Suppose $X, T$ are continuous random variables, and $X \sim \mathcal{N}(0, 1)$, $T$ have density function $f_T$. (But $X,T$ do not have joint density)
Is there any way to compute the following conditional CDF? $$ P(X \le A(T) \ | \ T= t) $$ where $A(T)$ is a function with variable $T$.
I was trying to rewrite above conditional CDF by using definition $$ P(X \le A(T) \ | \ T=t) = \frac{P( \{X \le A(T)\} \cap \{ T=t \})}{P(T=t)}= \frac{P(X \le A(t))}{P(T=t)} $$ but $P(T=t) =0$, and so the above formula is undefined and thus this approach failed.
Any comments or suggestion is appreciated.
By definition, $P(X\lt A(T)\mid T)=B(T)$, where, for every bounded measurable function $C$, $$ E(C(T)\,\mathbf 1_{X\lt A(T)})=E(C(T)B(T)). $$ With no further indication on the joint distribution of $(X,T)$, it is difficult to go further. If $(X,T)$ was independent, then $$ E(C(T)\,\mathbf 1_{X\lt A(T)})=E(C(T)\Phi(A(T))), $$ where $\Phi$ is the standard normal CDF, hence, unsurprisingly, $$ B=\Phi\circ A. $$ But you say this is not the case... At another extreme, if $T=X$ with full probability, then $$ B(t)=\left\{\begin{array}{ccc}1&\text{if}&A(t)\gt t,\\0&\text{if}&A(t)\leqslant t.\end{array}\right. $$