Continuous random variables: densities, CDFs, expectation and the uniform, exponential and normal

Worked solutions

Problem 1

1. $\int_0^3 cx^2\,dx = c\cdot\dfrac{27}{3} = 9c = 1$, so $c = \tfrac19$The density must integrate to $1$.
2. $F(x) = \int_0^x \dfrac{t^2}{9}\,dt = \dfrac{x^3}{27}$ for $0 \le x \le 3$$F = 0$ below $0$ and $F = 1$ above $3$; $F(3) = 1$ is a check on $c$.
3. $P(1 < X < 2) = \int_1^2 \dfrac{x^2}{9}\,dx = \dfrac{8 - 1}{27}$From the density.
4. $F(2) - F(1) = \dfrac{8}{27} - \dfrac{1}{27} = \dfrac{7}{27}$The same number from the CDF, with no integral left to do.
5. $c = \tfrac19$, $F(x) = \tfrac{x^3}{27}$ on $[0, 3]$, $P(1 < X < 2) = \tfrac{7}{27} \approx 0.2593$The middle third of the range holds about a quarter of the probability, because the density grows like $x^2$.

Problem 2

1. $f(x) = F'(x) = \dfrac{2}{x^3}$ for $x \ge 1$Differentiate the CDF.
2. $P(2 < X < 4) = F(4) - F(2) = \Big(1 - \dfrac1{16}\Big) - \Big(1 - \dfrac14\Big) = \dfrac{3}{16} = 0.1875$Differences of the CDF.
3. $1 - \dfrac{1}{m^2} = \dfrac12$, so $m^2 = 2$ and $m = \sqrt2 \approx 1.4142$The median solves $F(m) = \tfrac12$.
4. $E[X] = \int_1^\infty x\cdot\dfrac{2}{x^3}\,dx = \int_1^\infty \dfrac{2}{x^2}\,dx = 2$An improper integral that converges.
5. $f(x) = \tfrac{2}{x^3}$, $P(2 < X < 4) = 0.1875$, median $\sqrt2 \approx 1.414$, $E[X] = 2$, $E[X^2] = \infty$$E[X^2] = \int_1^\infty \tfrac2x\,dx$ diverges, so this heavy-tailed (Pareto) variable has a mean but no finite variance. The mean is pulled above the median by the long right tail.

Problem 3

1. $\int_0^1 2x\,dx = 1$Check it is a density.
2. $E[X] = \int_0^1 x\cdot2x\,dx = \dfrac23$$\int_0^1 2x^2\,dx = \tfrac23$.
3. $E[X^2] = \int_0^1 x^2\cdot2x\,dx = \dfrac24 = \dfrac12$Square the value, keep the density as the weight.
4. $\operatorname{Var}(X) = \dfrac12 - \dfrac49 = \dfrac{1}{18}$The same shortcut as in the discrete case.
5. $E[X] = \tfrac23$, $E[X^2] = \tfrac12$, $\operatorname{Var}(X) = \tfrac{1}{18} \approx 0.0556$, standard deviation $\approx 0.2357$, median $\tfrac{1}{\sqrt2} \approx 0.7071$$F(x) = x^2$, so $F(m) = \tfrac12$ gives $m = \tfrac{1}{\sqrt2}$. The density leans right, so the median is above the mean: the long tail is on the left.

Problem 4

1. $f(x) = \dfrac{1}{8 - 2} = \dfrac16$ on $(2, 8)$Constant height, area $1$.
2. $P(3 < X < 5) = \dfrac{5 - 3}{6} = \dfrac13$For a uniform, probability is length over total length.
3. $E[X] = \dfrac{2 + 8}{2} = 5$ and $\operatorname{Var}(X) = \dfrac{(8 - 2)^2}{12} = 3$From $E[X^2] = \int_2^8 \tfrac{x^2}{6}\,dx = \tfrac{512 - 8}{18} = 28$ and $28 - 25 = 3$.
4. $f_W(w) = \dfrac{1}{0.5} = 2$ on $(0, 0.5)$A density of $2$ is fine: the interval is only $0.5$ long, so the area is still $1$.
5. $f = \tfrac16$, $P(3 < X < 5) = \tfrac13$, $E[X] = 5$, $\operatorname{Var}(X) = 3$; $f_W = 2$, $P(0.3 < W < 0.31) = 0.02$$2 \times 0.01 = 0.02$: the density times a short width approximates the probability of a short interval (Mistake 1).

