{
  "deckName": "Exam P — Transformations & Order Statistics",
  "examCode": "Exam P",
  "cards": [
    {
      "front": "State the **CDF method** (distribution-function technique) for finding the distribution of $Y=g(X)$.",
      "back": "Write the cdf of $Y$ directly in terms of $X$: $F_Y(y)=\\Pr(Y\\leq y)=\\Pr(g(X)\\leq y)$. Solve the event $\\{g(X)\\leq y\\}$ for $X$ to express it through $F_X$, then differentiate to get the pdf: $f_Y(y)=\\frac{d}{dy}F_Y(y)$. It works for any $g$ (monotone or not) and is the safest general approach.",
      "tag": "CDF method"
    },
    {
      "front": "$X\\sim U(0,2)$ and $Y=X^{2}$. Use the CDF method to find $f_Y(y)$.",
      "back": "For $0<y<4$: $F_Y(y)=\\Pr(X^{2}\\leq y)=\\Pr(X\\leq\\sqrt{y})=\\frac{\\sqrt{y}}{2}$ since $F_X(x)=\\frac{x}{2}$.\nDifferentiate: $f_Y(y)=\\frac{d}{dy}\\frac{\\sqrt{y}}{2}=\\frac{1}{4\\sqrt{y}}$ for $0<y<4$ (and $0$ otherwise).",
      "tag": "CDF method"
    },
    {
      "front": "State the **change-of-variable (Jacobian) formula** for a one-to-one transformation $Y=g(X)$ in one dimension.",
      "back": "If $g$ is monotone with inverse $x=g^{-1}(y)$, then $f_Y(y)=f_X\\!\\big(g^{-1}(y)\\big)\\left|\\frac{dx}{dy}\\right|$. The factor $\\left|\\frac{dx}{dy}\\right|$ is the (one-dimensional) Jacobian; the absolute value keeps the density nonnegative. Forgetting this factor is the most common transformation error.",
      "tag": "Jacobian"
    },
    {
      "front": "$X\\sim U(0,1)$ and $Y=X^{3}$. Use the Jacobian method to find $f_Y(y)$ on $(0,1)$.",
      "back": "Invert: $x=y^{1/3}$, so $\\frac{dx}{dy}=\\frac{1}{3}y^{-2/3}$.\nSince $f_X(x)=1$ on $(0,1)$, $f_Y(y)=1\\cdot\\left|\\tfrac{1}{3}y^{-2/3}\\right|=\\frac{1}{3}y^{-2/3}$ for $0<y<1$.\nCheck: $\\int_{0}^{1}\\frac{1}{3}y^{-2/3}\\,dy=\\big[y^{1/3}\\big]_{0}^{1}=1$.",
      "tag": "Jacobian"
    },
    {
      "front": "$X\\sim \\text{Exp}(1)$ (so $f_X(x)=e^{-x}$, $x>0$) and $Y=e^{-X}$. Find the distribution of $Y$.",
      "back": "Here $Y\\in(0,1)$ and $x=-\\ln y$, so $\\left|\\frac{dx}{dy}\\right|=\\frac{1}{y}$.\nThen $f_Y(y)=f_X(-\\ln y)\\cdot\\frac{1}{y}=e^{\\ln y}\\cdot\\frac{1}{y}=y\\cdot\\frac{1}{y}=1$ for $0<y<1$.\nSo $Y\\sim U(0,1)$ — an instance of the probability integral transform since $e^{-X}=1-F_X(X)$.",
      "tag": "Jacobian"
    },
    {
      "front": "State the **probability integral transform**: what is the distribution of $F_X(X)$ when $X$ is continuous?",
      "back": "If $X$ is continuous with cdf $F_X$, then $U=F_X(X)\\sim U(0,1)$.\nConversely, if $U\\sim U(0,1)$ then $X=F_X^{-1}(U)$ has cdf $F_X$ — this is the basis of inverse-transform simulation.",
      "tag": "CDF method"
    },
    {
      "front": "$X\\sim U(-1,1)$ and $Y=X^{2}$. Why can't you just use $f_X(x)\\left|\\frac{dx}{dy}\\right|$ with one branch, and what is $f_Y(y)$?",
