Take the tour ×  I need to find $E[x\mid x>1]$ if $X \sim
\exp(\lambda)$.  I first tried: $$f(x|x>1) =
\frac{f(x)}{\int_{x=1}^{\infty}f(x) dx}.$$  share|improve this
question 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 show 1
more comment  1 Answer  up vote 6 down vote accepted  Hint: Use
the memorylessness property of the exponential distribution.
Given that you have waited $1$ hour, what is the distr