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Exam P — Continuous Distributions Flashcards

A comprehensive Exam P deck on continuous random variables: pdf/cdf/survival/hazard, and the uniform, exponential, gamma, normal, and beta distributions, with percentiles, modes, memorylessness, and worked probability-by-integration problems.

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Common Traps Conditional Probability & Bayes Conditional Probability & Bayes Practice Continuous Distributions Practice Covariance, Sums & CLT Covariance, Sums & CLT Practice

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  1. pdf/cdf
    For a continuous random variable, how is the cumulative distribution function $F(x)$ defined in terms of the pdf $f$?
    $F(x)=\Pr(X\leq x)=\int_{-\infty}^{x} f(t)\,dt$. Conversely the pdf is the derivative of the cdf, $f(x)=F'(x)$, wherever $F$ is differentiable.
  2. pdf/cdf
    What two conditions must a function $f$ satisfy to be a valid continuous pdf?
    It must be nonnegative everywhere, $f(x)\geq 0$, and it must integrate to one over the whole real line, $\int_{-\infty}^{\infty} f(x)\,dx=1$.
  3. pdf/cdf
    Define the survival function $S(x)$ and give its relationship to $F(x)$.
    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 $S'(x)=-f(x)$.
  4. hazard
    Define the hazard rate (failure rate) $h(x)$ of a continuous random variable.
    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 $S(0)=1$, $S(x)=\exp\!\left(-\int_{0}^{x} h(t)\,dt\right)$.
  5. hazard
    A continuous variable has constant hazard rate $h(x)=\lambda$ for $x\geq 0$. What is its distribution?
    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.
  6. pdf/cdf
    For a nonnegative continuous random variable, express $E[X]$ as an integral of the survival function.
    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.
  7. computing-probabilities
    Given the pdf $f(x)=c\,x^{2}$ on $0\leq x\leq 3$ and zero elsewhere, find $c$.
    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 $c=\dfrac{1}{9}$.
  8. computing-probabilities
    With pdf $f(x)=\tfrac{1}{9}x^{2}$ on $[0,3]$, compute $\Pr(1\leq X\leq 2)$.
    $\Pr(1\leq X\leq 2)=\int_{1}^{2}\tfrac{1}{9}x^{2}\,dx=\dfrac{1}{9}\cdot\dfrac{x^{3}}{3}\Big|_{1}^{2}=\dfrac{1}{27}(8-1)=\dfrac{7}{27}\approx 0.259$.
  9. percentiles
    How do you find the median of a continuous random variable from its cdf?
    The median $m$ solves $F(m)=0.5$, i.e. $\Pr(X\leq m)=\tfrac12$. More generally the $p$-th percentile $x_p$ solves $F(x_p)=p$.
  10. mode
    How is the mode of a continuous distribution found?
    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).
  11. uniform
    State the pdf and cdf of a continuous Uniform$(a,b)$ distribution.
    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 $b$.
  12. uniform
    Give the mean and variance of a continuous Uniform$(a,b)$ random variable.
    $E[X]=\dfrac{a+b}{2}$ and $\mathrm{Var}(X)=\dfrac{(b-a)^{2}}{12}$.
  13. uniform
    $X\sim$ Uniform$(2,10)$. Find $\Pr(X>7)$ and the variance of $X$.
    $\Pr(X>7)=\dfrac{10-7}{10-2}=\dfrac{3}{8}=0.375$. Variance: $\mathrm{Var}(X)=\dfrac{(10-2)^{2}}{12}=\dfrac{64}{12}=\dfrac{16}{3}\approx 5.33$.
  14. uniform
    $X\sim$ Uniform$(0,1)$. What is the distribution of $Y=-\tfrac{1}{\lambda}\ln X$?
    $Y$ 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.
  15. exponential
    State the pdf, cdf, and survival function of an Exponential distribution with rate $\lambda$.
    pdf: $f(x)=\lambda e^{-\lambda x}$ for $x\geq 0$. cdf: $F(x)=1-e^{-\lambda x}$. Survival: $S(x)=e^{-\lambda x}$.
  16. exponential
    Give the mean and variance of an Exponential distribution, both in rate form $\lambda$ and in mean (scale) form $\theta$.
    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 $\mathrm{Var}(X)=\theta^{2}$.
  17. exponential
    State the memoryless property of the exponential distribution.
    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.
  18. exponential
    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.
    By memorylessness the past $3$ years are irrelevant: $\Pr(X>3+4\mid X>3)=\Pr(X>4)=e^{-4/5}=e^{-0.8}\approx 0.449$.
  19. exponential
    Claim sizes are exponential with mean $\theta=200$. Find $\Pr(X>300)$.
    Rate $\lambda=1/200$. $\Pr(X>300)=S(300)=e^{-300/200}=e^{-1.5}\approx 0.223$.