      "back": "$g(x)=x^{2}$ is not monotone on $(-1,1)$: both $x=\\pm\\sqrt{y}$ map to the same $y$, so you must **sum over both branches**.\n$f_Y(y)=f_X(\\sqrt{y})\\cdot\\frac{1}{2\\sqrt{y}}+f_X(-\\sqrt{y})\\cdot\\frac{1}{2\\sqrt{y}}=\\frac{1}{2}\\cdot\\frac{1}{2\\sqrt{y}}+\\frac{1}{2}\\cdot\\frac{1}{2\\sqrt{y}}=\\frac{1}{2\\sqrt{y}}$ for $0<y<1$.",
      "tag": "Non-monotone"
    },
    {
      "front": "$X\\sim\\text{Exp}$ with mean $5$. Let $Y=2X$. Identify the distribution of $Y$ and compute $\\Pr(Y>15)$.",
      "back": "Scaling an exponential by a positive constant gives another exponential: $Y=2X\\sim\\text{Exp}$ with mean $2\\cdot 5=10$.\nThen $\\Pr(Y>15)=e^{-15/10}=e^{-1.5}\\approx 0.2231$.",
      "tag": "Linear transform"
    },
    {
      "front": "$U\\sim U(0,1)$ and $Y=-2\\ln U$. Show $Y$ is exponential and find $\\Pr(Y>4)$.",
      "back": "For $y>0$: $F_Y(y)=\\Pr(-2\\ln U\\leq y)=\\Pr\\!\\big(\\ln U\\geq -\\tfrac{y}{2}\\big)=\\Pr\\!\\big(U\\geq e^{-y/2}\\big)=1-e^{-y/2}$.\nThat is the cdf of $\\text{Exp}$ with mean $2$ (rate $\\tfrac{1}{2}$).\n$\\Pr(Y>4)=e^{-4/2}=e^{-2}\\approx 0.1353$.",
      "tag": "CDF method"
    },
    {
      "front": "If $Z\\sim N(0,1)$ and $X=\\mu+\\sigma Z$, use the Jacobian to confirm $X\\sim N(\\mu,\\sigma^{2})$.",
      "back": "Invert: $z=\\frac{x-\\mu}{\\sigma}$, $\\left|\\frac{dz}{dx}\\right|=\\frac{1}{\\sigma}$.\n$f_X(x)=\\frac{1}{\\sqrt{2\\pi}}e^{-z^{2}/2}\\cdot\\frac{1}{\\sigma}=\\frac{1}{\\sigma\\sqrt{2\\pi}}\\exp\\!\\Big(-\\frac{(x-\\mu)^{2}}{2\\sigma^{2}}\\Big)$,\nwhich is the $N(\\mu,\\sigma^{2})$ density.",
      "tag": "Linear transform"
    },
    {
      "front": "State the **bivariate change-of-variables formula** for $(U,V)=g(X,Y)$.",
      "back": "With inverse $x=x(u,v)$, $y=y(u,v)$, the joint density transforms as $f_{U,V}(u,v)=f_{X,Y}\\big(x(u,v),y(u,v)\\big)\\,|J|$, where the Jacobian determinant is $J=\\frac{\\partial x}{\\partial u}\\frac{\\partial y}{\\partial v}-\\frac{\\partial x}{\\partial v}\\frac{\\partial y}{\\partial u}$. Always take $|J|$ and re-express the support in the new variables.",
      "tag": "Bivariate Jacobian"
    },
    {
      "front": "$X,Y$ are independent $\\text{Exp}(1)$. Using $U=X+Y$, $V=X$, find the marginal pdf of $U=X+Y$.",
      "back": "Inverse: $x=v$, $y=u-v$; the Jacobian is $|J|=1$.\n$f_{U,V}(u,v)=e^{-v}e^{-(u-v)}\\cdot 1=e^{-u}$ on the region $0<v<u$.\nMarginalize over $v$: $f_U(u)=\\int_{0}^{u}e^{-u}\\,dv=u\\,e^{-u}$ for $u>0$ — the $\\text{Gamma}(2,1)$ density.",
      "tag": "Bivariate Jacobian"
    },
    {
      "front": "State the **convolution formula** for the pdf of $Z=X+Y$ when $X,Y$ are independent and continuous.",
      "back": "$f_Z(z)=\\int_{-\\infty}^{\\infty} f_X(x)\\,f_Y(z-x)\\,dx=\\int_{-\\infty}^{\\infty} f_X(z-y)\\,f_Y(y)\\,dy$.\nCare is needed with limits: integrate only where **both** densities are positive.",
      "tag": "Convolution"
    },
    {
      "front": "$X,Y$ are independent $U(0,1)$. Derive the pdf of $Z=X+Y$ by convolution.",
