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Exam P — Expectation, Variance & MGFs Practice Flashcards
Thirty-three original SOA/CAS Exam P-style multiple-choice problems covering expectation and LOTUS, variance, linear transforms, raw and central moments, skewness and kurtosis, the coefficient of variation, moment and probability generating functions, and the Markov and Chebyshev inequalities, each with a full worked solution.
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- ExpectationAn insurer's daily claim count $X$ has the probability mass function $p(0)=0.30$, $p(1)=0.35$, $p(2)=0.20$, $p(3)=0.10$, $p(4)=0.05$. Calculate $E[X]$. (A) $1.10$ (B) $1.25$ (C) $1.50$ (D) $1.75$ (E) $2.00$**Answer: (B).** Apply the definition $E[X]=\sum_x x\,p(x)$: $E[X]=0(0.30)+1(0.35)+2(0.20)+3(0.10)+4(0.05).$ $E[X]=0+0.35+0.40+0.30+0.20=1.25$. The answer is $1.25$, choice **(B)**.
- ExpectationA loss random variable $X$ has density $f(x)=\frac{3}{x^{4}}$ for $x>1$ and $0$ otherwise. Calculate $E[X]$. (A) $0.75$ (B) $1.00$ (C) $1.20$ (D) $1.50$ (E) $3.00$**Answer: (D).** Use $E[X]=\int_{1}^{\infty} x\,f(x)\,dx$: $E[X]=\int_{1}^{\infty} x\cdot\frac{3}{x^{4}}\,dx=\int_{1}^{\infty} 3x^{-3}\,dx.$ Integrate: $\int 3x^{-3}\,dx=-\tfrac{3}{2}x^{-2}$, so $E[X]=\left[-\tfrac{3}{2}x^{-2}\right]_{1}^{\infty}=0-\left(-\tfrac{3}{2}\right)=\frac{3}{2}=1.50.$ A candidate who forgets the extra $x$ from $E[X]$ (integrating $f$ alone) gets $1$, choice (B); using exponent $-3$ instead of $-4$ in $f$ gives $0.75$. The correct mean is $1.50$, choice **(D)**.
- ExpectationThe proportion $X$ of a risk pool that files a claim in a year has density $f(x)=6x(1-x)$ for $0<x<1$. An administrative cost equals $g(X)=100X^{2}$ thousand dollars. Calculate $E[g(X)]$ in thousands. (A) $25$ (B) $30$ (C) $50$ (D) $60$ (E) $100$**Answer: (B).** By LOTUS, $E[100X^{2}]=100\,E[X^{2}]$ with $E[X^{2}]=\int_{0}^{1} x^{2}\cdot 6x(1-x)\,dx=\int_{0}^{1}\left(6x^{3}-6x^{4}\right)dx.$ $E[X^{2}]=6\cdot\tfrac{1}{4}-6\cdot\tfrac{1}{5}=\tfrac{3}{2}-\tfrac{6}{5}=\tfrac{15-12}{10}=0.30.$ Thus $E[g(X)]=100(0.30)=30$. A candidate who computes $E[X]=0.5$ and then squares ($100\cdot0.25=25$) commits the $g(E[X])\ne E[g(X)]$ error, choice (A). The correct value is $30$, choice **(B)**.
- ExpectationAn auto policy has annual loss $X$ uniformly distributed on $[0,2000]$. The insurer applies an ordinary deductible of $500$, so the per-loss payment is $Y=\max(X-500,0)$. Calculate $E[Y]$. (A) $562.50$ (B) $750.00$ (C) $843.75$ (D) $1000.00$ (E) $1125.00$**Answer: (A).** With $X\sim\text{Uniform}(0,2000)$, $f(x)=\frac{1}{2000}$. Only losses above $500$ contribute: $E[Y]=\int_{500}^{2000}(x-500)\cdot\frac{1}{2000}\,dx.$ Let $u=x-500$ ranging $0$ to $1500$: $E[Y]=\frac{1}{2000}\int_{0}^{1500} u\,du=\frac{1}{2000}\cdot\frac{1500^{2}}{2}=\frac{2{,}250{,}000}{4000}=562.50.$ Forgetting to subtract the deductible inside the integral (computing $E[X\mid X>500]$-style or $E[X]-500$ over all losses) yields $750$, choice (B). The per-loss expected payment is $562.50$, choice **(A)**.
- VarianceA random variable $X$ has $E[X]=3$ and $E[X^{2}]=25$. Calculate $\operatorname{Var}(X)$. (A) $9$ (B) $16$ (C) $22$ (D) $25$ (E) $34$**Answer: (B).** Use the shortcut formula $\operatorname{Var}(X)=E[X^{2}]-(E[X])^{2}$: $\operatorname{Var}(X)=25-3^{2}=25-9=16.$ Subtracting in the wrong order or forgetting to square the mean ($25-3=22$) gives choice (C). The variance is $16$, choice **(B)**.
- VarianceA claim severity $X$ has density $f(x)=\frac{1}{2}e^{-x/2}$ for $x>0$ (exponential with mean $2$). Calculate $\operatorname{Var}(X)$. (A) $0.5$ (B) $1$ (C) $2$ (D) $4$ (E) $8$**Answer: (D).** For an exponential with mean $\theta$, $E[X]=\theta$ and $E[X^{2}]=2\theta^{2}$. Here $\theta=2$: $E[X]=2,\qquad E[X^{2}]=2(2)^{2}=8.$ $\operatorname{Var}(X)=E[X^{2}]-(E[X])^{2}=8-4=4.$ Equivalently $\operatorname{Var}=\theta^{2}=4$. A candidate who recalls $E[X^{2}]=\theta^{2}$ instead of $2\theta^{2}$ gets $\operatorname{Var}=0$ confusion or lands on $2$, choice (C). The variance is $4$, choice **(D)**.
- VarianceA discrete random variable $X$ takes value $-2$ with probability $0.25$, value $1$ with probability $0.50$, and value $3$ with probability $0.25$. Calculate $\operatorname{Var}(X)$. (A) $2.50$ (B) $3.19$ (C) $3.75$ (D) $4.00$ (E) $4.69$**Answer: (B).** First the mean: $E[X]=(-2)(0.25)+1(0.50)+3(0.25)=-0.5+0.5+0.75=0.75.$ Then the second raw moment: $E[X^{2}]=4(0.25)+1(0.50)+9(0.25)=1+0.5+2.25=3.75.$ $\operatorname{Var}(X)=3.75-(0.75)^{2}=3.75-0.5625=3.1875\approx 3.19.$ Forgetting to subtract $(E[X])^2$ gives $3.75$, choice (C). The variance is about $3.19$, choice **(B)**.
- VarianceA random variable $X$ has density $f(x)=2x$ for $0<x<1$. Calculate $\operatorname{Var}(X)$. (A) $0.0556$ (B) $0.0625$ (C) $0.1389$ (D) $0.2222$ (E) $0.5000$**Answer: (A).** Compute the first two moments: $E[X]=\int_{0}^{1} x\cdot 2x\,dx=\int_{0}^{1}2x^{2}\,dx=\frac{2}{3},$ $E[X^{2}]=\int_{0}^{1} x^{2}\cdot 2x\,dx=\int_{0}^{1}2x^{3}\,dx=\frac{1}{2}.$ $\operatorname{Var}(X)=\frac{1}{2}-\left(\frac{2}{3}\right)^{2}=\frac{1}{2}-\frac{4}{9}=\frac{9-8}{18}=\frac{1}{18}\approx 0.0556.$ The variance is $\frac{1}{18}\approx 0.0556$, choice **(A)**.