $s=1$ → $1-\frac{1}{12}=\frac{11}{12}\approx 0.916667$. $t=5$: $r=11$, $s=1$ → $1-\frac{1}{11}=\frac{10}{11}\approx 0.909091$. The two censored at $t=7$ leave without an event. Just before $t=9$: $12-1-1-2=8$ lives. $t=9$: $r=8$, $s=3$ → $1-\frac{3}{8}=0.625$. $\hat S(9)=\frac{11}{12}\times\frac{10}{11}\times 0.625 = \frac{10}{12}\times 0.625 = 0.833333\times 0.625 = 0.5208$. Ignoring the two censored lives (so $r=10$, factor $1-\frac{3}{10}=0.7$ at $t=9$) gives $0.833333\times 0.7\approx 0.5833$, near distractor (B); the partial product through $t=5$ alone is $0.8333$ — distractor (E).