1. **Problem statement:** Develop a formula for a random variate generator for a random variable $X$ with p.d.f. $$f(x) = \begin{cases} e^{2x} & -\infty < x \leq 0 \\ e^{-2x} & 0 < x < \infty \end{cases}$$ 2. **Step 1: Understand the problem and the p.d.f.** The p.d.f. is piecewise defined with two exponential parts, one for negative $x$ and one for positive $x$. We want to find a method to generate random values $X$ that follow this distribution. 3. **Step 2: Verify the p.d.f. is valid** Check if $f(x)$ integrates to 1: $$\int_{-\infty}^0 e^{2x} dx + \int_0^{\infty} e^{-2x} dx = \left[\frac{e^{2x}}{2}\right]_{-\infty}^0 + \left[-\frac{e^{-2x}}{2}\right]_0^{\infty} = \frac{1}{2} + \frac{1}{2} = 1$$ 4. **Step 3: Find the cumulative distribution function (CDF) $F(x)$** For $x \leq 0$: $$F(x) = \int_{-\infty}^x e^{2t} dt = \left[\frac{e^{2t}}{2}\right]_{-\infty}^x = \frac{e^{2x}}{2}$$ For $x > 0$: $$F(x) = F(0) + \int_0^x e^{-2t} dt = \frac{1}{2} + \left[-\frac{e^{-2t}}{2}\right]_0^x = \frac{1}{2} + \frac{1 - e^{-2x}}{2} = 1 - \frac{e^{-2x}}{2}$$ 5. **Step 4: Inverse transform method** To generate $X$, generate $U \sim \text{Uniform}(0,1)$ and solve $F(x) = U$ for $x$. - If $U < \frac{1}{2}$, then $$U = \frac{e^{2x}}{2} \implies e^{2x} = 2U \implies 2x = \ln(2U) \implies x = \frac{1}{2} \ln(2U)$$ - If $U \geq \frac{1}{2}$, then $$U = 1 - \frac{e^{-2x}}{2} \implies \frac{e^{-2x}}{2} = 1 - U \implies e^{-2x} = 2(1 - U) \implies -2x = \ln(2(1 - U)) \implies x = -\frac{1}{2} \ln(2(1 - U))$$ 6. **Final formula for the random variate generator:** $$x = \begin{cases} \frac{1}{2} \ln(2U) & U < \frac{1}{2} \\ -\frac{1}{2} \ln(2(1 - U)) & U \geq \frac{1}{2} \end{cases}$$ This formula allows generating random values $X$ with the given p.d.f. by sampling $U$ uniformly from $[0,1]$ and applying the piecewise inverse CDF.