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Exam P — Transformations & Order Statistics Practice Flashcards

Thirty-three original SOA/CAS Exam P-style multiple-choice problems on transformations of random variables (CDF method, Jacobian, bivariate change of variables, convolutions) and order statistics (minima, maxima, k-th order statistic, Beta and exponential cases).

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Browse all 33 problems as a list
  1. CDF method
    Let $X$ be uniformly distributed on $(0,3)$ and let $Y=X^{2}$. Determine $f_Y(y)$ for $0<y<9$. (A) $\dfrac{1}{6}$ (B) $\dfrac{1}{6\sqrt{y}}$ (C) $\dfrac{1}{3\sqrt{y}}$ (D) $\dfrac{\sqrt{y}}{3}$ (E) $\dfrac{1}{3}$
    **Answer: (B).** Use the CDF method. Since $X\sim U(0,3)$, $F_X(x)=\dfrac{x}{3}$ for $0<x<3$. For $0<y<9$: $F_Y(y)=\Pr(X^{2}\leq y)=\Pr(X\leq\sqrt{y})=\frac{\sqrt{y}}{3}.$ Differentiate: $f_Y(y)=\frac{d}{dy}\frac{\sqrt{y}}{3}=\frac{1}{3}\cdot\frac{1}{2\sqrt{y}}=\frac{1}{6\sqrt{y}},\qquad 0<y<9.$ Distractor (C) forgets the $\tfrac12$ from the chain rule; (E) forgets the Jacobian factor entirely.
  2. Jacobian
    Let $X$ have density $f_X(x)=2x$ for $0<x<1$ and let $Y=\sqrt{X}$. Determine $f_Y(y)$ for $0<y<1$. (A) $2y$ (B) $2y^{3}$ (C) $y^{3}$ (D) $4y^{2}$ (E) $4y^{3}$
    **Answer: (E).** The transformation $y=\sqrt{x}$ is one-to-one on $(0,1)$. Invert: $x=y^{2}$, so $\dfrac{dx}{dy}=2y$. By the Jacobian formula: $f_Y(y)=f_X(y^{2})\left|\frac{dx}{dy}\right|=2(y^{2})\cdot 2y=4y^{3},\qquad 0<y<1.$ Check: $\int_0^1 4y^{3}\,dy=1$. Distractor (B) drops the factor of $2$ in $\tfrac{dx}{dy}$; (A) forgets the Jacobian.
  3. Linear transform
    Let $X$ be an exponential random variable with mean $2$, so $f_X(x)=\tfrac{1}{2}e^{-x/2}$ for $x>0$. Let $Y=3X+1$. Determine $f_Y(y)$ for $y>1$. (A) $\dfrac{1}{6}e^{-(y-1)/6}$ (B) $\dfrac{1}{6}e^{-(y-1)/2}$ (C) $\dfrac{1}{2}e^{-(y-1)/2}$ (D) $\dfrac{1}{2}e^{-(y-1)/6}$ (E) $\dfrac{3}{2}e^{-3(y-1)/2}$
    **Answer: (A).** The map $y=3x+1$ is linear and one-to-one. Invert: $x=\dfrac{y-1}{3}$, so $\dfrac{dx}{dy}=\dfrac{1}{3}$. $f_Y(y)=f_X\!\left(\frac{y-1}{3}\right)\left|\frac{dx}{dy}\right|=\frac{1}{2}e^{-\frac{1}{2}\cdot\frac{y-1}{3}}\cdot\frac{1}{3}=\frac{1}{6}e^{-(y-1)/6},\qquad y>1.$ This is exponential (shifted by $1$) with mean $6=3\cdot 2$, as expected from scaling. Distractor (D) forgets the $\tfrac13$ Jacobian; (B) forgets to rescale the exponent.
  4. CDF method
    Let $U$ be uniformly distributed on $(0,1)$. Define $Y=-4\ln U$. Calculate $\Pr(Y>6)$. (A) $0.050$ (B) $0.135$ (C) $0.223$ (D) $0.472$ (E) $0.777$
    **Answer: (C).** Find the distribution of $Y$ by the CDF method. For $y>0$: $F_Y(y)=\Pr(-4\ln U\leq y)=\Pr\!\left(\ln U\geq-\frac{y}{4}\right)=\Pr\!\left(U\geq e^{-y/4}\right)=1-e^{-y/4}.$ So $Y\sim\text{Exp}$ with mean $4$ (rate $\tfrac14$). Then $\Pr(Y>6)=e^{-6/4}=e^{-1.5}\approx 0.2231.$ Distractor (B) $=e^{-2}$ uses mean $3$; (A) $=e^{-3}$ uses mean $2$.
