The problem occurs when reading Bombardini et al., 2023, "Did US Politicians Expect the China Shock?", American Economic Review, Vol.1, PP174-209.
The authors define $\xi_{it}$ to be a Gaussian term obeying $N(0,2\sigma_{\xi}^2)$ (see line 7, page 7 in the link), and $Y_{i,t} = \mathbf{1}\{a_{t}\theta_{i}+b_{t}+\delta_{t} E[S_{i,t+1}|\mathcal{I}_{i,t}]\geq\xi_{it}\}$ (see equation 4, page 8). To my understand, $a_{t}\theta_{i}+b_{t}+\delta_{t} E[S_{i,t+1}|\mathcal{I}_{i,t}]$ can be seen as a constant once $\mathcal{I}$ is given.
I do not know how, but the authors manage to show that
$$-E[Y_{i,t}\xi_{i,t}|\mathcal{I}_{i,t}]=(1-Y_{i,t})\frac{\phi(a_{t}\theta_{i}+b_{t}+\delta_{t} E[S_{i,t+1}|\mathcal{I}_{i,t}])}{1-\Phi(a_{t}\theta_{i}+b_{t}+\delta_{t} E[S_{i,t+1}|\mathcal{I}_{i,t}])},$$ where $\Phi$ and $\phi$ are respectively the distribution function and density function of the standard normal distribution (see equation (13), page 12 in the link).
Can anyone explain to me in detail? Thanks in advance!
I am not entirely sure if this proof is complete and correct, but I would like to share what I have attempted.
Let $z\in\mathbb{R}$, then it holds $$ \begin{aligned} \int_{\mathbb{R}}-\mathbb{1}_{\{z\ge x\}}x\ d\mu_{\mathcal{N}(0,1)}(x) &=\int_{\mathbb{R}}-\mathbb{1}_{\{z\ge x\}}x\phi(x)\ dx\\ &=\phi(z)\\ &=\frac{\phi(z)}{1-\Phi(z)}\int_\mathbb{R}\mathbb{1}_{\{z<x\}}\phi(x)\ dx\\ &=\int_\mathbb{R}(1-\mathbb{1}_{\{z\ge x\}})\frac{\phi(z)}{1-\Phi(z)}\ d\mu_{\mathcal{N}(0,1)}(x). \end{aligned} $$
Applying Fubini's theorem for conditional expectations yields
$$ \begin{aligned} \int_{\mathbb{R}}E[-\mathbb{1}_{\{Z\ge x\}}x|\mathcal{G}]\ d\mu_{\mathcal{N}(0,1)}(x) &=E[\int_{\mathbb{R}}-\mathbb{1}_{\{Z\ge x\}}x\ d\mu_{\mathcal{N}(0,1)}(x)|\mathcal{G}]\\ &=E[\int_\mathbb{R}(1-\mathbb{1}_{\{Z\ge x\}})\frac{\phi(Z)}{1-\Phi(Z)}\ d\mu_{\mathcal{N}(0,1)}(x)|\mathcal{G}]\\ &=\int_\mathbb{R}E[(1-\mathbb{1}_{\{Z\ge x\}})\frac{\phi(Z)}{1-\Phi(Z)}|\mathcal{G}]\ d\mu_{\mathcal{N}(0,1)}(x) \end{aligned} $$ for a random variable $Z$ and a $\sigma$-algebra $\mathcal{G}$.
You then might want to choose $Z:=a_{t}\theta_{i}+b_{t}+\delta_{t} E[S_{i,t+1}|\mathcal{I}_{i,t}]$ and $\mathcal{G}=\mathcal{I}_{i,t}$ to find $$ \begin{aligned} \int_{\mathbb{R}}E[-Y_{i,t}x|\mathcal{I}_{i,t}]\ d\mu_{\mathcal{N}(0,1)}(x) &= \int_\mathbb{R}(1-Y_{i,t})\frac{\phi(a_{t}\theta_{i}+b_{t}+\delta_{t} E[S_{i,t+1}|\mathcal{I}_{i,t}])}{1-\Phi(a_{t}\theta_{i}+b_{t}+\delta_{t} E[S_{i,t+1}|\mathcal{I}_{i,t}])}\ d\mu_{\mathcal{N}(0,1)}(x).\\ \end{aligned} $$
It would be greatly appreciated if anyone could provide additional feedback on whether this is sufficient and accurate. (The problem I am facing is that both $E[S_{i,t+1}|\mathcal{I}_{i,t}]$ and $\xi_{i,t}$ should end up having the same $\omega$ as an argument.)