Conditional expectation $E[1_{T\leq t}|F_s]1_{\{T>s\}}=1_{\{T>s\}}\frac{P(ss)}$ 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.
