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