{
  "deckName": "Exam P — Continuous Distributions",
  "examCode": "Exam P",
  "cards": [
    {
      "front": "For a continuous random variable, how is the cumulative distribution function $F(x)$ defined in terms of the pdf $f$?",
      "back": "$F(x)=\\Pr(X\\leq x)=\\int_{-\\infty}^{x} f(t)\\,dt$. Conversely the pdf is the derivative of the cdf, $f(x)=F'(x)$, wherever $F$ is differentiable.",
      "tag": "pdf/cdf"
    },
    {
      "front": "What two conditions must a function $f$ satisfy to be a valid continuous pdf?",
      "back": "It must be nonnegative everywhere, $f(x)\\geq 0$, and it must integrate to one over the whole real line, $\\int_{-\\infty}^{\\infty} f(x)\\,dx=1$.",
      "tag": "pdf/cdf"
    },
    {
      "front": "Define the survival function $S(x)$ and give its relationship to $F(x)$.",
      "back": "The survival function is $S(x)=\\Pr(X>x)=1-F(x)$. For a continuous variable $S(x)=\\int_{x}^{\\infty} f(t)\\,dt$, and $S'(x)=-f(x)$.",
      "tag": "pdf/cdf"
    },
    {
      "front": "Define the hazard rate (failure rate) $h(x)$ of a continuous random variable.",
      "back": "The hazard rate is $h(x)=\\dfrac{f(x)}{S(x)}=\\dfrac{f(x)}{1-F(x)}$. It equals $-\\dfrac{d}{dx}\\ln S(x)$, so for a nonnegative lifetime with $S(0)=1$, $S(x)=\\exp\\!\\left(-\\int_{0}^{x} h(t)\\,dt\\right)$.",
      "tag": "hazard"
    },
    {
      "front": "A continuous variable has constant hazard rate $h(x)=\\lambda$ for $x\\geq 0$. What is its distribution?",
      "back": "Constant hazard implies $S(x)=\\exp\\!\\left(-\\int_{0}^{x}\\lambda\\,dt\\right)=e^{-\\lambda x}$, which is the survival function of an **exponential** distribution with rate $\\lambda$. Constant hazard $\\Leftrightarrow$ exponential.",
      "tag": "hazard"
    },
    {
      "front": "For a nonnegative continuous random variable, express $E[X]$ as an integral of the survival function.",
      "back": "The tail (layer) formula: $E[X]=\\int_{0}^{\\infty} S(x)\\,dx=\\int_{0}^{\\infty}\\bigl(1-F(x)\\bigr)\\,dx$. This often avoids computing $\\int x f(x)\\,dx$ directly.",
      "tag": "pdf/cdf"
    },
    {
      "front": "Given the pdf $f(x)=c\\,x^{2}$ on $0\\leq x\\leq 3$ and zero elsewhere, find $c$.",
      "back": "Require $\\int_{0}^{3} c\\,x^{2}\\,dx=1$. Then $c\\cdot\\dfrac{x^{3}}{3}\\Big|_{0}^{3}=c\\cdot\\dfrac{27}{3}=9c=1$, so $c=\\dfrac{1}{9}$.",
      "tag": "computing-probabilities"
    },
    {
      "front": "With pdf $f(x)=\\tfrac{1}{9}x^{2}$ on $[0,3]$, compute $\\Pr(1\\leq X\\leq 2)$.",
      "back": "$\\Pr(1\\leq X\\leq 2)=\\int_{1}^{2}\\tfrac{1}{9}x^{2}\\,dx=\\dfrac{1}{9}\\cdot\\dfrac{x^{3}}{3}\\Big|_{1}^{2}=\\dfrac{1}{27}(8-1)=\\dfrac{7}{27}\\approx 0.259$.",
      "tag": "computing-probabilities"
    },
    {
