Banks Problem 8.1

\[ f(x) = \begin{cases} e^{2x} & -\infty \lt x \le 0 \\ e^{-2x} & 0 \lt x \lt \infty \end{cases} \]

\[ F(x) = \begin{cases} \frac{e^{2x}}{2} & -\infty \lt x \le 0 \\ \frac{-e^{-2x}}{2} & 0 \lt x \lt \infty \end{cases} \]

\[ X = \begin{cases} \frac{-ln(-2R)}{2} & \frac{-1}{2} \lt R \lt 0 \\ \frac{ln(2R)}{2} & 0 \lt R \lt \frac{1}{2} \end{cases} \]

x1 = seq(-5, 0, 0.01)
x2 = seq(0, 5, 0.01)
plot(c(exp(2*x1), exp(-2*x2)) ~ c(x1, x2))

R = runif(100000, -0.5, 0.5)
x = ifelse(R < 0, -log(-2*R)/2, log(2*R)/2)
## Warning in log(-2 * R): NaNs produced
## Warning in log(2 * R): NaNs produced
par(mfrow = c(1, 2))
hist(x)
plot(ecdf(x))

Banks Problem 8.5

\[ F(x) = \begin{cases} 0 & x \le -3 \\ \frac{1}{2} + \frac{x}{6} & -3 \lt x \le 0 \\ \frac{1}{2} + \frac{x^2}{32} & 0 \lt x \le 4 \\ 1 & x \gt 4 \end{cases} \]

\[ X = \begin{cases} 6R - 3 & 0 \lt R \le 1/2 \\ 4\sqrt{2R - 1 } & 1/2 \lt R \le 1 \end{cases} \]

R = runif(100000, 0, 1)
x = ifelse(R <= 0.5, 6*R - 3, 4*sqrt(2*R -1))
## Warning in sqrt(2 * R - 1): NaNs produced
par(mfrow = c(1, 2))
hist(x)
plot(ecdf(x))