# $E[x\mid x>1]$ if $X \sim \exp(\lambda)$

I need to find $E[x\mid x>1]$ if $X \sim \exp(\lambda)$.

I first tried: $$f(x|x>1) = \frac{f(x)}{\int_{x=1}^{\infty}f(x) dx}.$$

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and then... what? –  leonbloy Jun 19 '12 at 19:36
Calculated E[x|x>1] as the integral of x* f(x|x>1) from zero to infinity –  metroxylon Jun 19 '12 at 19:38
BE careful: your formula must add the support: $x>1$. The conditioned variable has zero density for $x<1$. So, your integral must go from 1 to infinity –  leonbloy Jun 19 '12 at 19:40
I am getting that the answer is 1/lambda + 1? –  metroxylon Jun 19 '12 at 19:55
If the expectation of the original was $1/\lambda$, then the expectation of the truncated is $1/\lambda + 1$. This happens to the exponential, only, because of the property mentioned in André Nicolas' answer. –  leonbloy Jun 19 '12 at 20:01
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Hint: Use the memorylessness property of the exponential distribution. Given that you have waited $1$ hour, what is the distribution of your additional waiting time? So what is the expectation of your additional waiting time? Now don't forget to add the hour already spent waiting.
So you want to derive the memorylessness property? Then I would suggest finding the probability that $X \ge a+t$ given that $X\ge a$. Things will look nicer. Let $A$ be the event $X\ge a$, $B$ the event $X\ge a+t$. Your conditional probability will be an integral from $a+t$ to infinity divided by an integral from $a$ to infinity. But since you already know the cdf of the exponential, you can just write down the answers. –  André Nicolas Jun 19 '12 at 19:49
@ravenea: Yes, a very important fact about the exponential. It can be derived simply as per my previous comment, since doing what I suggested we find that the probability that $X\gt a+t$ given $X\ge a$ is $\frac{e^{-\lambda(a+t)}}{e^{-\lambda a}}=e^{-\lambda t}$. But memorylessness may already have been proved for you, it is very basic. –  André Nicolas Jun 19 '12 at 19:55