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Exam P — Covariance, Sums & CLT Practice Flashcards

Thirty-three original SOA/CAS Exam P-style multiple-choice problems spanning covariance, correlation, variance of linear combinations, independence versus uncorrelatedness, distributions of sums of independent random variables, and the Central Limit Theorem.

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Common Traps Conditional Probability & Bayes Conditional Probability & Bayes Practice Continuous Distributions Continuous Distributions Practice Covariance, Sums & CLT

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Browse all 33 problems as a list
  1. Correlation
    For two random variables, $E[X]=2$, $E[Y]=5$, $E[XY]=12$, $\operatorname{Var}(X)=3$, and $\operatorname{Var}(Y)=8$. Calculate the correlation coefficient $\rho_{X,Y}$. (A) $0.083$ (B) $0.250$ (C) $0.408$ (D) $0.500$ (E) $0.707$
    **Answer: (C).** First the covariance, using the shortcut $\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=12-(2)(5)=2$. Then $\rho_{X,Y}=\dfrac{\operatorname{Cov}(X,Y)}{\sigma_X\,\sigma_Y}=\dfrac{2}{\sqrt{3}\,\sqrt{8}}=\dfrac{2}{\sqrt{24}}=\dfrac{2}{4.899}\approx 0.408$. Distractor (A) divides by $\operatorname{Var}(X)\operatorname{Var}(Y)=24$ instead of $\sigma_X\sigma_Y$; (B) divides covariance by $\operatorname{Var}(Y)=8$.
  2. Linear combinations
    Random variables $X$ and $Y$ satisfy $\operatorname{Var}(X)=4$, $\operatorname{Var}(Y)=9$, and $\operatorname{Cov}(X,Y)=-2$. Calculate $\operatorname{Var}(3X-2Y)$. (A) $48$ (B) $72$ (C) $84$ (D) $96$ (E) $120$
    **Answer: (D).** Use $\operatorname{Var}(aX+bY)=a^{2}\operatorname{Var}(X)+b^{2}\operatorname{Var}(Y)+2ab\operatorname{Cov}(X,Y)$ with $a=3$, $b=-2$. $\operatorname{Var}(3X-2Y)=9(4)+4(9)+2(3)(-2)(-2)=36+36+24=96$. Distractor (B) drops the cross term ($36+36$); (A) mishandles the sign as $36+36-24$; (C) uses $2ab=-12$ but $\operatorname{Cov}=-2$ with one sign error.
  3. Linear combinations
    An insurer's two lines of business produce annual profits $X$ and $Y$ (in \$ thousands) with $\sigma_X=20$, $\sigma_Y=30$, and correlation $\rho=-0.5$. Calculate the standard deviation of the combined profit $X+Y$. (A) $\$10.0$ (B) $\$26.5$ (C) $\$36.1$ (D) $\$43.6$ (E) $\$50.0$
    **Answer: (B).** Covariance: $\operatorname{Cov}(X,Y)=\rho\sigma_X\sigma_Y=(-0.5)(20)(30)=-300$. $\operatorname{Var}(X+Y)=20^{2}+30^{2}+2(-300)=400+900-600=700$. $\operatorname{SD}(X+Y)=\sqrt{700}\approx 26.46$, i.e. about $\$26.5$ thousand. Distractor (E) ignores correlation and adds SDs ($20+30$); (C) takes $\sqrt{1300}$, dropping the cross term; (D) adds the covariance with the wrong sign, $400+900+2(300)=1900$, giving $\sqrt{1900}\approx 43.6$.
  4. Correlation
    Let $X$ have $\operatorname{Var}(X)=25$ and define $Y=4-2X$. Calculate $\operatorname{Cov}(X,Y)$ and the correlation $\rho_{X,Y}$. (A) $\operatorname{Cov}=-50,\ \rho=-1$ (B) $\operatorname{Cov}=-50,\ \rho=-0.5$ (C) $\operatorname{Cov}=50,\ \rho=1$ (D) $\operatorname{Cov}=-100,\ \rho=-1$ (E) $\operatorname{Cov}=100,\ \rho=1$
