{ "deckName": "Exam P — Multivariate Distributions", "examCode": "Exam P", "cards": [ { "front": "For a continuous pair $(X,Y)$, how do you compute the probability that $(X,Y)$ falls in a region $R$ from the joint pdf $f(x,y)$?", "back": "Integrate the joint pdf over the region: $\\Pr((X,Y)\\in R)=\\iint_{R} f(x,y)\\,dx\\,dy$.\nA valid joint pdf satisfies $f(x,y)\\geq 0$ and $\\iint f(x,y)\\,dx\\,dy=1$ over the whole support.", "tag": "Joint & marginal" }, { "front": "The joint pdf is $f(x,y)=c(x+y)$ on $00$. Find the joint pdf, and are $X,Y$ independent?", "back": "Differentiate: $\\dfrac{\\partial}{\\partial x}F=e^{-x}(1-e^{-2y})$, then $\\dfrac{\\partial}{\\partial y}$ gives $f(x,y)=2e^{-x}e^{-2y}$.\nSince $f$ factors as $\\left(e^{-x}\\right)\\left(2e^{-2y}\\right)$ on a rectangular support, $X\\sim\\text{Exp}(1)$ and $Y\\sim\\text{Exp}(2)$ are **independent**.", "tag": "Joint & marginal" }, { "front": "Define the conditional density $f_{Y\\mid X}(y\\mid x)$ in terms of the joint and marginal densities.", "back": "$f_{Y\\mid X}(y\\mid x)=\\dfrac{f(x,y)}{f_X(x)}$, defined where $f_X(x)>0$.\nFor each fixed $x$, this is a genuine density in $y$: it is nonnegative and integrates to $1$ over $y$.", "tag": "Conditional" }, { "front": "For $f(x,y)=8xy$ on $00$.", "back": "Given the sum $X+Y=t$, the conditional distribution of $X$ is uniform on $(0,t)$ (a known property of two i.i.d. exponentials).\nTherefore $E[X\\mid X+Y=t]=\\tfrac{t}{2}$. By symmetry each of $X,Y$ carries half the total on average.", "tag": "Conditional" }, { "front": "State the criterion for two continuous random variables $X$ and $Y$ to be independent.", "back": "$X$ and $Y$ are independent iff $f(x,y)=f_X(x)\\,f_Y(y)$ for all $x,y$.\nEquivalently the joint pdf factors into a function of $x$ alone times a function of $y$ alone **and** the support is a rectangle (a product set).", "tag": "Independence" }, { "front": "The joint pdf is $f(x,y)=6xy^{2}$ on $00$. Integrate over $x+y\\leq 1$:\n$\\int_{0}^{1}\\int_{0}^{1-x} e^{-x}e^{-y}\\,dy\\,dx=\\int_{0}^{1} e^{-x}\\left(1-e^{-(1-x)}\\right)dx$.\n$=\\int_{0}^{1}\\left(e^{-x}-e^{-1}\\right)dx=(1-e^{-1})-e^{-1}=1-2e^{-1}\\approx 0.264$.", "tag": "Region & uniform" }, { "front": "$(X,Y)$ is uniform on the unit square $0X^{2})$.", "back": "For the uniform on the unit square $f(x,y)=1$, so the probability equals the area where $y>x^{2}$.\nArea $=\\int_{0}^{1}\\left(1-x^{2}\\right)dx=1-\\tfrac{1}{3}=\\tfrac{2}{3}$.\nThus $\\Pr(Y>X^{2})=\\tfrac{2}{3}$.", "tag": "Region & uniform" }, { "front": "$(X,Y)$ has joint pdf $f(x,y)=x+y$ on the unit square. Find $\\Pr(XY)$, and the diagonal has probability $0$, so each is $\\tfrac{1}{2}$.\nDirectly: $\\int_{0}^{1}\\int_{0}^{y}(x+y)\\,dx\\,dy=\\int_{0}^{1}\\left(\\tfrac{y^{2}}{2}+y^{2}\\right)dy=\\int_{0}^{1}\\tfrac{3y^{2}}{2}\\,dy=\\tfrac{1}{2}$.", "tag": "Region & uniform" }, { "front": "$(X,Y)$ is uniform over the triangle with vertices $(0,0)$, $(2,0)$, $(0,2)$. What is the joint pdf?", "back": "The region is $\\{x>0,\\,y>0,\\,x+y<2\\}$, a triangle of area $\\tfrac{1}{2}(2)(2)=2$.