      "front": "How do you find the median of a continuous random variable from its cdf?",
      "back": "The median $m$ solves $F(m)=0.5$, i.e. $\\Pr(X\\leq m)=\\tfrac12$. More generally the $p$-th percentile $x_p$ solves $F(x_p)=p$.",
      "tag": "percentiles"
    },
    {
      "front": "How is the mode of a continuous distribution found?",
      "back": "The mode is the value of $x$ that **maximizes the pdf** $f(x)$. Find it by setting $f'(x)=0$ and checking it is a maximum (or by inspecting endpoints of the support).",
      "tag": "mode"
    },
    {
      "front": "State the pdf and cdf of a continuous Uniform$(a,b)$ distribution.",
      "back": "pdf: $f(x)=\\dfrac{1}{b-a}$ for $a\\leq x\\leq b$. cdf: $F(x)=\\dfrac{x-a}{b-a}$ on $[a,b]$, with $F(x)=0$ below $a$ and $F(x)=1$ above $b$.",
      "tag": "uniform"
    },
    {
      "front": "Give the mean and variance of a continuous Uniform$(a,b)$ random variable.",
      "back": "$E[X]=\\dfrac{a+b}{2}$ and $\\mathrm{Var}(X)=\\dfrac{(b-a)^{2}}{12}$.",
      "tag": "uniform"
    },
    {
      "front": "$X\\sim$ Uniform$(2,10)$. Find $\\Pr(X>7)$ and the variance of $X$.",
      "back": "$\\Pr(X>7)=\\dfrac{10-7}{10-2}=\\dfrac{3}{8}=0.375$. Variance: $\\mathrm{Var}(X)=\\dfrac{(10-2)^{2}}{12}=\\dfrac{64}{12}=\\dfrac{16}{3}\\approx 5.33$.",
      "tag": "uniform"
    },
    {
      "front": "$X\\sim$ Uniform$(0,1)$. What is the distribution of $Y=-\\tfrac{1}{\\lambda}\\ln X$?",
      "back": "$Y$ is **exponential with rate $\\lambda$**. Check: $\\Pr(Y\\leq y)=\\Pr\\!\\left(-\\tfrac1\\lambda\\ln X\\leq y\\right)=\\Pr(X\\geq e^{-\\lambda y})=1-e^{-\\lambda y}$, the exponential cdf. This is the inverse-cdf (probability integral) method.",
      "tag": "uniform"
    },
    {
      "front": "State the pdf, cdf, and survival function of an Exponential distribution with rate $\\lambda$.",
      "back": "pdf: $f(x)=\\lambda e^{-\\lambda x}$ for $x\\geq 0$. cdf: $F(x)=1-e^{-\\lambda x}$. Survival: $S(x)=e^{-\\lambda x}$.",
      "tag": "exponential"
    },
    {
      "front": "Give the mean and variance of an Exponential distribution, both in rate form $\\lambda$ and in mean (scale) form $\\theta$.",
      "back": "With rate $\\lambda$: $E[X]=\\dfrac{1}{\\lambda}$ and $\\mathrm{Var}(X)=\\dfrac{1}{\\lambda^{2}}$. With scale $\\theta=1/\\lambda$ (so $\\theta$ is the mean): $E[X]=\\theta$ and $\\mathrm{Var}(X)=\\theta^{2}$.",
      "tag": "exponential"
    },
    {
      "front": "State the memoryless property of the exponential distribution.",
      "back": "For $s,t\\geq 0$, $\\Pr(X>s+t\\mid X>s)=\\Pr(X>t)$. Having already survived $s$ units gives no information about the remaining lifetime. The exponential is the only continuous distribution with this property.",
      "tag": "exponential"
    },
    {
