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

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#### Best Answer

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 $A={win Omega: X(w)<bar x}$, 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.

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