Chapter 10: Q. 10.3 (page 430)

Give a technique for simulating a random variable having the probability density function
$f (x) \{\begin{cases} \frac{1}{2} (x - 2) & 2 \leq x \leq 3 \\ \frac{1}{2} (2 - \frac{x}{3}) & 3 < x \leq 6 \\ 0 & o t h e r w i s e \end{cases}$

Short Answer

Expert verified

The technique for simulating a random variable is the universality of unform.

Step by step solution

Given Information

We have given the function

$f (x) \{\begin{cases} \frac{1}{2} (x - 2) & 2 \leq x \leq 3 \\ \frac{1}{2} (2 - \frac{x}{3}) & 3 < x \leq 6 \\ 0 & o t h e r w i s e \end{cases}$

Simplify

We have

$F (x) = \int_{2}^{x} \frac{1}{2} (s - 2) d s = \frac{{(x - 2)}^{2}}{4}$

and for $3 < x \leq 6$ we have

$F (x) = \int_{2}^{3} \frac{1}{2} (s - 2) d s + \int_{3}^{x} \frac{1}{2} (2 - \frac{s}{3} d s = \frac{1}{4} - \frac{(x - 9) (x - 3)}{12}$

Let's find $F^{- 1}$ . For $y \in (0, 1 / 4)$ we have

$y = \frac{{(x - 2)}^{2}}{4} x = 2 \sqrt{y} + 2$

and for $y \in (1 / 4, 1)$ we have

$y = \frac{1}{4} - \frac{(x - 9) (x - 3)}{12} x = 6 - 2 \sqrt{3 - 3 y}$

Now, choose a random number (call it $y$ ) from the interval $(0, 1)$ . If it is less than $\frac{1}{4}$ , declare $x = 2 \sqrt{y} + 2 .$ if it is greater than $\frac{1}{4}$ , declare $x = 6 - 2 \sqrt{3 - 3 y}$ . From the universality of the uniform, we have the $x$ follows the distribution of $X$ .

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Give a technique for simulating a random variable having the probability density function
$f (x) \{\begin{cases} \frac{1}{2} (x - 2) & 2 \leq x \leq 3 \\ \frac{1}{2} (2 - \frac{x}{3}) & 3 < x \leq 6 \\ 0 & o t h e r w i s e \end{cases}$

Short Answer

Step by step solution

Given Information

Simplify

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Give a technique for simulating a random variable having the probability density functionf(x)12(x-2)2≤x≤312(2-x3)3<x≤60otherwise

Short Answer

Step by step solution

Given Information

Simplify

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Most popular questions from this chapter

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Give a technique for simulating a random variable having the probability density function
$f (x) \{\begin{cases} \frac{1}{2} (x - 2) & 2 \leq x \leq 3 \\ \frac{1}{2} (2 - \frac{x}{3}) & 3 < x \leq 6 \\ 0 & o t h e r w i s e \end{cases}$