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Exam P — Covariance, Sums & CLT Flashcards

Covariance and correlation, variance of sums and linear combinations, independence versus uncorrelatedness, distributions of sums of independent random variables, the Central Limit Theorem, and the sampling distribution of the mean for SOA/CAS Exam P.

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  1. Covariance
    Define the covariance $\mathrm{Cov}(X,Y)$ and give the computational ('shortcut') form.
    Covariance measures the linear co-movement of two random variables: $\mathrm{Cov}(X,Y)=E[(X-\mu_X)(Y-\mu_Y)]$. The shortcut form used in almost every problem is $\mathrm{Cov}(X,Y)=E[XY]-E[X]\,E[Y]$.
  2. Covariance
    What is $\mathrm{Cov}(X,X)$, and what does $\mathrm{Cov}(X,Y)$ equal when $X$ and $Y$ are independent?
    $\mathrm{Cov}(X,X)=\mathrm{Var}(X)$, since covariance of a variable with itself is its variance. If $X$ and $Y$ are independent then $E[XY]=E[X]E[Y]$, so $\mathrm{Cov}(X,Y)=0$.
  3. Correlation
    Define the correlation coefficient $\rho_{X,Y}$ and state its bounds.
    $\rho_{X,Y}=\dfrac{\mathrm{Cov}(X,Y)}{\sigma_X\,\sigma_Y}$, the covariance normalized by the two standard deviations. It always satisfies $-1\leq\rho_{X,Y}\leq 1$. The value $\rho=\pm 1$ occurs exactly when $Y$ is a perfect linear function of $X$.
  4. Covariance
    State the bilinearity rule: $\mathrm{Cov}(aX+b,\;cY+d)=?$
    Covariance is bilinear and ignores additive constants: $\mathrm{Cov}(aX+b,\;cY+d)=ac\,\mathrm{Cov}(X,Y)$. The shifts $b$ and $d$ drop out entirely; only the multiplicative factors $a$ and $c$ survive.
  5. Covariance
    Expand $\mathrm{Cov}(X+Y,\;Z)$ and, more generally, the covariance of two sums.
    Covariance distributes over sums: $\mathrm{Cov}(X+Y,\;Z)=\mathrm{Cov}(X,Z)+\mathrm{Cov}(Y,Z)$. In general $\mathrm{Cov}\!\left(\sum_i a_iX_i,\;\sum_j b_jY_j\right)=\sum_i\sum_j a_i b_j\,\mathrm{Cov}(X_i,Y_j)$.
  6. Linear combinations
    Give the formula for $\mathrm{Var}(X+Y)$ when $X$ and $Y$ may be dependent.
    $\mathrm{Var}(X+Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)+2\mathrm{Cov}(X,Y)$. The cross term $2\mathrm{Cov}(X,Y)$ vanishes only when $X$ and $Y$ are uncorrelated.
  7. Linear combinations
    Give the formula for $\mathrm{Var}(X-Y)$.
    $\mathrm{Var}(X-Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)-2\mathrm{Cov}(X,Y)$. Note the variances still **add**; only the sign on the covariance term flips. A common trap is to subtract the variances.
  8. Linear combinations
    Write the general formula for $\mathrm{Var}\!\left(\sum_{i=1}^{n}a_iX_i\right)$.
    $\mathrm{Var}\!\left(\sum_{i=1}^{n}a_iX_i\right)=\sum_{i=1}^{n}a_i^{2}\mathrm{Var}(X_i)+2\sum_{i<j}a_i a_j\,\mathrm{Cov}(X_i,X_j)$. If the $X_i$ are pairwise uncorrelated this reduces to $\sum_i a_i^{2}\mathrm{Var}(X_i)$.
  9. Linear combinations
    For independent random variables $X_1,\dots,X_n$, what is $E\!\left[\sum a_iX_i\right]$ and $\mathrm{Var}\!\left(\sum a_iX_i\right)$?