      "back": "$f_Z(z)=\\int f_X(x)f_Y(z-x)\\,dx$ with both factors equal to $1$ when $0<x<1$ and $0<z-x<1$.\nFor $0\\leq z\\leq 1$: limits $0<x<z$ give $f_Z(z)=z$.\nFor $1<z\\leq 2$: limits $z-1<x<1$ give $f_Z(z)=2-z$.\nThis is the triangular density on $(0,2)$ peaking at $z=1$.",
      "tag": "Convolution"
    },
    {
      "front": "$X,Y$ independent $U(0,1)$. Find $\\Pr(X+Y\\leq 0.5)$ geometrically.",
      "back": "The event $\\{X+Y\\leq 0.5\\}$ is the triangular region in the unit square below the line $x+y=0.5$.\nArea $=\\frac{1}{2}(0.5)(0.5)=0.125$. Since the joint density is $1$ on the square, $\\Pr(X+Y\\leq 0.5)=0.125$.",
      "tag": "Convolution"
    },
    {
      "front": "$X,Y$ independent $U(0,1)$. Find $\\Pr(X+Y>1.5)$.",
      "back": "The region $\\{x+y>1.5\\}$ inside the unit square is the small triangle with vertices $(0.5,1),(1,0.5),(1,1)$, with legs of length $0.5$.\nArea $=\\frac{1}{2}(0.5)(0.5)=0.125$, so $\\Pr(X+Y>1.5)=0.125$.",
      "tag": "Convolution"
    },
    {
      "front": "State the **discrete convolution** for the pmf of $S=X+Y$ with $X,Y$ independent and integer-valued, then use it to find $\\Pr(S=5)$ for two independent fair dice.",
      "back": "Discrete convolution: $\\Pr(S=s)=\\sum_{k}\\Pr(X=k)\\,\\Pr(Y=s-k)$, summing over all $k$ with both terms positive.\nFor two fair dice each have $\\Pr=\\tfrac{1}{6}$ on $\\{1,\\dots,6\\}$. For $s=5$ the valid pairs are $k\\in\\{1,2,3,4\\}$, so $\\Pr(S=5)=\\sum_{k=1}^{4}\\tfrac{1}{6}\\cdot\\tfrac{1}{6}=\\frac{4}{36}=\\frac{1}{9}$.\nThe sum's pmf is the triangular pattern $\\tfrac{1}{36},\\tfrac{2}{36},\\dots,\\tfrac{6}{36},\\dots,\\tfrac{1}{36}$ for $s=2,\\dots,12$ — the discrete analogue of the continuous triangular density.",
      "tag": "Convolution"
    },
    {
      "front": "$X\\sim\\text{Poisson}(\\lambda_1)$ and $Y\\sim\\text{Poisson}(\\lambda_2)$ are independent. Use the pmf convolution to show $S=X+Y\\sim\\text{Poisson}(\\lambda_1+\\lambda_2)$.",
      "back": "$\\Pr(S=s)=\\sum_{k=0}^{s}\\Pr(X=k)\\Pr(Y=s-k)=\\sum_{k=0}^{s}\\frac{e^{-\\lambda_1}\\lambda_1^{k}}{k!}\\cdot\\frac{e^{-\\lambda_2}\\lambda_2^{\\,s-k}}{(s-k)!}$.\nFactor out $e^{-(\\lambda_1+\\lambda_2)}$ and multiply/divide by $s!$: $\\Pr(S=s)=\\frac{e^{-(\\lambda_1+\\lambda_2)}}{s!}\\sum_{k=0}^{s}\\binom{s}{k}\\lambda_1^{k}\\lambda_2^{\\,s-k}=\\frac{e^{-(\\lambda_1+\\lambda_2)}(\\lambda_1+\\lambda_2)^{s}}{s!}$ by the binomial theorem.\nThat is the $\\text{Poisson}(\\lambda_1+\\lambda_2)$ pmf, so Poisson rates add under independent sums.",
      "tag": "Convolution"
    },
    {
      "front": "What is the distribution of a sum of $n$ independent $\\text{Exp}(\\lambda)$ random variables, and why?",
      "back": "The sum is $\\text{Gamma}(\\alpha=n,\\lambda)$, also called Erlang-$n$.\nReason: each exponential has MGF $\\frac{\\lambda}{\\lambda-t}$, and independent MGFs multiply, giving $\\big(\\frac{\\lambda}{\\lambda-t}\\big)^{n}$ — the $\\text{Gamma}(n,\\lambda)$ MGF. Mean $=\\frac{n}{\\lambda}$, variance $=\\frac{n}{\\lambda^{2}}$.",
      "tag": "Convolution"
    },
    {