      "front": "A machine's lifetime is exponential with mean $5$ years. Given it has run $3$ years, find the probability it runs at least $4$ more years.",
      "back": "By memorylessness the past $3$ years are irrelevant: $\\Pr(X>3+4\\mid X>3)=\\Pr(X>4)=e^{-4/5}=e^{-0.8}\\approx 0.449$.",
      "tag": "exponential"
    },
    {
      "front": "Claim sizes are exponential with mean $\\theta=200$. Find $\\Pr(X>300)$.",
      "back": "Rate $\\lambda=1/200$. $\\Pr(X>300)=S(300)=e^{-300/200}=e^{-1.5}\\approx 0.223$.",
      "tag": "exponential"
    },
    {
      "front": "If $X_1,\\dots,X_n$ are independent exponentials with rates $\\lambda_1,\\dots,\\lambda_n$, what is the distribution of $\\min(X_1,\\dots,X_n)$?",
      "back": "The minimum is **exponential with rate $\\sum_{i=1}^{n}\\lambda_i$**. Reason: $\\Pr(\\min>x)=\\prod_i \\Pr(X_i>x)=\\prod_i e^{-\\lambda_i x}=e^{-(\\sum\\lambda_i)x}$.",
      "tag": "exponential"
    },
    {
      "front": "Three independent components have exponential lifetimes with means $10$, $20$, and $40$ hours. What is the mean time until the first failure?",
      "back": "Rates are $0.1,\\ 0.05,\\ 0.025$ per hour. The minimum is exponential with rate $0.1+0.05+0.025=0.175$, so the mean time to first failure is $\\dfrac{1}{0.175}\\approx 5.71$ hours.",
      "tag": "exponential"
    },
    {
      "front": "For an exponential loss with mean $\\theta$, what is the expected payment per loss under an ordinary deductible $d$?",
      "back": "$E[(X-d)_+]=\\int_{d}^{\\infty} S(x)\\,dx=\\int_{d}^{\\infty} e^{-x/\\theta}\\,dx=\\theta\\,e^{-d/\\theta}$. The memoryless property makes the answer the survival probability $e^{-d/\\theta}$ times the full mean $\\theta$.",
      "tag": "exponential"
    },
    {
      "front": "State the pdf of a Gamma distribution with shape $\\alpha$ and rate $\\lambda$.",
      "back": "$f(x)=\\dfrac{\\lambda^{\\alpha}}{\\Gamma(\\alpha)}\\,x^{\\alpha-1}e^{-\\lambda x}$ for $x>0$, where $\\Gamma(\\alpha)=\\int_{0}^{\\infty} t^{\\alpha-1}e^{-t}\\,dt$ is the gamma function.",
      "tag": "gamma"
    },
    {
      "front": "Give the mean and variance of a Gamma$(\\alpha,\\lambda)$ distribution (rate parameterization).",
      "back": "$E[X]=\\dfrac{\\alpha}{\\lambda}$ and $\\mathrm{Var}(X)=\\dfrac{\\alpha}{\\lambda^{2}}$. With scale $\\theta=1/\\lambda$: $E[X]=\\alpha\\theta$ and $\\mathrm{Var}(X)=\\alpha\\theta^{2}$.",
      "tag": "gamma"
    },
    {
      "front": "How does the Gamma distribution relate to the exponential and the Erlang distributions?",
      "back": "Gamma with shape $\\alpha=1$ is exponential with rate $\\lambda$. For integer shape $\\alpha=n$, the Gamma is an **Erlang**, equal in distribution to the sum of $n$ i.i.d. exponentials each with rate $\\lambda$.",