    Expectation is always linear (independence not required): $E\!\left[\sum a_iX_i\right]=\sum a_i E[X_i]$. With independence all covariances are $0$, so $\mathrm{Var}\!\left(\sum a_iX_i\right)=\sum a_i^{2}\mathrm{Var}(X_i)$.
  10. Independence
    Does $\mathrm{Cov}(X,Y)=0$ imply $X$ and $Y$ are independent? Explain.
    No. Independence $\Rightarrow$ uncorrelated, but **not** the converse. Uncorrelated means no *linear* relationship; variables can be uncorrelated yet strongly dependent nonlinearly. The one notable exception is the bivariate normal, where $\rho=0$ does imply independence.
  11. Linear combinations
    Given $\mathrm{Var}(X)=9$, $\mathrm{Var}(Y)=16$, and $\rho_{X,Y}=0.5$, find $\mathrm{Var}(X+Y)$.
    First recover the covariance: $\mathrm{Cov}(X,Y)=\rho\,\sigma_X\sigma_Y=0.5\cdot 3\cdot 4=6$. Then $\mathrm{Var}(X+Y)=9+16+2(6)=\boxed{37}$.
  12. Linear combinations
    $X$ and $Y$ have $\sigma_X=2$, $\sigma_Y=5$, $\mathrm{Cov}(X,Y)=-3$. Find $\rho_{X,Y}$ and $\mathrm{Var}(2X-Y)$.
    Correlation: $\rho=\dfrac{-3}{2\cdot 5}=-0.3$. For the linear combination, $\mathrm{Var}(2X-Y)=2^{2}\mathrm{Var}(X)+(-1)^{2}\mathrm{Var}(Y)+2(2)(-1)\mathrm{Cov}(X,Y)$. $=4(4)+1(25)-4(-3)=16+25+12=\boxed{53}$.
  13. Covariance
    A joint pmf gives $E[X]=1.5$, $E[Y]=2$, and $E[XY]=3.6$. Compute $\mathrm{Cov}(X,Y)$.
    Use the shortcut: $\mathrm{Cov}(X,Y)=E[XY]-E[X]E[Y]=3.6-(1.5)(2)=3.6-3=\boxed{0.6}$.
  14. Linear combinations
    Two assets have returns with $\sigma_X=0.10$, $\sigma_Y=0.20$, $\rho=0.25$. A portfolio is $W=0.6X+0.4Y$. Find $\mathrm{SD}(W)$.
    Covariance: $\mathrm{Cov}(X,Y)=0.25(0.10)(0.20)=0.005$. $\mathrm{Var}(W)=0.6^{2}(0.10)^{2}+0.4^{2}(0.20)^{2}+2(0.6)(0.4)(0.005)$ $=0.36(0.01)+0.16(0.04)+0.0024=0.0036+0.0064+0.0024=0.0124$. $\mathrm{SD}(W)=\sqrt{0.0124}\approx\boxed{0.1114}$.
  15. Correlation
    What is the maximum possible value of $|\mathrm{Cov}(X,Y)|$ given $\mathrm{Var}(X)=4$ and $\mathrm{Var}(Y)=9$?
    The Cauchy–Schwarz bound gives $|\mathrm{Cov}(X,Y)|\leq\sigma_X\sigma_Y=\sqrt{4}\sqrt{9}=6$, attained only when $|\rho|=1$. So any reported covariance with $|\mathrm{Cov}|>6$ (e.g. $7$) is impossible — a quick sanity check on computed answers.
  16. Sums of independent
    Let $S=X_1+X_2+\cdots+X_n$ where the $X_i$ are i.i.d. with mean $\mu$ and variance $\sigma^{2}$. Give $E[S]$ and $\mathrm{Var}(S)$.
    By linearity, $E[S]=n\mu$. By independence the variances add: $\mathrm{Var}(S)=n\sigma^{2}$, so $\mathrm{SD}(S)=\sigma\sqrt{n}$.
  17. CLT
    If $\bar{X}=\frac{1}{n}\sum_{i=1}^{n}X_i$ for i.i.d. $X_i$ with mean $\mu$ and variance $\sigma^{2}$, give $E[\bar{X}]$, $\mathrm{Var}(\bar{X})$, and the standard error.
