1. **State the problem:** We have a random variable $X$ with pdf $$ f(x) = \begin{cases} e^{2x} & \text{if } x < 0 \\ e^{-2x} & \text{if } x \geq 0 \end{cases} $$ We want to: (i) Develop a random variate generator for $X$. (ii) Use this generator to transform given random numbers $0.129, 0.974, 0.477, 0.852, 0.508, 0.021$. 2. **Find the CDF $F(x)$:** For $x < 0$: $$ F(x) = P(X \leq x) = \int_{-\infty}^x e^{2t} dt = \left[ \frac{e^{2t}}{2} \right]_{-\infty}^x = \frac{e^{2x}}{2} $$ For $x \geq 0$: $$ F(x) = P(X < 0) + \int_0^x e^{-2t} dt = \frac{1}{2} + \left[-\frac{e^{-2t}}{2} \right]_0^x = \frac{1}{2} + \frac{1 - e^{-2x}}{2} = 1 - \frac{e^{-2x}}{2} $$ 3. **Inverse CDF method:** Let $U \sim \text{Uniform}(0,1)$. - If $U < 0.5$, then $$ U = F(x) = \frac{e^{2x}}{2} \implies e^{2x} = 2U \implies 2x = \ln(2U) \implies x = \frac{1}{2} \ln(2U) $$ - If $U \geq 0.5$, then $$ U = 1 - \frac{e^{-2x}}{2} \implies \frac{e^{-2x}}{2} = 1 - U \implies e^{-2x} = 2(1-U) \implies -2x = \ln(2(1-U)) \implies x = -\frac{1}{2} \ln(2(1-U)) $$ 4. **Random variate generator:** Given $U \sim \text{Uniform}(0,1)$, $$ x = \begin{cases} \frac{1}{2} \ln(2U) & \text{if } U < 0.5 \\ -\frac{1}{2} \ln(2(1-U)) & \text{if } U \geq 0.5 \end{cases} $$ 5. **Transform the given random numbers:** For each $U_i$: - $U_1 = 0.129 < 0.5$: $$ x_1 = \frac{1}{2} \ln(2 \times 0.129) = \frac{1}{2} \ln(0.258) \approx \frac{1}{2} \times (-1.354) = -0.677 $$ - $U_2 = 0.974 \geq 0.5$: $$ x_2 = -\frac{1}{2} \ln(2(1 - 0.974)) = -\frac{1}{2} \ln(0.052) \approx -\frac{1}{2} \times (-2.956) = 1.478 $$ - $U_3 = 0.477 < 0.5$: $$ x_3 = \frac{1}{2} \ln(2 \times 0.477) = \frac{1}{2} \ln(0.954) \approx \frac{1}{2} \times (-0.047) = -0.0235 $$ - $U_4 = 0.852 \geq 0.5$: $$ x_4 = -\frac{1}{2} \ln(2(1 - 0.852)) = -\frac{1}{2} \ln(0.296) \approx -\frac{1}{2} \times (-1.217) = 0.609 $$ - $U_5 = 0.508 \geq 0.5$: $$ x_5 = -\frac{1}{2} \ln(2(1 - 0.508)) = -\frac{1}{2} \ln(0.984) \approx -\frac{1}{2} \times (-0.016) = 0.008 $$ - $U_6 = 0.021 < 0.5$: $$ x_6 = \frac{1}{2} \ln(2 \times 0.021) = \frac{1}{2} \ln(0.042) \approx \frac{1}{2} \times (-3.169) = -1.585 $$ **Final transformed values:** $$ \boxed{ -0.677, 1.478, -0.0235, 0.609, 0.008, -1.585 } $$