      "tag": "gamma"
    },
    {
      "front": "Why is the sum of $n$ independent Exponential$(\\lambda)$ variables a Gamma, and what are its parameters?",
      "back": "Each exponential has MGF $\\lambda/(\\lambda-t)$. Independent sums multiply MGFs, giving $\\bigl[\\lambda/(\\lambda-t)\\bigr]^{n}$, which is the MGF of a Gamma with shape $\\alpha=n$ and rate $\\lambda$. So the sum is Gamma$(n,\\lambda)$ with mean $n/\\lambda$.",
      "tag": "gamma"
    },
    {
      "front": "Key gamma-function facts to memorize for Exam P.",
      "back": "$\\Gamma(\\alpha+1)=\\alpha\\,\\Gamma(\\alpha)$ (recursion); $\\Gamma(n)=(n-1)!$ for a positive integer $n$; and $\\Gamma\\!\\left(\\tfrac12\\right)=\\sqrt{\\pi}$.",
      "tag": "gamma"
    },
    {
      "front": "Claims arrive in a Poisson process at rate $2$ per day. What is the distribution and mean of the waiting time until the $3$rd claim?",
      "back": "The waiting time to the $3$rd event is a sum of $3$ i.i.d. exponential interarrival times, i.e. Gamma (Erlang) with shape $\\alpha=3$ and rate $\\lambda=2$. Its mean is $\\dfrac{\\alpha}{\\lambda}=\\dfrac{3}{2}=1.5$ days.",
      "tag": "gamma"
    },
    {
      "front": "How is the chi-square distribution related to the Gamma distribution?",
      "back": "A chi-square with $k$ degrees of freedom is a Gamma with shape $\\alpha=k/2$ and rate $\\lambda=\\tfrac12$ (scale $\\theta=2$). Hence its mean is $k$ and its variance is $2k$.",
      "tag": "gamma"
    },
    {
      "front": "$X\\sim$ Gamma with shape $\\alpha=2$ and rate $\\lambda=\\tfrac13$. Compute $E[X]$ and $\\mathrm{SD}(X)$.",
      "back": "$E[X]=\\dfrac{\\alpha}{\\lambda}=\\dfrac{2}{1/3}=6$. $\\mathrm{Var}(X)=\\dfrac{\\alpha}{\\lambda^{2}}=\\dfrac{2}{(1/3)^{2}}=2\\cdot 9=18$, so $\\mathrm{SD}(X)=\\sqrt{18}\\approx 4.24$.",
      "tag": "gamma"
    },
    {
      "front": "State the pdf of a Normal$(\\mu,\\sigma^{2})$ distribution.",
      "back": "$f(x)=\\dfrac{1}{\\sigma\\sqrt{2\\pi}}\\exp\\!\\left(-\\dfrac{(x-\\mu)^{2}}{2\\sigma^{2}}\\right)$ for all real $x$, where $\\mu=E[X]$ and $\\sigma^{2}=\\mathrm{Var}(X)$.",
      "tag": "normal"
    },
    {
      "front": "How do you standardize a Normal$(\\mu,\\sigma^{2})$ variable, and what does the normal table give?",
      "back": "Set $Z=\\dfrac{X-\\mu}{\\sigma}\\sim N(0,1)$. The table gives $\\Phi(z)=\\Pr(Z\\leq z)$. Note you divide by the standard deviation $\\sigma$, not the variance $\\sigma^{2}$.",
      "tag": "normal"
    },
    {
      "front": "State the empirical (68–95–99.7) rule for a normal distribution.",
      "back": "Approximately $68\\%$ of the probability lies within $\\mu\\pm\\sigma$, about $95\\%$ within $\\mu\\pm 2\\sigma$, and about $99.7\\%$ within $\\mu\\pm 3\\sigma$.",
      "tag": "normal"