    $E[\bar{X}]=\mu$ (the sample mean is unbiased). $\mathrm{Var}(\bar{X})=\dfrac{\sigma^{2}}{n}$, so the standard error is $\mathrm{SD}(\bar{X})=\dfrac{\sigma}{\sqrt{n}}$.
  18. Sums of independent
    $X\sim N(3,4)$ and $Y\sim N(5,9)$ are independent. What is the distribution of $X+Y$?
    Sums of independent normals are normal, with means and variances adding. $X+Y\sim N(3+5,\;4+9)=N(8,\,13)$. (Here the second parameter is the variance.)
  19. Sums of independent
    $X\sim N(10,4)$ and $Y\sim N(7,9)$ are independent. Find the distribution of $X-Y$ and $\Pr(X>Y)$.
    $X-Y\sim N(10-7,\;4+9)=N(3,13)$; note the **variances add** even for a difference. $\Pr(X>Y)=\Pr(X-Y>0)=\Pr\!\left(Z>\dfrac{0-3}{\sqrt{13}}\right)=\Pr(Z>-0.832)=\Phi(0.832)\approx\boxed{0.797}$.
  20. Sums of independent
    If $X_1\sim\text{Poisson}(2)$ and $X_2\sim\text{Poisson}(3)$ are independent, what is the distribution of $X_1+X_2$?
    Independent Poissons add: the sum is $\text{Poisson}$ with the rates summed. $X_1+X_2\sim\text{Poisson}(2+3)=\text{Poisson}(5)$.
  21. Sums of independent
    What is the distribution of the sum of $n$ independent $\text{Exponential}(\lambda)$ random variables?
    The sum is $\text{Gamma}(\alpha=n,\;\text{rate}=\lambda)$, also called an Erlang distribution. Its mean is $n/\lambda$ and its variance is $n/\lambda^{2}$.
  22. CLT
    State the Central Limit Theorem for the sum $S_n=\sum_{i=1}^{n}X_i$ of i.i.d. variables with mean $\mu$ and variance $\sigma^{2}$.
    For large $n$, $S_n$ is approximately normal: $S_n\approx N(n\mu,\;n\sigma^{2})$. Equivalently the standardized sum $\dfrac{S_n-n\mu}{\sigma\sqrt{n}}\to N(0,1)$ in distribution, regardless of the original distribution's shape.
  23. CLT
    State the CLT for the sample mean $\bar{X}$ of $n$ i.i.d. variables with mean $\mu$ and variance $\sigma^{2}$.
    For large $n$, $\bar{X}\approx N\!\left(\mu,\;\dfrac{\sigma^{2}}{n}\right)$. The standardized form is $\dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}}\approx N(0,1)$ — note the standard error uses $\sigma/\sqrt{n}$, not $\sigma$.
  24. CLT
    A policy's annual losses have mean $\mu=500$ and SD $\sigma=300$. For $n=100$ independent policies, approximate $\Pr(\text{total losses}>52{,}000)$.
    Total $S\approx N(n\mu,\,n\sigma^{2})=N(50{,}000,\;100\cdot 90{,}000)$, so $\mathrm{SD}(S)=300\sqrt{100}=3000$. $\Pr(S>52{,}000)=\Pr\!\left(Z>\dfrac{52{,}000-50{,}000}{3000}\right)=\Pr(Z>0.667)=1-\Phi(0.667)\approx\boxed{0.2525}$.
  25. CLT
    Claim sizes are i.i.d. with mean $1{,}000$ and SD $400$. For a sample of $64$ claims, approximate $\Pr(\bar{X}<950)$ using the CLT.
    Standard error $=\dfrac{\sigma}{\sqrt{n}}=\dfrac{400}{\sqrt{64}}=\dfrac{400}{8}=50$. $\Pr(\bar{X}<950)=\Pr\!\left(Z<\dfrac{950-1000}{50}\right)=\Pr(Z<-1)=1-\Phi(1)\approx\boxed{0.1587}$.
