1. **Problem a**: Develop a formula for a random variate generator for the random variable $X$ with p.d.f. $$f(x) = \begin{cases} e^{2x} & -\infty < x \leq 0 \\ e^{-2x} & 0 < x < \infty \end{cases}$$ - Step 1: Find the CDF $F(x)$ by integrating $f(x)$. 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 $x > 0$: $$F(x) = F(0) + \int_0^x e^{-2t} dt = \frac{1}{2} + \left[-\frac{e^{-2t}}{2} \right]_0^x = \frac{1}{2} + \frac{1 - e^{-2x}}{2} = 1 - \frac{e^{-2x}}{2}$$ - Step 2: The CDF is $$F(x) = \begin{cases} \frac{e^{2x}}{2} & x \leq 0 \\ 1 - \frac{e^{-2x}}{2} & x > 0 \end{cases}$$ - Step 3: To generate $X$ from uniform $U \sim \text{Uniform}(0,1)$, invert $F$: If $U < 0.5$, solve $U = \frac{e^{2x}}{2} \Rightarrow e^{2x} = 2U \Rightarrow x = \frac{1}{2} \ln(2U)$. If $U \geq 0.5$, solve $U = 1 - \frac{e^{-2x}}{2} \Rightarrow e^{-2x} = 2(1-U) \Rightarrow x = -\frac{1}{2} \ln(2(1-U))$. **Random variate generator formula:** $$X = \begin{cases} \frac{1}{2} \ln(2U) & U < 0.5 \\ -\frac{1}{2} \ln(2(1-U)) & U \geq 0.5 \end{cases}$$ 2. **Problem b**: Triangular distribution with range (1,10) and mode at 4. - Step 1: The triangular distribution CDF is piecewise: For $x$ in $[1,4]$: $$F(x) = \frac{(x-1)^2}{(10-1)(4-1)} = \frac{(x-1)^2}{27}$$ For $x$ in $[4,10]$: $$F(x) = 1 - \frac{(10 - x)^2}{(10-1)(10-4)} = 1 - \frac{(10 - x)^2}{54}$$ - Step 2: To generate $X$ from $U \sim \text{Uniform}(0,1)$: If $U < F(4) = \frac{(4-1)^2}{27} = \frac{9}{27} = \frac{1}{3}$, $$X = 1 + \sqrt{27 U}$$ If $U \geq \frac{1}{3}$, $$X = 10 - \sqrt{54 (1 - U)}$$ - Step 3: Generate 10 values by sampling $U_i$ and applying above. - Step 4: Compute sample mean $\bar{X} = \frac{1}{10} \sum_{i=1}^{10} X_i$. - Step 5: True mean of triangular distribution: $$\mu = \frac{1 + 10 + 4}{3} = 5$$ Compare sample mean to 5. 3. **Problem c**: Exponentially distributed lead times with mean 3.7 days. - Step 1: Exponential distribution parameter $\lambda = \frac{1}{3.7}$. - Step 2: Generate 5 random values $X_i$ from $U_i \sim \text{Uniform}(0,1)$: $$X_i = -\frac{1}{\lambda} \ln(1 - U_i) = -3.7 \ln(1 - U_i)$$ 4. **Problem d**: L’ecuyer combined generator with parameters: $a=157, m=32363, a_2=146, m_2=31727, a_3=142, m_3=31657$ Initial seeds: $X_{1,0}=100, X_{2,0}=300, X_{3,0}=500$ - Step 1: For each step $n$, compute: $$X_{1,n} = (a X_{1,n-1}) \bmod m$$ $$X_{2,n} = (a_2 X_{2,n-1}) \bmod m_2$$ $$X_{3,n} = (a_3 X_{3,n-1}) \bmod m_3$$ - Step 2: Combine: $$Z_n = (X_{1,n} - X_{2,n} + X_{3,n}) \bmod (m - 1)$$ - Step 3: Normalize: $$U_n = \frac{Z_n}{m}$$ - Step 4: Generate 5 values by iterating from $n=1$ to 5. 5. **Problem e**: Given CDF $$F(x) = \frac{x^4}{16}, \quad 0 \leq x \leq 2$$ - Step 1: Invert CDF for $U \sim \text{Uniform}(0,1)$: $$U = \frac{x^4}{16} \Rightarrow x = (16 U)^{1/4}$$ - Step 2: Generate sample values $X_i = (16 U_i)^{1/4}$. - Step 3: Compute sample mean $\bar{X} = \frac{1}{n} \sum X_i$. - Step 4: True mean: $$E[X] = \int_0^2 x f(x) dx = \int_0^2 x \frac{d}{dx}F(x) dx = \int_0^2 x \frac{4x^3}{16} dx = \int_0^2 \frac{x^4}{4} dx = \frac{1}{4} \left[ \frac{x^5}{5} \right]_0^2 = \frac{1}{4} \times \frac{32}{5} = \frac{8}{5} = 1.6$$ Compare sample mean to 1.6.