I am trying to derive following property:
Consider the following filtration $F_t=\sigma\{1_{\{T\leq s\}},\,\forall\,s\leq t \}$ where $T$ is a random variable. Let $s<t$. Then $$ E[1_{T\leq t}|F_s]1_{\{T>s\}}=1_{\{T>s\}}\frac{P(s<T\leq t)}{P(T>s)}$$. I would appreciate any possible idea. Thank you in advance.
We consider event in $F_s$ of the form: $A=\{T \le u\}$ for $u\le s$, and the event $A=\{T >s\}$. If $A=\{T \le u\}$ for $u\le s$, then \begin{align*} \int_A E(1_{T \le t} \mid F_s) 1_{T>s} dP &= \int_A E(1_{T \le t} 1_{T>s}\mid F_s) dP\\ &=\int_A 1_{T \le t} 1_{T>s} dP\\ &=E(1_A1_{T \le t} 1_{T>s})\\ &=0, \end{align*} and \begin{align*} \int_A 1_{T>s}\frac{P(s<T\le t)}{P(T>s)} dP &=\frac{P(s<T\le t)}{P(T>s)}E(1_A 1_{T>s})\\ &=0. \end{align*} That is, \begin{align*} \int_A E(1_{T \le t} \mid F_s) 1_{T>s} dP &= \int_A 1_{T>s}\frac{P(s<T\le t)}{P(T>s)} dP. \end{align*} Moreover, if $A=\{T >s\}$, \begin{align*} \int_A E(1_{T \le t} \mid F_s) 1_{T>s} dP &= E(1_{T>s} 1_{T\le t})\\ &=P(T>s)\frac{P(s<T\le t)}{P(T>s)} \\ &=P(A)\frac{P(s<T\le t)}{P(T>s)} \\ &=\int_{\Omega} 1_A\frac{P(s<T\le t)}{P(T>s)} dP\\ &=\int_A 1_A\frac{P(s<T\le t)}{P(T>s)} dP\\ &=\int_A 1_{T>s}\frac{P(s<T\le t)}{P(T>s)} dP. \end{align*} That is, \begin{align*} E(1_{T \le t} \mid F_s) 1_{T>s} &= 1_{T>s}\frac{P(s<T\le t)}{P(T>s)}. \end{align*} See also Chapter 4 of the book Credit Risk.