Question

Let X have an exponential distribution, i.e., $f(x) = e^{-x}$, $0 < x < infty$ For a > 0 and b > 0, show that $P(X > a + b | X > a) = P(X > b)$

          Let X have an exponential distribution, i.e.,
$f(x) = e^{-x}$, $0 < x < infty$
For a > 0 and b > 0, show that
$P(X > a + b | X > a) = P(X > b)$

Added by Emily F.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Sri K

08/01/2023

Step 1

This can be written as: P(X > a + b | X > a) Using the conditional probability formula, we can rewrite this as: P(X > a + b and X > a) / P(X > a) Since X is an exponential distribution, we know that the probability density function is: f(x) = λe^(-λx) where Show more…

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