  26. CLT
    Explain the continuity correction and when it is used in CLT approximations.
    When approximating a **discrete** integer-valued sum by a continuous normal, adjust the boundary by $\pm 0.5$ to span the integer's full interval. For $\Pr(S\leq k)$ use $\Pr(S\leq k+0.5)$; for $\Pr(S\geq k)$ use $\Pr(S\geq k-0.5)$; for $\Pr(S=k)$ use $\Pr(k-0.5\leq S\leq k+0.5)$.
  27. CLT
    A fair coin is tossed $100$ times. Using the normal approximation with continuity correction, approximate $\Pr(\text{heads}\leq 45)$.
    $X\sim\text{Binomial}(100,0.5)$ has $\mu=50$, $\sigma=\sqrt{100(0.5)(0.5)}=5$. With continuity correction, $\Pr(X\leq 45)\approx\Pr\!\left(Z\leq\dfrac{45.5-50}{5}\right)=\Pr(Z\leq-0.9)=1-\Phi(0.9)\approx\boxed{0.1841}$.
  28. CLT
    The number of claims per month is $\text{Poisson}(36)$. Approximate $\Pr(N\geq 30)$ using a normal approximation with continuity correction.
    For $\text{Poisson}(36)$, $\mu=36$ and $\sigma=\sqrt{36}=6$. $\Pr(N\geq 30)\approx\Pr\!\left(Z\geq\dfrac{29.5-36}{6}\right)=\Pr(Z\geq-1.083)=\Phi(1.083)\approx\boxed{0.8606}$.
  29. Linear combinations
    $X$ and $Y$ have $\mathrm{Var}(X)=5$, $\mathrm{Var}(Y)=7$, and $\mathrm{Var}(X+Y)=20$. Find $\mathrm{Cov}(X,Y)$ and $\rho_{X,Y}$.
    From $\mathrm{Var}(X+Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)+2\mathrm{Cov}(X,Y)$: $20=5+7+2\mathrm{Cov}(X,Y)\Rightarrow\mathrm{Cov}(X,Y)=4$. $\rho=\dfrac{4}{\sqrt{5}\sqrt{7}}=\dfrac{4}{\sqrt{35}}\approx\boxed{0.676}$.
  30. Covariance
    Find $\mathrm{Cov}(X+Y,\;X-Y)$ in terms of the variances of $X$ and $Y$.
    Expand using bilinearity: $\mathrm{Cov}(X+Y,X-Y)=\mathrm{Cov}(X,X)-\mathrm{Cov}(X,Y)+\mathrm{Cov}(Y,X)-\mathrm{Cov}(Y,Y)$. The two cross terms cancel, leaving $\mathrm{Var}(X)-\mathrm{Var}(Y)$. (So $X+Y$ and $X-Y$ are uncorrelated iff $\mathrm{Var}(X)=\mathrm{Var}(Y)$.)
  31. Covariance
    A joint pmf places probability $0.3$ at $(0,0)$, $0.2$ at $(1,0)$, $0.1$ at $(0,1)$, and $0.4$ at $(1,1)$. Find $\mathrm{Cov}(X,Y)$.
    Marginals: $E[X]=\Pr(X{=}1)=0.2+0.4=0.6$; $E[Y]=\Pr(Y{=}1)=0.1+0.4=0.5$. $E[XY]=1\cdot 1\cdot\Pr(1,1)=0.4$. $\mathrm{Cov}(X,Y)=0.4-(0.6)(0.5)=0.4-0.30=\boxed{0.10}$.
  32. CLT
    $X_1,X_2,X_3$ are i.i.d. with variance $\sigma^{2}=4$. Find $\mathrm{Var}(X_1+X_2+X_3)$ and $\mathrm{Var}(\bar{X})$ for this sample of $3$.