    },
    {
      "front": "$X\\sim N(\\mu=50,\\ \\sigma^{2}=16)$. Find $\\Pr(X\\leq 58)$ using $\\Phi(2)=0.9772$.",
      "back": "$\\sigma=4$, so $Z=\\dfrac{58-50}{4}=2$. Thus $\\Pr(X\\leq 58)=\\Phi(2)=0.9772$.",
      "tag": "normal"
    },
    {
      "front": "$X\\sim N(\\mu=100,\\ \\sigma=15)$. Find $\\Pr(85\\leq X\\leq 130)$ using $\\Phi(1)=0.8413,\\ \\Phi(2)=0.9772$.",
      "back": "Standardize: lower $z=\\dfrac{85-100}{15}=-1$, upper $z=\\dfrac{130-100}{15}=2$. Then $\\Pr=\\Phi(2)-\\Phi(-1)=0.9772-(1-0.8413)=0.9772-0.1587=0.8185$.",
      "tag": "normal"
    },
    {
      "front": "$X\\sim N(\\mu=20,\\ \\sigma=5)$. Find the $90$th percentile, given $z_{0.90}=1.2816$.",
      "back": "The $90$th percentile satisfies $\\Phi(z)=0.90$ at $z=1.2816$. Unstandardize: $x=\\mu+z\\sigma=20+1.2816\\cdot 5=20+6.41=26.41$.",
      "tag": "normal"
    },
    {
      "front": "If $X\\sim N(\\mu_1,\\sigma_1^{2})$ and $Y\\sim N(\\mu_2,\\sigma_2^{2})$ are independent, what is the distribution of $X-Y$?",
      "back": "Linear combinations of independent normals are normal: $X-Y\\sim N(\\mu_1-\\mu_2,\\ \\sigma_1^{2}+\\sigma_2^{2})$. Means subtract but **variances add**.",
      "tag": "normal"
    },
    {
      "front": "Two independent normal claim amounts have $X\\sim N(100,400)$ and $Y\\sim N(120,225)$. Find $\\Pr(X>Y)$ using $\\Phi(-0.8)=0.2119$.",
      "back": "Let $D=X-Y\\sim N(100-120,\\ 400+225)=N(-20,625)$, so $\\sigma_D=25$. Then $\\Pr(X>Y)=\\Pr(D>0)=\\Pr\\!\\left(Z>\\dfrac{0-(-20)}{25}\\right)=\\Pr(Z>0.8)=1-\\Phi(0.8)=0.2119$.",
      "tag": "normal"
    },
    {
      "front": "State the pdf of a Beta$(\\alpha,\\beta)$ distribution on $[0,1]$.",
      "back": "$f(x)=\\dfrac{\\Gamma(\\alpha+\\beta)}{\\Gamma(\\alpha)\\Gamma(\\beta)}\\,x^{\\alpha-1}(1-x)^{\\beta-1}$ for $0<x<1$. The constant is $1/B(\\alpha,\\beta)$ where $B(\\alpha,\\beta)=\\dfrac{\\Gamma(\\alpha)\\Gamma(\\beta)}{\\Gamma(\\alpha+\\beta)}$.",
      "tag": "beta"
    },
    {
      "front": "Give the mean and variance of a Beta$(\\alpha,\\beta)$ distribution.",
      "back": "$E[X]=\\dfrac{\\alpha}{\\alpha+\\beta}$ and $\\mathrm{Var}(X)=\\dfrac{\\alpha\\beta}{(\\alpha+\\beta)^{2}(\\alpha+\\beta+1)}$.",
      "tag": "beta"
    },
    {
      "front": "What special distribution is Beta$(1,1)$, and why?",
      "back": "Beta$(1,1)$ is the **Uniform$(0,1)$** distribution. With $\\alpha=\\beta=1$ the pdf is $\\dfrac{\\Gamma(2)}{\\Gamma(1)\\Gamma(1)}x^{0}(1-x)^{0}=1$ on $[0,1]$, a flat density.",
      "tag": "beta"
    },
    {
      "front": "A proportion $X$ follows Beta$(\\alpha=2,\\beta=3)$. Compute $E[X]$ and $\\mathrm{Var}(X)$.",
      "back": "$E[X]=\\dfrac{2}{2+3}=\\dfrac{2}{5}=0.4$. $\\mathrm{Var}(X)=\\dfrac{2\\cdot 3}{(5)^{2}(5+1)}=\\dfrac{6}{25\\cdot 6}=\\dfrac{6}{150}=0.04$.",