      "front": "Ten independent claim delays are each $\\text{Exp}$ with mean $\\tfrac{1}{3}$. Find the mean and variance of the total delay $S$.",
      "back": "Each delay has rate $\\lambda=3$, so $S\\sim\\text{Gamma}(10,3)$.\nMean $=\\frac{n}{\\lambda}=\\frac{10}{3}\\approx 3.33$.\nVariance $=\\frac{n}{\\lambda^{2}}=\\frac{10}{9}\\approx 1.11$.",
      "tag": "Convolution"
    },
    {
      "front": "$X,Y$ are independent $U(0,1)$. Find $\\Pr\\!\\big(\\frac{X}{Y}<1\\big)$.",
      "back": "$\\frac{X}{Y}<1\\iff X<Y$. By symmetry of the unit square about the line $x=y$, $\\Pr(X<Y)=\\frac{1}{2}$ (the diagonal contributes zero probability).",
      "tag": "Ratio / functions"
    },
    {
      "front": "$X\\sim U(0,1)$. Find $f_Y(y)$ where $Y=-\\ln(1-X)$, and name the distribution.",
      "back": "$Y\\geq 0$; invert $x=1-e^{-y}$, $\\frac{dx}{dy}=e^{-y}$.\n$f_Y(y)=f_X(1-e^{-y})\\cdot e^{-y}=1\\cdot e^{-y}=e^{-y}$, $y>0$.\nSo $Y\\sim\\text{Exp}(1)$. (Again the inverse-cdf transform, since $1-e^{-y}$ is the $\\text{Exp}(1)$ cdf.)",
      "tag": "Functions / min-max"
    },
    {
      "front": "For $n$ iid continuous variables with cdf $F$ and pdf $f$, give the cdf and pdf of the **maximum** $X_{(n)}$.",
      "back": "$F_{(n)}(x)=\\Pr(\\text{all}\\leq x)=[F(x)]^{n}$.\nDifferentiate: $f_{(n)}(x)=n[F(x)]^{n-1}f(x)$.",
      "tag": "Order statistics"
    },
    {
      "front": "For $n$ iid continuous variables with cdf $F$ and pdf $f$, give the survival function and pdf of the **minimum** $X_{(1)}$.",
      "back": "$\\Pr(X_{(1)}>x)=\\Pr(\\text{all}>x)=[1-F(x)]^{n}$, so $F_{(1)}(x)=1-[1-F(x)]^{n}$.\nDifferentiate: $f_{(1)}(x)=n[1-F(x)]^{n-1}f(x)$.",
      "tag": "Order statistics"
    },
    {
      "front": "Write the pdf of the **$k$-th order statistic** $X_{(k)}$ from $n$ iid continuous variables with cdf $F$, pdf $f$.",
      "back": "$f_{(k)}(x)=\\frac{n!}{(k-1)!\\,(n-k)!}\\,[F(x)]^{k-1}\\,[1-F(x)]^{n-k}\\,f(x)$.\nInterpretation: $k-1$ values fall below $x$, one is at $x$, and $n-k$ exceed $x$. Dropping the multinomial coefficient is a classic error.",
      "tag": "Order statistics"
    },
    {
      "front": "If $X_1,\\dots,X_n$ are independent with $X_i\\sim\\text{Exp}(\\lambda_i)$, what is the distribution of $\\min_i X_i$?",
      "back": "$\\Pr(\\min>x)=\\prod_i e^{-\\lambda_i x}=e^{-(\\sum_i\\lambda_i)x}$, so $\\min_i X_i\\sim\\text{Exp}\\!\\big(\\sum_i\\lambda_i\\big)$.\nThe rate is the **sum** of the individual rates (not the average). For iid $\\text{Exp}(\\lambda)$, the minimum is $\\text{Exp}(n\\lambda)$.",
      "tag": "Min of exponentials"
    },
    {
      "front": "Three components have independent lifetimes that are exponential with means $4$, $5$, and $10$ years. Find the expected time until the first failure.",
      "back": "Rates: $\\lambda_1=\\tfrac{1}{4},\\ \\lambda_2=\\tfrac{1}{5},\\ \\lambda_3=\\tfrac{1}{10}$.\nFirst failure $=\\min\\sim\\text{Exp}(\\lambda_1+\\lambda_2+\\lambda_3)$, rate $=0.25+0.20+0.10=0.55$.\nExpected time $=\\frac{1}{0.55}\\approx 1.818$ years.",