\nUniform means constant density equal to $1/\\text{area}$: $f(x,y)=\\tfrac{1}{2}$ on the triangle, and $0$ elsewhere.", "tag": "Region & uniform" }, { "front": "$(X,Y)$ is uniform on the triangle $0Y\\}]=2\\int_{0}^{1}\\int_{0}^{x}(x-y)\\,dy\\,dx$.\nInner: $\\int_{0}^{x}(x-y)\\,dy=\\tfrac{x^{2}}{2}$.\nOuter: $2\\int_{0}^{1}\\tfrac{x^{2}}{2}\\,dx=2\\cdot\\tfrac{1}{6}=\\tfrac{1}{3}$.\nThe expected gap between two independent uniforms is $\\tfrac{1}{3}$.", "tag": "Region & uniform" }, { "front": "For $f(x,y)=x+y$ on the unit square, the marginals give $E[X]=E[Y]=\\tfrac{7}{12}$ and $E[XY]=\\tfrac{1}{3}$. Find $\\operatorname{Cov}(X,Y)$.", "back": "$\\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=\\tfrac{1}{3}-\\left(\\tfrac{7}{12}\\right)^{2}=\\tfrac{1}{3}-\\tfrac{49}{144}$.\n$=\\tfrac{48}{144}-\\tfrac{49}{144}=-\\tfrac{1}{144}\\approx -0.00694$.\nThe small negative covariance reflects that $X$ and $Y$ are mildly negatively associated here.", "tag": "Covariance & correlation" }, { "front": "Show that $\\operatorname{Cov}(X,Y)=0$ does not imply independence. Take $X$ uniform on $\\{-1,0,1\\}$ and $Y=X^{2}$.", "back": "Here $E[X]=0$ and $E[XY]=E[X^{3}]=0$ (by symmetry), so $\\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=0$.\nBut $Y$ is completely determined by $X$ ($Y=X^{2}$), so they are clearly **dependent**. Zero covariance only rules out *linear* association.", "tag": "Covariance & correlation" }, { "front": "$(X_1,X_2,X_3)$ is multinomial (trinomial): $n$ trials, each landing in category $i$ with probability $p_i$ (with $p_1+p_2+p_3=1$). Give the marginal of each $X_i$ and $\\operatorname{Cov}(X_i,X_j)$ for $i\\neq j$.", "back": "Each count is binomial: $X_i\\sim\\text{Bin}(n,p_i)$, so $E[X_i]=np_i$ and $\\operatorname{Var}(X_i)=np_i(1-p_i)$.\nDifferent categories compete for the same $n$ trials, so they are **negatively** correlated: $\\operatorname{Cov}(X_i,X_j)=-np_i p_j$ for $i\\neq j$.\nThe negative sign is intuitive — every trial in category $i$ is a trial that cannot land in category $j$.", "tag": "Multinomial" }, { "front": "$n=10$ policyholders are independently classified as low/medium/high risk with probabilities $p_1=0.5$, $p_2=0.3$, $p_3=0.2$. Let $(X_1,X_2,X_3)$ be the counts. Find $\\operatorname{Var}(X_1)$ and $\\operatorname{Cov}(X_1,X_2)$.", "back": "Marginal $X_1\\sim\\text{Bin}(10,0.5)$: $\\operatorname{Var}(X_1)=np_1(1-p_1)=10(0.5)(0.5)=2.5$.\nCross-covariance: $\\operatorname{Cov}(X_1,X_2)=-np_1 p_2=-10(0.5)(0.3)=-1.5$.\nNegative, as expected — more low-risk policies means fewer slots left for medium-risk ones.", "tag": "Multinomial" }, { "front": "List the five parameters of the bivariate normal distribution of $(X,Y)$, and state what kind of distribution any linear combination $aX+bY$ has.", "back": "Parameters: the two means $\\mu_X,\\mu_Y$, the two standard deviations $\\sigma_X,\\sigma_Y$, and the correlation $\\rho$ (with $-1<\\rho<1$).\nKey property: every linear combination $aX+bY$ is itself **(univariate) normal**, with mean $a\\mu_X+b\\mu_Y$ and variance $a^{2}\\sigma_X^{2}+b^{2}\\sigma_Y^{2}+2ab\\,\\rho\\sigma_X\\sigma_Y$.