Let $X$ be random variable and $f$ it's density. How can one calculate $E(X\vert X<a)$?
From definition we have:
$$E(X\vert X<a)=\frac{E\left(X \mathbb{1}_{\{X<a\}}\right)}{P(X<a)}$$
Is this equal to:
$$\frac{\int_{\{X<a\}}xf(x)dx}{P(X<a)}$$
? If yes, then how one justify it? Thanks. I'm conditional expectation noob.
Also, what is $E(X|X=x_0)$? In discrete case it is $x_0$...
On
Under condition $X<a$ you are "observing" the 'same' rv, but somehow it is 'forbidden' for $X$ to take values $\geq a$.
That asks for a PDF that maintains the quantitive relations when it comes to probabilities on subsets of $\{X<a\}$. We can start with naively taking a function equal to the original PDF on $(-\infty,a)$ and taking value $0$ on $[a,\infty)$. But this function is not a PDF because taking the integral over it we get $P(X<a)$ as value, and not $1$ as we should.
This is solved by dividing the function by $P(X<a)$. The new PDF is: $$\frac{f(x)1_{(-\infty,a)}(x)}{P(X<a)}$$
By definition, $\int_BXdP=\int_\Omega X1_BdP$. If $X$ has density $f$, this is equal to $\int x1_B(x)f(x)dx=\int_Bxf(x)dx$.