# Compute $P(X>40\; |\; X>10)$ where $X$ has an exponential distribution

Let $X$ have an exponential distribution with a mean of $i = 20$

(1) Compute $P(X>40 \;| \;X>10)$

I believe the correct solution here is to find $P(X>40)$ because of the inclusion / exclusion principle. That is if we define the event $A = P(X>40)$ and the event $B = P(X>10)$. In addition, define $A \cap B$ as $P(X>10)$. Then to compute (1) we find the following $P(X>40) \cup P(X>10) - P(X>10)$ which is

$$1- (1-e^{-\frac{30}{20}})=0.2231$$

EDIT Due to the "memorylessness" of the probability distribution

$$P(X>40 \mid X>10)= P(X>30)$$

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First, I think you meant to define $A=\{X>40\}$ and $B=\{X>10\}$. Then you can't define $A\cap B$, because it is already decided by $A$ and $B$. In particular $A\cap B=A$, because $B\subseteq A$. Second, unions are to be put between sets and not numbers, i.e. $P(X>40)\cup P(X>10)$ does not make sense. Third, note that the exponential distribution has the memorylessness property. – Stefan Hansen Jan 11 '13 at 12:46
+1 Stefan ..Thanks for the heads up about memorylessness – bosra Jan 11 '13 at 12:52
Are you having trouble calculating $P(X>30)$? – Stefan Hansen Jan 11 '13 at 12:55
If you do, look at the cumulative distribution function of the exponential function. – fabee Jan 11 '13 at 13:09
It is correct.  – Did Jan 14 '13 at 11:02

Yes, due to "memorylessness" of an exponential distribution, if $s>0$, $t>0$, then $$P(X>s+t\mid X>s) = P(X>t) = e^{-\lambda t}.$$
\begin{align} P(X>s+t\mid X>s) &= \frac{P\big(X>s+t\cap X>s\big)}{P(X>s)} = \frac{P(X>s+t)}{P(X>s)} \\ &= \frac{\int_{s+t}^{\infty}\lambda e^{-\lambda x}dx}{\int_{s}^{\infty}\lambda e^{-\lambda x}dx} = e^{-\lambda t} = P(X>t),\quad (s>0,t>0). \end{align}
$\cap$ is used between sets, not numbers and hence $P(X>s+t)\cap P(X>s)$ does not make sense. The correct notation is $P(\{X>s+t\}\cap\{X>s\})$ or in short-hand notation $P(X>s+t,X>s)$. – Stefan Hansen Apr 30 '13 at 12:46