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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? –  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

1 Answer 1

up vote 5 down vote accepted

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? –  metroxylon Jun 19 '12 at 19:41
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
So, in answer to your hint, the distribution of the additional waiting time is the same, an exponential with parameter lambda? –  metroxylon Jun 19 '12 at 19:51
@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
So I am getting the expected value should be 1/lambda + 1, but it is supposed to be 2/lambda –  metroxylon Jun 19 '12 at 19:57

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