  20. exponential
    If $X_1,\dots,X_n$ are independent exponentials with rates $\lambda_1,\dots,\lambda_n$, what is the distribution of $\min(X_1,\dots,X_n)$?
    The minimum is **exponential with rate $\sum_{i=1}^{n}\lambda_i$**. Reason: $\Pr(\min>x)=\prod_i \Pr(X_i>x)=\prod_i e^{-\lambda_i x}=e^{-(\sum\lambda_i)x}$.
  21. exponential
    Three independent components have exponential lifetimes with means $10$, $20$, and $40$ hours. What is the mean time until the first failure?
    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.
  22. exponential
    For an exponential loss with mean $\theta$, what is the expected payment per loss under an ordinary deductible $d$?
    $E[(X-d)_+]=\int_{d}^{\infty} S(x)\,dx=\int_{d}^{\infty} e^{-x/\theta}\,dx=\theta\,e^{-d/\theta}$. The memoryless property makes the answer the survival probability $e^{-d/\theta}$ times the full mean $\theta$.
  23. gamma
    State the pdf of a Gamma distribution with shape $\alpha$ and rate $\lambda$.
    $f(x)=\dfrac{\lambda^{\alpha}}{\Gamma(\alpha)}\,x^{\alpha-1}e^{-\lambda x}$ for $x>0$, where $\Gamma(\alpha)=\int_{0}^{\infty} t^{\alpha-1}e^{-t}\,dt$ is the gamma function.
  24. gamma
    Give the mean and variance of a Gamma$(\alpha,\lambda)$ distribution (rate parameterization).
    $E[X]=\dfrac{\alpha}{\lambda}$ and $\mathrm{Var}(X)=\dfrac{\alpha}{\lambda^{2}}$. With scale $\theta=1/\lambda$: $E[X]=\alpha\theta$ and $\mathrm{Var}(X)=\alpha\theta^{2}$.
  25. gamma
    How does the Gamma distribution relate to the exponential and the Erlang distributions?
    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 $\lambda$.
  26. gamma
    Why is the sum of $n$ independent Exponential$(\lambda)$ variables a Gamma, and what are its parameters?
    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 $n/\lambda$.
  27. gamma
    Key gamma-function facts to memorize for Exam P.
    $\Gamma(\alpha+1)=\alpha\,\Gamma(\alpha)$ (recursion); $\Gamma(n)=(n-1)!$ for a positive integer $n$; and $\Gamma\!\left(\tfrac12\right)=\sqrt{\pi}$.
  28. gamma
    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?
    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.
  29. gamma
    How is the chi-square distribution related to the Gamma distribution?
    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 $2k$.
  30. gamma
    $X\sim$ Gamma with shape $\alpha=2$ and rate $\lambda=\tfrac13$. Compute $E[X]$ and $\mathrm{SD}(X)$.
    $E[X]=\dfrac{\alpha}{\lambda}=\dfrac{2}{1/3}=6$. $\mathrm{Var}(X)=\dfrac{\alpha}{\lambda^{2}}=\dfrac{2}{(1/3)^{2}}=2\cdot 9=18$, so $\mathrm{SD}(X)=\sqrt{18}\approx 4.24$.
  31. normal
    State the pdf of a Normal$(\mu,\sigma^{2})$ distribution.
    $f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\exp\!\left(-\dfrac{(x-\mu)^{2}}{2\sigma^{2}}\right)$ for all real $x$, where $\mu=E[X]$ and $\sigma^{2}=\mathrm{Var}(X)$.
  32. normal
    How do you standardize a Normal$(\mu,\sigma^{2})$ variable, and what does the normal table give?
    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 $\sigma^{2}$.
  33. normal
    State the empirical (68–95–99.7) rule for a normal distribution.
    Approximately $68\%$ of the probability lies within $\mu\pm\sigma$, about $95\%$ within $\mu\pm 2\sigma$, and about $99.7\%$ within $\mu\pm 3\sigma$.
  34. normal
    $X\sim N(\mu=50,\ \sigma^{2}=16)$. Find $\Pr(X\leq 58)$ using $\Phi(2)=0.9772$.
    $\sigma=4$, so $Z=\dfrac{58-50}{4}=2$. Thus $\Pr(X\leq 58)=\Phi(2)=0.9772$.
  35. normal
    $X\sim N(\mu=100,\ \sigma=15)$. Find $\Pr(85\leq X\leq 130)$ using $\Phi(1)=0.8413,\ \Phi(2)=0.9772$.
    Standardize: lower $z=\dfrac{85-100}{15}=-1$, upper $z=\dfrac{130-100}{15}=2$. Then $\Pr=\Phi(2)-\Phi(-1)=0.9772-(1-0.8413)=0.9772-0.1587=0.8185$.
  36. normal
    $X\sim N(\mu=20,\ \sigma=5)$. Find the $90$th percentile, given $z_{0.90}=1.2816$.
