Expected value of a random variable in a range I have a random variable $x$ with $E(x) = \mu$ and PDF $f(x)$ and CDF of $F(x)$.
Is there any way to represent the $E(x|x<\bar{x})$ in terms of $\mu$ and $f(x)$ or $F(x)$? 
i.e. to write the following integral only in terms of $\mu$, $f$ and $F$:
$$
\int_{-\infty}^\bar{x} x f(x) dx
$$
Thanks
 A: For fixed sets $A$ the definition of the conditional expectation is the following:
$$E(X|A)=\frac{1}{P(A)}EX1_{A},$$
with $1_A(w)$ being indicator function of the set. Now your set $A$ is defined as $A=\{w\in \Omega: X(w)<\bar x\}$, so $P(A)=F(\bar x)$. Plugging this into the formula we get:
$$E(X|X<\bar x)=\frac{1}{F(\bar x)}\int_{-\infty}^{\bar x}xf(x)dx$$
Note that $\mu$ does not play any part in this formula, unless $\bar x$ has special meaning.  I've assumed that bar over $x$ is only your choice of typographical differentiation of different quantities.
A: In general, it is not possible for just $\mu$, $f$ and $F$. However, in at least some cases pleasant expressions seem to be derivable, using the general
$$
E \{ x | x < \bar{x} \} = \frac{1}{F(\bar{x})} \int_{-\infty}^{\bar{x}} x f(x) dx \enspace,
$$
from the answer by mpiktas.
For instance, if $f(x)$ a normal distribution with mean $\mu$ and variance $\sigma^2$, I get:
$$
E \{ x | x < \bar{x} \} = \mu - \sigma^2 \frac{ f(\bar{x})}{F(\bar{x})} \enspace.
$$
Note that in this case we need $\sigma^2$ in addition to $\mu$, $f$ and $F$.
For a uniform distribution between $a$ and $b$, I get (assuming $\bar{x} \in [a,b]$):
$$
E \{ x | x < \bar{x} \} = \mu - \frac{1 - F(\bar{x})}{2 f( \bar{x} ) } \enspace,
$$
from which it is clear that no general expression will exist, since the expression differs from the one for the normal distribution.
A: I don't think there is a general answer. But you could calculate an answer for each specific distribution. For example, for the normal distribution, see http://en.wikipedia.org/wiki/Truncated_normal_distribution. 