\nBoth marginals are normal: $X\\sim N(\\mu_X,\\sigma_X^{2})$ and $Y\\sim N(\\mu_Y,\\sigma_Y^{2})$.", "tag": "Bivariate normal" }, { "front": "For a bivariate normal $(X,Y)$, give the conditional mean and variance of $Y$ given $X=x$. Then evaluate them when $\\mu_X=10$, $\\mu_Y=20$, $\\sigma_X=2$, $\\sigma_Y=5$, $\\rho=0.6$, at $x=12$.", "back": "The conditional distribution is normal with the regression-line mean\n$E[Y\\mid X=x]=\\mu_Y+\\rho\\dfrac{\\sigma_Y}{\\sigma_X}(x-\\mu_X)$ and constant variance $\\operatorname{Var}(Y\\mid X=x)=\\sigma_Y^{2}(1-\\rho^{2})$.\nHere $E[Y\\mid X=12]=20+0.6\\cdot\\dfrac{5}{2}(12-10)=20+0.6(2.5)(2)=23$.\n$\\operatorname{Var}(Y\\mid X=12)=5^{2}(1-0.6^{2})=25(0.64)=16$, so $\\operatorname{SD}=4$ (note: it does not depend on $x$).", "tag": "Bivariate normal" }, { "front": "For a bivariate normal $(X,Y)$, why does $\\rho=0$ imply $X$ and $Y$ are independent? Is this true for general random variables?", "back": "Set $\\rho=0$ in the joint density: the cross term vanishes and the density factors as $f(x,y)=f_X(x)\\,f_Y(y)$ (a product of two univariate normals), so $X\\perp Y$. Conversely independence always gives $\\rho=0$. Hence **for the bivariate normal, $\\rho=0\\iff$ independence.**\nThis equivalence is **special to the bivariate normal**. In general $\\rho=0$ (zero covariance) does *not* imply independence — e.g. $X$ uniform on $\\{-1,0,1\\}$ with $Y=X^{2}$ has $\\rho=0$ yet $Y$ is a function of $X$.", "tag": "Bivariate normal" }, { "front": "$X$ and $Y$ have $\\operatorname{Var}(X)=4$, $\\operatorname{Var}(Y)=9$, $\\operatorname{Cov}(X,Y)=-2$. Find $\\operatorname{Var}(X-Y)$.", "back": "$\\operatorname{Var}(X-Y)=\\operatorname{Var}(X)+\\operatorname{Var}(Y)-2\\operatorname{Cov}(X,Y)$.\n$=4+9-2(-2)=4+9+4=17$.\nNote the sign: subtracting a variable still adds its variance, and $-2\\operatorname{Cov}$ here becomes $+4$.", "tag": "Linear combinations" }, { "front": "An insurer holds $n=3$ i.i.d. losses, each with mean $200$ and standard deviation $50$. Find the variance and SD of the total $S=X_1+X_2+X_3$.", "back": "Independent, so variances add: $\\operatorname{Var}(S)=3\\cdot 50^{2}=3\\cdot 2500=7500$.\nStandard deviation: $\\operatorname{SD}(S)=\\sqrt{7500}\\approx 86.6$.\n(The mean is $3\\cdot 200=600$, but SD grows like $\\sqrt{n}$, not $n$.)", "tag": "Linear combinations" }, { "front": "$X$ and $Y$ are bivariate with $\\operatorname{Var}(X)=4$, $\\operatorname{Var}(Y)=9$, $\\rho=0.5$. Find $\\operatorname{Var}(X+Y)$.", "back": "First recover the covariance: $\\operatorname{Cov}(X,Y)=\\rho\\,\\sigma_X\\sigma_Y=0.5\\cdot 2\\cdot 3=3$.\n$\\operatorname{Var}(X+Y)=\\operatorname{Var}(X)+\\operatorname{Var}(Y)+2\\operatorname{Cov}(X,Y)=4+9+2(3)=19$.", "tag": "Linear combinations" }, { "front": "Three losses have equal variance $\\sigma^{2}=10$ and equal pairwise covariance $\\operatorname{Cov}=4$. Find $\\operatorname{Var}(X_1+X_2+X_3)$.", "back": "$\\operatorname{Var}\\!\\left(\\sum X_i\\right)=\\sum\\operatorname{Var}(X_i)+2\\sum_{i