Conditional Expectation of Continuous Random Variable

Question

Let $Y$ be a random variable of density g. How could I compute the expectation

$E[Y|Y<a]$ ? Thank you in advance!

Answer 1

If $P(Y<a)>0$ and $EY$ exists. $$E[Y|Y<a] = \frac{E[Y\cdot1_{\{Y<a\}}]}{P(Y<a)} = \frac{\int\limits_{-\infty}^a xg(x)dx}{\int\limits_{-\infty}^a g(x)dx}$$

Answer 2

I know that it's been over 2 years, but I suggest this answer $$ E(Y|Y<a) = \frac{E(Y \times (Y < a))}{P(Y<a)} = \frac{ \int_{-\infty}^{+\infty} (t \times c(t)) f_{Y}(t) dt }{ \int_{-\infty}^{a}{ f_{Y}(y) dy } } $$ where $$ c(t) = \left\{\begin{array}{cc} 1, & t < a \\ 0, & {\rm otherwise} \end{array}\right. $$ Hence, we have $$ E(Y|Y<a) = \frac{ \int_{-\infty}^{a} t f_{Y}(t) dt }{ \int_{-\infty}^{a}{ f_{Y}(y) dy } }. $$