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](/img/relate-questions.png)
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:
\(CDF = \mathop \smallint \limits_{ - \infty }^x a{e^{ - b\left| u \right|}}du\)
\(CDF = \mathop \smallint \limits_{ - \infty }^0 a{e^{bu}}du + \mathop \smallint \limits_0^x a{e^{ - bu}}du\)
\( = \left[ {\frac{a}{b}{e^{bu}}} \right]_{ - \infty }^0 + \left[ { - \frac{a}{b}{e^{ - bu}}} \right]_0^x\)
\( = \frac{a}{b}\left( {1 - 0} \right) + \left[ { - \frac{a}{b}\left( {{e^{ - bx}} - 1} \right)} \right]\)
\( = \frac{a}{b}\left( {2 - {e^{ - bx}}} \right)\)