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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?2012-06-19
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    Calculated E[x|x>1] as the integral of x* f(x|x>1) from zero to infinity2012-06-19
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    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 infinity2012-06-19
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    I am getting that the answer is 1/lambda + 1?2012-06-19
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    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.2012-06-19

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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.

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    That's what I tried to use when calculating f(x|x>1) as the quotient in the first post. Is that correct?2012-06-19
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    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.2012-06-19
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    So, in answer to your hint, the distribution of the additional waiting time is the same, an exponential with parameter lambda?2012-06-19
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    @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.2012-06-19
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    So I am getting the expected value should be 1/lambda + 1, but it is supposed to be 2/lambda2012-06-19
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    I do not see the $2/\lambda$. There seems to be either a typo in the answer, or in the problem, or else the original problem is different and you reduced it incorrectly to the problem you posted.2012-06-19
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    No, I was checking the result against my notes, which for some reason said it was lambda/2.2012-06-19
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    @ravenea: If the *parameter* in the usual sense is $\lambda$, as in density $\lambda e^{-\lambda x}$, then $\lambda/2$ is utterly impossible. It does not even have the right units, the unit for $\lambda$ is $t^{-1}$ (inverse of time).2012-06-19