Question

1 3. (25 P) Account balance of customers in a bank have the following probability density function: 2 3 (0.05, 0<x< 5 4 f(x) = a, 5 <x< 10 5 0, otherwise 6 7 a. Develop a random variate generator for the distribution 8 9 b. Generate 3 values of the random variate using R 1 = 0.1. R 2 = 0.2. R 3 = 0.95. 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Answer 1

$a.\\ \text{Note that }\int_{0}^{10}f(x)dx=(a+0.05)\times5=1\\ \implies a+0.05=0.2\implies a = 0.15\\ \text{This gives the CDF as:}\\ F(x)=\left\{\begin{matrix} 0.05x, 0\leq x< 5\\ 0.25+0.15(x-5),5\leq x<10\\ 1,x\geq 10 \end{matrix}\right.\\ \text{Thus, for }r\in (0,1),\text{ the inverse is defined as: }\\ F^{-1}(r)=\left\{\begin{matrix} \frac{r}{0.05},0\leq r\leq 0.25\\ 5+\frac{r-0.25}{0.15},0.25<r\leq 1 \end{matrix}\right. \\ \text{Thus if }R\sim U(0,1),\text{ a generated observation is: }\\ X=F^{-1}(R),\\ \text{Where }F^{-1}\text{ is defined above.}$$b.\\ \text{Using the derived formula, the generated values are: }\\ X_{1}=R_{1}/0.05=0.1/0.05=2\\ X_{2}=R_{2}/0.05=0.2/0.05=4\\ X_{3}=5+(R_{3}-0.25)/0.15=5+(0.95-0.25)/0.15=9.67\text{ (Ans.)}$