  5. Non-monotone
    Let $X$ be uniformly distributed on $(-2,2)$ and let $Y=X^{2}$. Determine $f_Y(y)$ for $0<y<4$. (A) $\dfrac{1}{8\sqrt{y}}$ (B) $\dfrac{1}{4\sqrt{y}}$ (C) $\dfrac{1}{2\sqrt{y}}$ (D) $\dfrac{1}{4}$ (E) $\dfrac{\sqrt{y}}{4}$
    **Answer: (B).** Here $g(x)=x^{2}$ is **not** monotone on $(-2,2)$; both $x=\pm\sqrt{y}$ map to the same $y$, so sum over both branches. The density of $X$ is $f_X(x)=\dfrac{1}{4}$ on $(-2,2)$, and $\left|\dfrac{dx}{dy}\right|=\dfrac{1}{2\sqrt{y}}$. $f_Y(y)=f_X(\sqrt{y})\frac{1}{2\sqrt{y}}+f_X(-\sqrt{y})\frac{1}{2\sqrt{y}}=\frac{1}{4}\cdot\frac{1}{2\sqrt{y}}+\frac{1}{4}\cdot\frac{1}{2\sqrt{y}}=\frac{1}{4\sqrt{y}},\qquad 0<y<4.$ Distractor (A) uses only one branch.
  6. Jacobian
    Let $X$ have density $f_X(x)=\dfrac{1}{x^{2}}$ for $x>1$ and let $Y=\ln X$. Determine $f_Y(y)$ for $y>0$. (A) $e^{-2y}$ (B) $e^{-y}$ (C) $2e^{-2y}$ (D) $\dfrac{1}{y^{2}}$ (E) $y e^{-y}$
    **Answer: (B).** The map $y=\ln x$ is one-to-one for $x>1$ (giving $y>0$). Invert: $x=e^{y}$, so $\dfrac{dx}{dy}=e^{y}$. $f_Y(y)=f_X(e^{y})\left|\frac{dx}{dy}\right|=\frac{1}{(e^{y})^{2}}\cdot e^{y}=e^{-2y}\cdot e^{y}=e^{-y},\qquad y>0.$ So $Y\sim\text{Exp}(1)$. Distractor (A) forgets the Jacobian factor $e^{y}$; (D) substitutes naively without transforming.
  7. CDF method
    An insurance loss $X$ (in thousands) has cdf $F_X(x)=1-\dfrac{16}{(x+4)^{2}}$ for $x>0$. An actuary simulates losses by drawing $U\sim U(0,1)$ and setting $X=F_X^{-1}(U)$. Which formula correctly generates $X$? (A) $X=\dfrac{4}{\sqrt{1-U}}-4$ (B) $X=\dfrac{4}{\sqrt{U}}-4$ (C) $X=4\sqrt{1-U}-4$ (D) $X=\dfrac{16}{1-U}-4$ (E) $X=\sqrt{\dfrac{16}{1-U}}+4$
    **Answer: (A).** By the inverse-transform (probability integral transform) method, set $U=F_X(X)$ and solve for $X$: $U=1-\frac{16}{(X+4)^{2}}\implies \frac{16}{(X+4)^{2}}=1-U\implies (X+4)^{2}=\frac{16}{1-U}.$ Take the positive root (since $X+4>0$): $X+4=\frac{4}{\sqrt{1-U}}\implies X=\frac{4}{\sqrt{1-U}}-4.$ Distractor (B) sets $U=1-F_X(X)$ by mistake; (D) forgets to take the square root.
  8. Jacobian
    Let $Z\sim N(0,1)$ and let $Y=e^{Z}$ (so $Y$ is lognormal). Which expression gives $f_Y(y)$ for $y>0$? (A) $\dfrac{1}{\sqrt{2\pi}}e^{-(\ln y)^{2}/2}$ (B) $\dfrac{1}{y\sqrt{2\pi}}e^{-(\ln y)^{2}/2}$ (C) $\dfrac{1}{y\sqrt{2\pi}}e^{-y^{2}/2}$ (D) $\dfrac{y}{\sqrt{2\pi}}e^{-(\ln y)^{2}/2}$ (E) $\dfrac{1}{\sqrt{2\pi}}e^{-y^{2}/2}$
    **Answer: (B).** The map $y=e^{z}$ is one-to-one. Invert: $z=\ln y$, so $\dfrac{dz}{dy}=\dfrac{1}{y}$. $f_Y(y)=f_Z(\ln y)\left|\frac{dz}{dy}\right|=\frac{1}{\sqrt{2\pi}}e^{-(\ln y)^{2}/2}\cdot\frac{1}{y}=\frac{1}{y\sqrt{2\pi}}e^{-(\ln y)^{2}/2},\qquad y>0.$ Distractor (A) forgets the $\tfrac{1}{y}$ Jacobian; (C) and (E) keep $z^{2}=y^{2}$ instead of $(\ln y)^{2}$.