      "tag": "Min of exponentials"
    },
    {
      "front": "$X_i$ are independent $\\text{Exp}(\\lambda_i)$. What is $\\Pr(X_1=\\min)$, i.e. the probability $X_1$ is the smallest?",
      "back": "$\\Pr(X_1<\\text{all others})=\\dfrac{\\lambda_1}{\\lambda_1+\\lambda_2+\\cdots+\\lambda_n}$.\nExample: if $\\lambda_1=2$ and $\\lambda_2=3$, then $\\Pr(X_1\\text{ first})=\\frac{2}{2+3}=0.4$.",
      "tag": "Min of exponentials"
    },
    {
      "front": "Five identical relays each have independent $\\text{Exp}(0.2)$ failure times (per year). Find the probability the **first** relay failure occurs after $3$ years.",
      "back": "The minimum of five iid $\\text{Exp}(0.2)$ is $\\text{Exp}(5\\cdot 0.2)=\\text{Exp}(1)$.\n$\\Pr(\\min>3)=e^{-1\\cdot 3}=e^{-3}\\approx 0.0498$.",
      "tag": "Min of exponentials"
    },
    {
      "front": "For $n$ iid $U(0,1)$ variables, find $\\Pr\\!\\big(X_{(n)}<t\\big)$, $E\\big[X_{(n)}\\big]$, and evaluate $\\Pr(X_{(3)}<0.7)$ for $n=3$.",
      "back": "Since $F(t)=t$: $\\Pr(X_{(n)}<t)=t^{n}$, and $E[X_{(n)}]=\\frac{n}{n+1}$.\nFor $n=3$, $t=0.7$: $\\Pr(X_{(3)}<0.7)=0.7^{3}=0.343$.",
      "tag": "Max — uniform"
    },
    {
      "front": "For $n=4$ iid $U(0,1)$ variables, find $E[X_{(1)}]$ (the minimum) and $E[X_{(4)}]$ (the maximum).",
      "back": "For $U(0,1)$, $E[X_{(k)}]=\\frac{k}{n+1}$.\nMinimum: $E[X_{(1)}]=\\frac{1}{5}=0.2$.\nMaximum: $E[X_{(4)}]=\\frac{4}{5}=0.8$.",
      "tag": "Min/Max — uniform"
    },
    {
      "front": "Show that the $k$-th order statistic of $n$ iid $U(0,1)$ variables is $\\text{Beta}(k,\\,n-k+1)$, and give $E\\big[X_{(k)}\\big]$ and $\\operatorname{Var}\\big(X_{(k)}\\big)$.",
      "back": "Substituting $F(x)=x$, $f(x)=1$ into the $k$-th order-statistic pdf gives $f_{(k)}(x)=\\frac{n!}{(k-1)!(n-k)!}x^{k-1}(1-x)^{n-k}$ — the $\\text{Beta}(k,n-k+1)$ density.\nHence $E[X_{(k)}]=\\frac{k}{n+1}$ and $\\operatorname{Var}(X_{(k)})=\\frac{k(n-k+1)}{(n+1)^{2}(n+2)}$.",
      "tag": "Order statistics"
    },
    {
      "front": "A sample of $5$ iid $U(0,1)$ values is taken. Find the mean and variance of the sample **median** $X_{(3)}$.",
      "back": "Here $n=5$, $k=3$, so $X_{(3)}\\sim\\text{Beta}(3,3)$.\nMean $=\\frac{k}{n+1}=\\frac{3}{6}=0.5$.\nVariance $=\\frac{k(n-k+1)}{(n+1)^{2}(n+2)}=\\frac{3\\cdot 3}{6^{2}\\cdot 7}=\\frac{9}{252}\\approx 0.0357$.",
      "tag": "Order statistics"
    },
    {
      "front": "Two machines have independent lifetimes, each $\\text{Exp}$ with mean $10$ hours. The system fails only when **both** have failed. Find the expected time to system failure.",
      "back": "System failure $=\\max(X_1,X_2)$ of two iid $\\text{Exp}(\\lambda)$, $\\lambda=\\tfrac{1}{10}$.\nUsing $E[\\max]=\\frac{1}{\\lambda}\\big(1+\\tfrac{1}{2}\\big)=\\frac{3}{2\\lambda}$ (mean is sum of independent $\\text{Exp}$ gaps): $E[\\max]=10\\cdot 1.5=15$ hours.",
      "tag": "Max of exponentials"
    }
  ]
}