I'm having trouble composing the proof for a random variable with conditional probability. Would like some help on proof construction. I'm still pretty new to the forum, the post format maybe a bit messy.
Problem:
Suppose $Y$ is a continuous random variable with distribution function $F(y)$, and density $f(y)$. Show that the conditional probability that $Y\lt y$, given that $Y$ exceeds a constant, $c$ is
$$ \mathbb{P}(Y\lt y | Y\ge c) = \frac {F(y)-F(c)}{1-F(c)}, $$ $$1-F(c)\neq0 $$
Proof Attempt:
$$\mathbb{P}(c \le Y \lt y )= P(Y \lt y)-P(Y \le c)=F(y)-F(c)$$
And
$$\mathbb{P}(Y \ge c )= 1- P(Y\lt c)= 1-F(c)$$
Therefore
$$\mathbb{P}(Y\lt y | Y\ge c) = \frac {F(y)-F(c)}{1-F(c)}$$
Recall that in general $\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}$. I do understand that $\Pr(B) = 1- P(Y\lt c) = 1-F(c)$, but is having trouble prove $F(y) - F(c) = \Pr(A\cap B)$ .It is not hard to see the overlap/intersection by drawing a number line, but how should I construct a proof that validates the relationship?
Hint: We know that the event A and B is $c \leq Y < y$. How can you write this using only the CDF?