The probability density function F(x) = ae -b |x|, where x is a r

The probability density function F(x) = ae -b |x|, where x is a r
| The probability density function F(x) = ae-b |x|, where x is a random variable whose allowable value range is from x = -∞ to x = +∞. The CDF for this function for x ≥ 0 is

A. <span class="math-tex">\(\frac{a}{b}{e^{bx}}\)</span>

B. <span class="math-tex">\(\frac{a}{b}\left( {2 - {e^{ - bx}}} \right)\)</span>

C. <span class="math-tex">\( - \frac{a}{b}{e^{bx}}\)</span>

D. <span class="math-tex">\( - \frac{a}{b}\left( {2 + {e^{ - bx}}} \right)\)</span>

Please scroll down to see the correct answer and solution guide.

Right Answer is: B

SOLUTION

Concept:

To analyse the Random Variable ‘x’ two functions are used.

1) PDF (Probability Distribution Function)

2) pdf (Probability Density Function)

CDF and pdf are related as:

\(CDF = \mathop \smallint \limits_{ - \infty }^\infty pdfdx\) 

Calculation:

Given pdf is F(x) = ae-b |x|

CDF is given by

\(CDF = \mathop \smallint \limits_{ - \infty }^\infty a{e^{ - b\left| x \right|}}dx\) 

For x ≥ 0 the CDF is defined as:

\(CDF = \mathop \smallint \limits_{ - \infty }^x a{e^{ - b\left| u \right|}}du\) 

\(CDF = \mathop \smallint \limits_{ - \infty }^0 a{e^{bu}}du + \mathop \smallint \limits_0^x a{e^{ - bu}}du\) 

\( = \left[ {\frac{a}{b}{e^{bu}}} \right]_{ - \infty }^0 + \left[ { - \frac{a}{b}{e^{ - bu}}} \right]_0^x\) 

\( = \frac{a}{b}\left( {1 - 0} \right) + \left[ { - \frac{a}{b}\left( {{e^{ - bx}} - 1} \right)} \right]\) 

\( = \frac{a}{b}\left( {2 - {e^{ - bx}}} \right)\)