CDF of Exponential Distribution

The probability density function of a certain random variable X X is:

f X ( x ) = λ e λ x , f_X (x) = \lambda e^{-\lambda x},

where x x takes values in [ 0 , ) [0,\infty) .

Find the probability that x < 100 x < 100 .

1 e 100 λ 1-e^{-100 \lambda} λ e 100 λ \lambda e^{-100 \lambda} e 100 λ e^{100 \lambda} e 100 λ -e^{-100 \lambda}

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1 solution

Matt DeCross
Apr 20, 2016

This probability is found by computing the CDF and evaluating at 100 100 . Given the PDF above, the CDF is:

F X ( x ) = x λ e λ x d x = e λ x + C . F_X (x) = \int^x \lambda e^{-\lambda x} \,dx = -e^{-\lambda x} + C.

Since F X ( ) = 1 F_X (\infty) = 1 , the integration constant C = 1 C = 1 , and the CDF is thus:

F X ( x ) = 1 e λ x . F_X (x) = 1 - e^{-\lambda x}.

The probability that x < 100 x < 100 is therefore:

F X ( 100 ) = 1 e 100 λ . F_X (100) = 1-e^{-100 \lambda}.

For any reasonable-sized ( O ( 1 ) \mathcal{O}(1) ) λ \lambda , this probability is incredibly close to 1 1 !

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