I'm going through preparatory tasks for an exam, but I'm stuck on a particular problem.
pdf(x) = $c\cdot e^{-3x}$, where $x \ge 0$ and $0$ for any other value of $x$
I worked out the constant to 3.
A) $Pr(X > a), a > 0$
B) $Pr(X > a+b\mid X > b)$
I worked out A) to be $e^{-3a}$, and since B) is an exponential distribution problem, I simplified it to $Pr(X > a)$, giving me the same answer as A). Am I on the right track?
The probability you want is $P(X>a+b|X>b)$ where $X$ has an exponential distribution.
Using the CDF of the exponential distribution...
$$ P(X\leq a+b|X>b) = \frac{P(X\leq a+b\cap X > b) }{P(X>b)} = \frac{P(b< X \leq a+b)}{P(X>b)} = \frac{ e^{-b\lambda} - e^{(a+b)\lambda} }{e^{-b\lambda}}$$ $$=1 - e^{-a\lambda}$$
This means $$P(X>a+b|X>b) = 1- P(X \leq a+b | X> b ) = e^{-a \lambda}$$
This is true because the exponential has the property of being memoryless, i.e. previous events have no bearing on the current event. I've always found an interesting application of this in car batteries. Many studies have suggested car batteries to very closely approximate the exponential.