The probability density function $F (x) = a e^{- b | x |},$ where x is a random variable whose allowable value range is from $x = - \infty t o x = + \infty .$ The CDF for this function for $x \geq 0$ is

\frac{a}{b} e^{b x}

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\frac{a}{b} (2 - e^{- b x})

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- \frac{a}{b} e^{b x}

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- \frac{a}{b} (2 + e^{- b x})

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Solution

The correct option is B $\frac{a}{b} (2 - e^{- b x})$
Given PDF $F (x) = a e^{- b | x |}$

for $x \geq 0$

$C D F = \int_{- \infty}^{x} f (u) d u = \int_{- i n f t y}^{x} a e^{- b | u |} d u$

$= \int_{- \infty}^{0} a e^{b u} d u + \int_{0}^{x} e^{- b u} d u$

$= {[\frac{a}{b} e^{b u}]}_{- \infty}^{b u} + {[\frac{- a}{b} e^{- b u}]}_{0}^{- b u}$

$= \frac{a}{b} (1 - 0) + [\frac{- a}{b} (e^{- b x} - 1)]$

$= \frac{a}{b} (2 - e^{- b x})$

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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

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