    The $90$th percentile satisfies $\Phi(z)=0.90$ at $z=1.2816$. Unstandardize: $x=\mu+z\sigma=20+1.2816\cdot 5=20+6.41=26.41$.
  37. normal
    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 $X-Y$?
    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**.
  38. normal
    Two independent normal claim amounts have $X\sim N(100,400)$ and $Y\sim N(120,225)$. Find $\Pr(X>Y)$ using $\Phi(-0.8)=0.2119$.
    Let $D=X-Y\sim N(100-120,\ 400+225)=N(-20,625)$, so $\sigma_D=25$. Then $\Pr(X>Y)=\Pr(D>0)=\Pr\!\left(Z>\dfrac{0-(-20)}{25}\right)=\Pr(Z>0.8)=1-\Phi(0.8)=0.2119$.
  39. beta
    State the pdf of a Beta$(\alpha,\beta)$ distribution on $[0,1]$.
    $f(x)=\dfrac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\,x^{\alpha-1}(1-x)^{\beta-1}$ for $0<x<1$. The constant is $1/B(\alpha,\beta)$ where $B(\alpha,\beta)=\dfrac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$.
  40. beta
    Give the mean and variance of a Beta$(\alpha,\beta)$ distribution.
    $E[X]=\dfrac{\alpha}{\alpha+\beta}$ and $\mathrm{Var}(X)=\dfrac{\alpha\beta}{(\alpha+\beta)^{2}(\alpha+\beta+1)}$.
  41. beta
    What special distribution is Beta$(1,1)$, and why?
    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.
  42. beta
    A proportion $X$ follows Beta$(\alpha=2,\beta=3)$. Compute $E[X]$ and $\mathrm{Var}(X)$.
    $E[X]=\dfrac{2}{2+3}=\dfrac{2}{5}=0.4$. $\mathrm{Var}(X)=\dfrac{2\cdot 3}{(5)^{2}(5+1)}=\dfrac{6}{25\cdot 6}=\dfrac{6}{150}=0.04$.
  43. beta
    Using $\int_0^1 x^{\alpha-1}(1-x)^{\beta-1}\,dx=B(\alpha,\beta)$, evaluate $\int_{0}^{1} x^{2}(1-x)^{3}\,dx$.
    Here $\alpha-1=2$ and $\beta-1=3$, so $\alpha=3,\ \beta=4$. Then $B(3,4)=\dfrac{\Gamma(3)\Gamma(4)}{\Gamma(7)}=\dfrac{2!\cdot 3!}{6!}=\dfrac{2\cdot 6}{720}=\dfrac{12}{720}=\dfrac{1}{60}$.
  44. pdf/cdf
    A continuous variable has cdf $F(x)=1-e^{-x^{2}/2}$ for $x\geq 0$ (a Rayleigh distribution). Find its pdf.
    Differentiate: $f(x)=F'(x)=\dfrac{d}{dx}\bigl(1-e^{-x^{2}/2}\bigr)=x\,e^{-x^{2}/2}$ for $x\geq 0$.
  45. percentiles
    A loss $X$ has pdf $f(x)=\tfrac{1}{18}x$ on $0\leq x\leq 6$. Find the median.
    First the cdf: $F(x)=\int_0^x \tfrac{1}{18}t\,dt=\dfrac{x^{2}}{36}$. Set $\dfrac{m^{2}}{36}=0.5\Rightarrow m^{2}=18\Rightarrow m=\sqrt{18}\approx 4.24$.
  46. mode
    $X$ has pdf $f(x)=6x(1-x)$ on $[0,1]$ (this is Beta$(2,2)$). Find the mode.
    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 $0.5$.)
  47. percentiles
    A lifetime $X$ has pdf $f(x)=3x^{2}$ on $[0,1]$. Compute $E[X]$ and the $25$th percentile.
    $E[X]=\int_0^1 x\cdot 3x^{2}\,dx=3\cdot\dfrac{x^{4}}{4}\Big|_0^1=\dfrac34=0.75$. cdf $F(x)=x^{3}$; set $x^{3}=0.25\Rightarrow x=0.25^{1/3}\approx 0.630$.
  48. exponential
    A device lifetime is exponential with mean $4$ years. Find the median lifetime.
    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.
  49. uniform
    $X\sim$ Uniform$(0,10)$. Compute $E[X^{2}]$ and use it to confirm the variance.
    $E[X^{2}]=\int_0^{10} x^{2}\cdot\tfrac{1}{10}\,dx=\dfrac{1}{10}\cdot\dfrac{x^{3}}{3}\Big|_0^{10}=\dfrac{1000}{30}=\dfrac{100}{3}$. Then $\mathrm{Var}(X)=\dfrac{100}{3}-5^{2}=\dfrac{100}{3}-25=\dfrac{25}{3}\approx 8.33$, matching $\dfrac{(10-0)^{2}}{12}$.
  50. pdf/cdf
    How does a linear shift and scale affect the mean and standard deviation of a continuous variable?
    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.