Mean of a random variable from a reservation value to infinity

Question

say $a$ is a random variable from an $F(.)$ distribution and $\tilde{a} = E(a|a>a^r)$

I was wondering which law allows us to write the following : $$\tilde{a}= \frac{\int_{a^r}^\infty adF(a)}{1-F(a^r)} $$

Do you have any idea? Thank you

Answer 1

I presume the reservation value $a^r$ is a constant. Note that $$\int_{a^r}^\infty \tilde adF(\tilde a)=E[a{\bf1}_{a>a^r}]\ \\=E[E[a{\bf1}_{a>a^r}|{\bf1}_{a>a^r}]]\quad \text{(LIE)}\\ =E[a{\bf1}_{a>a^r}|{\bf1}_{a>a^r}=1]P({\bf1}_{a>a^r}=1)+\underbrace{E[a{\bf1}_{a>a^r}|{\bf1}_{a>a^r}=0]}_{=0}P({\bf1}_{a>a^r}=0)\\ =E[a|a>a^r]P(a>a^r)\\ =E[a|a>a^r](1-F(a^r)),$$

and rearranging gets you the result.