Problem 5

1. $E[X] = \dfrac{1}{0.5} = 2$ minutesHalf an arrival per minute is one every two minutes on average.
2. $P(X > 4) = e^{-0.5\cdot4} = e^{-2} \approx 0.1353$The survival function $e^{-\lambda x}$.
3. $1 - e^{-0.5m} = \dfrac12$, so $m = \dfrac{\ln 2}{0.5} = 2\ln2 \approx 1.3863$ minutesThe median solves $F(m) = \tfrac12$.
4. $P(X > 6 \mid X > 2) = \dfrac{e^{-3}}{e^{-1}} = e^{-2} \approx 0.1353$$X > 6$ is contained in $X > 2$, so the intersection is $X > 6$.
5. Mean $2$ minutes, $P(X > 4) \approx 0.1353$, median $\approx 1.386$ minutes, $P(X > 6 \mid X > 2) \approx 0.1353$Memorylessness: after two minutes with nothing, the chance of waiting four more is the same as it was at the start. The continuous version of the geometric distribution.

Problem 6

1. $E[X] = \int_0^\infty x\lambda e^{-\lambda x}\,dx = \Big[-xe^{-\lambda x}\Big]_0^\infty + \int_0^\infty e^{-\lambda x}\,dx$Parts with $u = x$ and $dv = \lambda e^{-\lambda x}\,dx$, so $v = -e^{-\lambda x}$.
2. $E[X] = 0 + \dfrac1\lambda$$xe^{-\lambda x} \to 0$ as $x \to \infty$, and $\int_0^\infty e^{-\lambda x}\,dx = \tfrac1\lambda$.
3. $E[X^2] = \int_0^\infty x^2\lambda e^{-\lambda x}\,dx = 0 + \int_0^\infty 2xe^{-\lambda x}\,dx = \dfrac2\lambda E[X] = \dfrac{2}{\lambda^2}$Parts again with $u = x^2$; the remaining integral is $\tfrac2\lambda$ times the one in step 1.
4. $\operatorname{Var}(X) = \dfrac{2}{\lambda^2} - \dfrac{1}{\lambda^2} = \dfrac{1}{\lambda^2}$The shortcut.
5. $E[X] = \tfrac1\lambda$ and $\operatorname{Var}(X) = \tfrac{1}{\lambda^2}$The standard deviation equals the mean. For $\lambda = 0.5$ in Problem 5, both are $2$ minutes.

Problem 7

1. $z = \dfrac{130 - 100}{15} = 2$Standardize: subtract the mean, divide by the standard deviation.
2. $P(X > 130) = 1 - \Phi(2) = 1 - 0.97725 \approx 0.0228$About $2.3\%$.
3. $P(85 < X < 115) = \Phi(1) - \Phi(-1) = 2\Phi(1) - 1 \approx 0.6827$$z = \pm1$; by symmetry $\Phi(-1) = 1 - \Phi(1)$.
4. $x = 100 + 15 \times 1.2816 \approx 119.22$Solve $\Phi(z) = 0.9$ for $z = 1.2816$, then undo the standardization: $x = \mu + \sigma z$.
5. $P(X > 130) \approx 0.0228$, $P(85 < X < 115) \approx 0.6827$, 90th percentile $\approx 119.2$Within one standard deviation of the mean lies about $68\%$ of a normal distribution, and beyond two standard deviations on one side lies about $2.3\%$.

