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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).

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What are $S$ and $Z$, one of them is just a typo? What is $I_{A|C}$? – Ilya Jan 30 '13 at 19:28
@Ilya: yes, that was a typo -just corrected it. $I_{A|C}=I(A|C)$ Not quite sure if that makes sense (the notation is new to me) – EOO Jan 30 '13 at 19:35
I have never seen such notation. Conditioning is only defined for probabilities/expectations usually. Unfortunately unless your clarify what does $I(A|C)$ mean, it's hard to help you – Ilya Jan 30 '13 at 19:38
@EOO : You wrote "A|\text{ }", using \text{ } for spacing. But "A\mid B" does it: $\displaystyle A\mid B$. – Michael Hardy Jan 30 '13 at 19:47
The first and the third expressions coincide but the second one is a mystery, with respect to notations as well as to its role in the proof. – Did Feb 1 '13 at 16:43
up vote 1 down vote accepted

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, $$ \mathbb E(T\mathbf 1_A\mathbf 1_B\mid H)=\mathbb P(B)\frac{\displaystyle\int_a^{+\infty}t\pi(t)p(t)\mathrm dt}{\displaystyle\int_{-\infty}^{+\infty}t\pi(t)p(t)\mathrm dt}. $$

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Thanks. I tried going this way but I got stuck on your second line. Why is $\mathbb E(T\mathbf 1_A\mathbf 1_H)=\mathbb E(T\mathbf 1_A\mathbb P(H\mid T))$? – EOO Feb 1 '13 at 20:09
For every sigma-algebra $G$ and random variables $U$ and $V$, if $U$ is $G$-measurable then $E(UV)=E(UE(V\mid G))$. Use this with $G=\sigma(T)$, $U=T\mathbf 1_A$ and $V=\mathbb 1_H$. – Did Feb 2 '13 at 9:28
Thanks for spelling it out – EOO Feb 2 '13 at 12:34

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