# Could someone explain this probability proof to me?

For a continuous random variable X, $X\sim EXP(\theta) \iff$ $P(X > a + t | X > a) = P(X > t)$

Proof:

$P(X > a + t | X > a) = \dfrac{P(X > a + t \cap X >a)}{P(X > a)}$

$= \dfrac{P(X > a + t)}{P(X > a)}$

$= \dfrac{e^{-(a+t)/\theta}}{e^{-a/\theta}}$

$= P(X>t)$

Can someone explain to me why $P(X > a + t \cap X >a) = P(X > a + t)$

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Just FYI, this is called memoryless property – Alex Oct 14 '12 at 23:13
You didn't say $t>0$, which is required for this step. – GEdgar Oct 14 '12 at 23:31
Well, I think my answer is at least more simply expressed than the others. – Michael Hardy Oct 15 '12 at 2:06

$(X > a + t) \bigcap (X > a)$ means that the random variable $X$ is both greater than $a + t$ and also greater than $a$. But, if $X > a + t$, then it is already greater than $a$. So, that statement is equivalent to $X > a + t$.

A simpler statement that is similar would be $(X > 1) \bigcap (X > 0)$. If $X > 1$, then we already know $X > 0$, so that extra statement doesn't add any new information. So, here, this is equivalent to just $X > 1$.

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Because, if $t\geqslant0$, then $[X\gt a+t]\subseteq[X\gt a]$, hence $$[X\gt a+t,X\gt a]=[X\gt a+t]\cap[X\gt a]\subseteq[X\gt a].$$

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Since $a$ and $t$ are both positive saying "$X>a+t$ and $X>a$" is the same as saying $X>a+t$.

For example saying the price of an item is more than $\$6$and also more than$\$8$, is the same as saying it's more than $\$8\$.

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