Use of indicator function for conditional expectation

Question

Consider three random variables: $T \in [0,1] $, $S\in \{ l,h \}$, and $K\in [0,1]$.

We know that (i) $K$ is independent of $T$ and $S$; (ii) $T \sim U[0,1]$; and (iii) $\Pr[S=h \mid T=t]=\pi(t)$ with $\pi'(\cdot)>0$.

In this context, I came across the following statement (where $I$ is the indicator function):

$$\begin{align} &E[T \times I(T>a,K<b) \mid S=h]\\[8pt] &=E[T \times I(T>a \mid S=h)]\cdot\Pr[K<b)]\\[8pt] &=\left(\frac{\int^1_a t\pi(t) dt}{\int^1_0 \pi(u) du}\right)\Pr[K<b]\end{align}$$

I want to know if the first step makes sense and if there is an abuse of notation. I understand from the last equality that $E[T \times I(T>a \mid S=h)]$ refers to the expectation of a random variable with the distribution of $T\mid S\!=\!s$ truncated at $T>a$, (and multiplied by the probability of truncation).

Answer 1

Introduce the events $A=[T\gt a]$, $B=[K\lt b]$ and $H=[S=h]$. Then, by definition, $$ \mathbb E(T\mathbf 1_A\mathbf 1_B\mid H)=\frac{\mathbb E(T\mathbf 1_A\mathbf 1_B\mathbf 1_H)}{\mathbb P(H)}. $$ Since $K$ is independent of $(T,S)$, $\mathbf 1_B$ is independent of $T\mathbf 1_A\mathbf 1_H$ hence the numerator is $\mathbb E(T\mathbf 1_A\mathbf 1_H)\mathbb P(B)$. Furthermore, $\mathbb P(H\mid T)=\pi(T)$, hence $$ \mathbb E(T\mathbf 1_A\mathbf 1_H)=\mathbb E(T\mathbf 1_A\mathbb P(H\mid T))=\mathbb E(T\mathbf 1_A\pi(T)). $$ Assuming that the distribution of $T$ has density $p$, $$ \mathbb E(T\mathbf 1_A\pi(T))=\int_a^{+\infty}t\pi(t)p(t)\mathrm dt. $$ Likewise, $$ \mathbb P(H)=\mathbb E(\mathbb P(H\mid T))=\mathbb E(\pi(T))=\int_{-\infty}^{+\infty}t\pi(t)p(t)\mathrm dt. $$ Finally, indeed, $$ \mathbb E(T\mathbf 1_A\mathbf 1_B\mid H)=\mathbb P(B)\frac{\displaystyle\int_a^{+\infty}t\pi(t)p(t)\mathrm dt}{\displaystyle\int_{-\infty}^{+\infty}t\pi(t)p(t)\mathrm dt}. $$