Problem 8

1. $F_Y(y) = P(X^2 \le y) = P(X \le \sqrt y) = \sqrt y$ for $0 \le y \le 1$$X \ge 0$, so $X^2 \le y$ exactly when $X \le \sqrt y$.
2. $f_Y(y) = \dfrac{d}{dy}\sqrt y = \dfrac{1}{2\sqrt y}$ on $(0, 1)$Equivalently $f_X(\sqrt y)\,\big|\tfrac{dx}{dy}\big| = 1\cdot\tfrac{1}{2\sqrt y}$. It is unbounded near $0$ and still integrates to $1$.
3. $P(Y < 0.25) = F_Y(0.25) = 0.5$$Y < 0.25$ exactly when $X < 0.5$.
4. $E[Y] = \int_0^1 y\cdot\dfrac{1}{2\sqrt y}\,dy = \int_0^1 \dfrac{\sqrt y}{2}\,dy = \dfrac13$From the density of $Y$.
5. $F_Y(y) = \sqrt y$, $f_Y(y) = \tfrac{1}{2\sqrt y}$ on $(0, 1)$, $P(Y < 0.25) = 0.5$, $E[Y] = \tfrac13$Check without $f_Y$: $E[X^2] = \int_0^1 x^2\,dx = \tfrac13$. Squaring numbers in $(0, 1)$ crowds them toward $0$, which is where the density piles up.

Problem 9

1. $P(Y > y) = P\big(-\ln U > \lambda y\big) = P\big(U < e^{-\lambda y}\big)$Multiply by $-1$ and exponentiate; $e^x$ is increasing, so the inequality keeps its direction.
2. $P(Y > y) = e^{-\lambda y}$ for $y \ge 0$For a uniform, $P(U < u) = u$. That is the exponential survival function.
3. $U = 0.3$, $\lambda = 2$: $y = -\dfrac{\ln 0.3}{2} \approx 0.6020$One uniform random number becomes one exponential sample.
4. In general $X = F^{-1}(U)$ has CDF $F$: $P(F^{-1}(U) \le x) = P(U \le F(x)) = F(x)$Inverse-CDF sampling. For the exponential, $F^{-1}(u) = -\tfrac{\ln(1 - u)}{\lambda}$, and $1 - U$ is uniform too, which gives the formula above.
5. $Y = -\tfrac{\ln U}{\lambda}$ is Exponential$(\lambda)$, with $y \approx 0.602$ for $U = 0.3$; for Problem 2, $X = \tfrac{1}{\sqrt{1 - U}}$, which is $2$ at $U = 0.75$Solve $u = 1 - \tfrac{1}{x^2}$ for $x$. This is how libraries turn a uniform random number generator into samples from other distributions.

Problem 10

1. $P(T > t) = P(X_1 > t)P(X_2 > t)P(X_3 > t) = e^{-t}e^{-2t}e^{-3t} = e^{-6t}$The minimum is above $t$ only if all three are; independence lets the probabilities multiply.
2. $T \sim \operatorname{Exponential}(6)$, so $E[T] = \dfrac16$ secondThe rates add.
3. $P(T > 0.5) = e^{-3} \approx 0.0498$The survival function at $0.5$.
4. $P(X_1 = T) = \int_0^\infty e^{-t}\cdot e^{-5t}\,dt = \dfrac{1}{6}$Replica 1 answers at $t$ (density $e^{-t}$) and the other two are still busy (probability $e^{-2t}e^{-3t}$).
5. $T \sim \operatorname{Exponential}(6)$, $E[T] = \tfrac16 \approx 0.167$ s, $P(T > 0.5) \approx 0.0498$, $P(\text{replica 1 first}) = \tfrac16$In general each replica wins with probability its rate over the total, $\tfrac{\lambda_i}{\lambda_1 + \lambda_2 + \lambda_3}$, so the others win with $\tfrac26$ and $\tfrac36$.