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I am given that X is a continuous variable with the pdf, $f(x)= 2 e^{-2x}$ if $x \geq 0$ and asked to find the $P(X \geq 3 | X \geq 2)$.

I am having trouble with the intuition behind the answer, my understanding is that the conditional probability could be written as:

$P(X \geq 3 \cap X \geq 2)/P(X \geq 2) $

Wouldn't the intersection simply be the smaller set, namely $P(X \geq 2) $? But this doesn't make sense because then all you would have is:

$P(X\geq 2)/P(X \geq 2) $

Answers I am seeing online give the probability by:

$P(X \geq 3)/P(X \geq 2) $

But I don't understand how you arrive there. I feel like the answer is probably obvious and I'm just missing it.

Thank you.

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  • $\begingroup$ The event $X \gt 2$ is strictly larger than the event $X \gt 3$, not smaller. $\endgroup$ – whuber Sep 13 '16 at 18:16
  • $\begingroup$ oh duh - thank you so much, this was really bugging me! $\endgroup$ – Edna Sep 13 '16 at 18:17
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    $\begingroup$ I think your question needs a self-study tag as this is clearly an exercise you need to solve by yourself. $\endgroup$ – Xi'an Sep 13 '16 at 19:15
  • $\begingroup$ It is worth reading our wiki to see how we handle self-study questions - this is just the type of question that the self-study tag is designed for. $\endgroup$ – Silverfish Sep 13 '16 at 19:17
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The distribution is an exponential distribution:

$ f(x;\lambda) = \lambda e^{-\lambda x } x \ge 0, $

with parameter $\lambda=2$.

An exponentially distributed random variable $x$ is a memory-less probability distributions that obeys the relation

$P( x > s + t | x > s) = P(x> t), \qquad \forall s, t \ge 0 $.

Intuitively this means that if you have waited $s$ seconds for a event the probability that that event will occur $t$ seconds later is the same as if you spent no time waiting i.e $s=0$.

This is simple to demonstrate:

$P( x > s + t | x > s)=\frac{P( x > s + t ,x>s)}{P( x > s)},$

because $x > s + t>s$ then $P( x > s + t ,x>s)=P( x > s + t)$ we simplifies he expression to:

$\frac{P( x > s + t ,x>s)}{P( x > s)}=\frac{P( x > s + t)}{P( x > s)}$

$=\frac{1-P( x < s + t)}{1-P( x < s)} (1)$

where the term $P( x < u)=1-e^{-\lambda u } (2)$ and $P( x >u)=e^{-\lambda u } (3)$

substituting (2) into (1) you get :

$\frac{1-P( x < s + t)}{1-P( x < s)}=\frac{1-(1-e^{-\lambda (s + t)}) }{1-(1-e^{-\lambda s )})}=\frac{e^{-\lambda (s + t)} }{e^{-\lambda s }} =e^{-\lambda (t)}$

This expression is the same as (3) i.e $P(x>t)$.

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