Consider three independent normally distributed variables: A, B, C. How would you calculate the distribution $p(A | A + B > C)$?
I know that the distribution $p(A + B | A + B > C) = p(A+B) \times CDF(C)/s$, where s is a normalising factor, but I'm not sure whether it is possible to use this to calculate what A is.
Hint: Since the r.v. are independent then:
$$ p_A(u\mid A+B>C) \; = \; \dfrac { p_A(u)\; \mathsf P(B+u > C) }{ \mathsf P(B+A>C)} $$
Can you calculate the two probabilities ?: $\mathsf P(B+u > C), \mathsf P(B+A > C)$