I recently came across and exercise from a past exam and I was wondering how to solve it.
Two independent random variables $A$ and $B$ are given and they both follow the exponential distribution but have different parameters.
$$A \sim Exp(\lambda_a), B \sim Exp(\lambda_b)$$
We next consider another r.v. $C = min(A,B)$ and we are asked to find the conditional probability :
$$ P(C > c | B < b) $$
My reasoning was as follow :
$$P(C > c| B > b) = 1 - P(C <= c | B > b) = 1 - F_{C|B}(c,b) = 1 - \int_{-\infty}^{c}P_{C|B}(v,b)dv$$
Yet I can't figure out how to calculate $P_{C|B}(c,b)$
I figured that $F_{C}(c)$ is equal to $1 - e^{-\lambda_{c}c}$ where $\lambda_{c} = \lambda_{a}+ \lambda_{b}$ but I'm unsure how this could be helpfull in computing $F_{C|B}(c,b)$ or even $P_{C|B}(c,b)$.
Any hint?
Edit : I've tried another aproach, here it is.
We distinguish two cases : c <= b and c > b.
In the first case since c <= b and B > b, C will be grater than c in all those cases where either A is greater than B or A is smaller than B but greater than c. This is also the complement of the event that A is smaller than c
In other words :
$$P(C>c | B>b) = 1 - P(A <= c)$$
In the second case since c>b and B>b we would have to require that A > c and A < b wich is unreasonable, thus the probability is 0.
Is this somehow correct?
Hint:
$$P(C > c | B < b) = \frac{P(C > c , B < b)}{P(B<b)}=\frac{P(A > c , B>c, B < b)}{P(B<b)}$$