    Sum: by independence, $\mathrm{Var}(X_1+X_2+X_3)=3\sigma^{2}=3(4)=12$. Mean: $\mathrm{Var}(\bar{X})=\dfrac{\sigma^{2}}{n}=\dfrac{4}{3}\approx\boxed{1.333}$.
  33. Linear combinations
    Insurer A has $\mathrm{Var}=100$, insurer B has $\mathrm{Var}=144$, and the correlation of their losses is $\rho=-0.5$. Find $\mathrm{Var}(A+B)$ (the combined book).
    $\mathrm{Cov}(A,B)=\rho\sigma_A\sigma_B=-0.5(10)(12)=-60$. $\mathrm{Var}(A+B)=100+144+2(-60)=244-120=\boxed{124}$. Negative correlation reduces total variance — the diversification benefit.
  34. Correlation
    $X$ has variance $\mathrm{Var}(X)=16$, and $Y=3X-2$. Find $\mathrm{Cov}(X,Y)$ and $\rho_{X,Y}$.
    $\mathrm{Cov}(X,Y)=\mathrm{Cov}(X,\,3X-2)=3\mathrm{Cov}(X,X)=3\mathrm{Var}(X)=48$. Since $Y$ is a positive linear function of $X$, the correlation is exactly $\rho=+1$. (Check: $\sigma_Y=|3|\sigma_X=3\cdot 4=12$, so $\rho=48/(4\cdot 12)=1$.)
  35. CLT
    A random sample of $n=144$ losses has population mean $\mu=800$ and SD $\sigma=120$. Approximate $\Pr(770<\bar{X}<830)$.
    Standard error $=\dfrac{120}{\sqrt{144}}=\dfrac{120}{12}=10$. Lower $z=\dfrac{770-800}{10}=-3$; upper $z=\dfrac{830-800}{10}=3$. $\Pr(-3<Z<3)=\Phi(3)-\Phi(-3)=2\Phi(3)-1\approx 2(0.99865)-1=\boxed{0.9973}$.
  36. Sums of independent
    $X$ and $Y$ are independent $\text{Exponential}$ with means $E[X]=2$ and $E[Y]=2$. What is the distribution of $X+Y$, and find $\Pr(X+Y\leq 4)$.
    Each is $\text{Exponential}(\lambda=0.5)$, so $X+Y\sim\text{Gamma}(\alpha=2,\lambda=0.5)$ (an Erlang-2). For Erlang-2, $\Pr(X+Y\leq t)=1-e^{-\lambda t}(1+\lambda t)$. At $t=4$, $\lambda t=2$: $\Pr=1-e^{-2}(1+2)=1-3e^{-2}\approx 1-0.4060=\boxed{0.5940}$.
  37. Sums of independent
    A compound model has $N$ claims with $E[N]=10$, $\mathrm{Var}(N)=10$ (Poisson), and i.i.d. claim sizes $X$ with $E[X]=200$, $\mathrm{Var}(X)=5000$. Find $E[S]$ and $\mathrm{Var}(S)$ for $S=\sum_{i=1}^{N}X_i$.
    Compound mean: $E[S]=E[N]\,E[X]=10(200)=2000$. Compound variance: $\mathrm{Var}(S)=E[N]\mathrm{Var}(X)+(E[X])^{2}\mathrm{Var}(N)=10(5000)+200^{2}(10)=50{,}000+400{,}000=\boxed{450{,}000}$.
  38. Covariance
    Let $U=X+Z$ and $V=Y+Z$, where $X,Y,Z$ are mutually independent with $\mathrm{Var}(Z)=4$. Find $\mathrm{Cov}(U,V)$.
    By bilinearity, $\mathrm{Cov}(U,V)=\mathrm{Cov}(X+Z,\,Y+Z)=\mathrm{Cov}(X,Y)+\mathrm{Cov}(X,Z)+\mathrm{Cov}(Z,Y)+\mathrm{Cov}(Z,Z)$. All cross terms among independent variables vanish, leaving $\mathrm{Cov}(Z,Z)=\mathrm{Var}(Z)=\boxed{4}$. A shared component induces positive covariance.