    **Answer: (A).** Additive constants drop out of covariance: $\operatorname{Cov}(X,Y)=\operatorname{Cov}(X,\,4-2X)=-2\operatorname{Cov}(X,X)=-2\operatorname{Var}(X)=-2(25)=-50$. Because $Y$ is an exact linear function of $X$ with negative slope, $\rho_{X,Y}=-1$ exactly. Check: $\sigma_Y=|-2|\sigma_X=2(5)=10$, so $\rho=\dfrac{-50}{(5)(10)}=-1$. Distractor (D) squares the slope ($-4\cdot 25$); (B) miscomputes $\rho$ as half the slope.
  5. Covariance
    A joint probability table gives $\Pr(X=0,Y=0)=0.25$, $\Pr(X=1,Y=0)=0.25$, $\Pr(X=0,Y=1)=0.15$, $\Pr(X=1,Y=1)=0.35$. Calculate $\operatorname{Cov}(X,Y)$. (A) $0.05$ (B) $0.10$ (C) $0.15$ (D) $0.20$ (E) $0.30$
    **Answer: (A).** Marginals (Bernoulli indicators): $E[X]=\Pr(X=1)=0.25+0.35=0.60$; $E[Y]=\Pr(Y=1)=0.15+0.35=0.50$. $E[XY]=(1)(1)\Pr(X=1,Y=1)=0.35$. $\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=0.35-(0.60)(0.50)=0.35-0.30=0.05$. Distractor (B) uses the wrong marginal $E[Y]=0.25$; (E) forgets to subtract $E[X]E[Y]$.
  6. Sums of independent
    Independent random variables satisfy $X\sim N(8,9)$ and $Y\sim N(3,16)$, where the second parameter is the variance. Calculate $\Pr(X>Y+10)$. (A) $0.16$ (B) $0.20$ (C) $0.31$ (D) $0.50$ (E) $0.84$
    **Answer: (A).** Let $D=X-Y$. Then $D\sim N(8-3,\ 9+16)=N(5,25)$ — variances **add** for a difference of independents. $\Pr(X>Y+10)=\Pr(D>10)=\Pr\!\left(Z>\dfrac{10-5}{5}\right)=\Pr(Z>1)=1-\Phi(1)\approx 1-0.8413=0.1587\approx 0.16$. Distractor (E) reports $\Phi(1)$ instead of the upper tail; using variance $16-9=7$ or $\sqrt 9+\sqrt{16}=7$ gives the other wrong $z$-values.
  7. CLT
    Annual aggregate losses on a single policy have mean $400$ and standard deviation $250$. An insurer holds $n=625$ independent such policies. Using the CLT, approximate the $95$th percentile of total losses $S$. (Use $z_{0.95}=1.645$.) (A) $\$250{,}000$ (B) $\$256{,}300$ (C) $\$260{,}281$ (D) $\$260{,}411$ (E) $\$410{,}281$
    **Answer: (C).** Total $S\approx N(n\mu,\ n\sigma^{2})$. Mean $=625(400)=250{,}000$. Variance $=625(250)^{2}=625(62{,}500)=39{,}062{,}500$, so $\operatorname{SD}(S)=\sqrt{39{,}062{,}500}=6{,}250$. $95$th percentile $=250{,}000+1.645(6{,}250)=250{,}000+10{,}281=260{,}281$. Distractor (A) omits the percentile shift; (E) wrongly takes $\operatorname{SD}=250\sqrt{625}\cdot 25$ scale; (D) uses $z=1.96$ then mis-rounds.
  8. Sums of independent
    The number of claims $N_1$ and $N_2$ from two independent books are Poisson with means $1$ and $4$, respectively. Calculate $\Pr(N_1+N_2=4)$. (A) $0.073$ (B) $0.156$ (C) $0.175$ (D) $0.195$ (E) $0.224$
    **Answer: (C).** Independent Poissons add: $N_1+N_2\sim\text{Poisson}(1+4)=\text{Poisson}(5)$. $\Pr(N_1+N_2=4)=\dfrac{e^{-5}\,5^{4}}{4!}=\dfrac{e^{-5}(625)}{24}=\dfrac{0.006738(625)}{24}=\dfrac{4.2113}{24}\approx 0.1755$. Distractor (A) evaluates $\text{Poisson}(5)$ at $k=7$; (D) uses $\text{Poisson}(4)$ at $k=4$, forgetting to add the second rate.