      "tag": "beta"
    },
    {
      "front": "Using $\\int_0^1 x^{\\alpha-1}(1-x)^{\\beta-1}\\,dx=B(\\alpha,\\beta)$, evaluate $\\int_{0}^{1} x^{2}(1-x)^{3}\\,dx$.",
      "back": "Here $\\alpha-1=2$ and $\\beta-1=3$, so $\\alpha=3,\\ \\beta=4$. Then $B(3,4)=\\dfrac{\\Gamma(3)\\Gamma(4)}{\\Gamma(7)}=\\dfrac{2!\\cdot 3!}{6!}=\\dfrac{2\\cdot 6}{720}=\\dfrac{12}{720}=\\dfrac{1}{60}$.",
      "tag": "beta"
    },
    {
      "front": "A continuous variable has cdf $F(x)=1-e^{-x^{2}/2}$ for $x\\geq 0$ (a Rayleigh distribution). Find its pdf.",
      "back": "Differentiate: $f(x)=F'(x)=\\dfrac{d}{dx}\\bigl(1-e^{-x^{2}/2}\\bigr)=x\\,e^{-x^{2}/2}$ for $x\\geq 0$.",
      "tag": "pdf/cdf"
    },
    {
      "front": "A loss $X$ has pdf $f(x)=\\tfrac{1}{18}x$ on $0\\leq x\\leq 6$. Find the median.",
      "back": "First the cdf: $F(x)=\\int_0^x \\tfrac{1}{18}t\\,dt=\\dfrac{x^{2}}{36}$. Set $\\dfrac{m^{2}}{36}=0.5\\Rightarrow m^{2}=18\\Rightarrow m=\\sqrt{18}\\approx 4.24$.",
      "tag": "percentiles"
    },
    {
      "front": "$X$ has pdf $f(x)=6x(1-x)$ on $[0,1]$ (this is Beta$(2,2)$). Find the mode.",
      "back": "Maximize $f$: $f'(x)=6(1-2x)=0\\Rightarrow x=\\tfrac12$. Since $f''<0$, the mode is $x=0.5$. (By symmetry the mean and median are also $0.5$.)",
      "tag": "mode"
    },
    {
      "front": "A lifetime $X$ has pdf $f(x)=3x^{2}$ on $[0,1]$. Compute $E[X]$ and the $25$th percentile.",
      "back": "$E[X]=\\int_0^1 x\\cdot 3x^{2}\\,dx=3\\cdot\\dfrac{x^{4}}{4}\\Big|_0^1=\\dfrac34=0.75$. cdf $F(x)=x^{3}$; set $x^{3}=0.25\\Rightarrow x=0.25^{1/3}\\approx 0.630$.",
      "tag": "percentiles"
    },
    {
      "front": "A device lifetime is exponential with mean $4$ years. Find the median lifetime.",
      "back": "Rate $\\lambda=\\tfrac14$. Set $F(m)=1-e^{-m/4}=0.5\\Rightarrow e^{-m/4}=0.5\\Rightarrow m=-4\\ln(0.5)=4\\ln 2\\approx 2.77$ years. Note the median $<$ mean for this right-skewed distribution.",
      "tag": "exponential"
    },
    {
      "front": "$X\\sim$ Uniform$(0,10)$. Compute $E[X^{2}]$ and use it to confirm the variance.",
      "back": "$E[X^{2}]=\\int_0^{10} x^{2}\\cdot\\tfrac{1}{10}\\,dx=\\dfrac{1}{10}\\cdot\\dfrac{x^{3}}{3}\\Big|_0^{10}=\\dfrac{1000}{30}=\\dfrac{100}{3}$. Then $\\mathrm{Var}(X)=\\dfrac{100}{3}-5^{2}=\\dfrac{100}{3}-25=\\dfrac{25}{3}\\approx 8.33$, matching $\\dfrac{(10-0)^{2}}{12}$.",
      "tag": "uniform"
    },
    {
      "front": "How does a linear shift and scale affect the mean and standard deviation of a continuous variable?",
      "back": "For constants $a,b$: $E[aX+b]=aE[X]+b$ and $\\mathrm{Var}(aX+b)=a^{2}\\mathrm{Var}(X)$, so $\\mathrm{SD}(aX+b)=|a|\\,\\sigma$. A pure shift $b$ leaves variance unchanged.",
      "tag": "pdf/cdf"
    }
  ]
}