probability of event on two random variables and their sum

Question

Let $x \sim N(0,1)$ and $y \sim N(0,1)$ be two independent random variables that follow a standard normal. I'm trying to find the probability of the following event: $$\Pr(x > a, y > b, x + y > c)$$ We have that $x + y \sim N(0,2)$, but besides that, I'm a bit stuck. I've been thinking of something along the lines of $$\Pr(x > a, y > b, x + y > c) = \Pr(x > a) \Pr(y > b) \Pr(x+y > c | x>a, y>b)$$ So we'd have $$\Pr(x > a, y > b, x + y > c) = [1-\Phi(a)] [1 - \Phi(b)] \Pr(x+y > c | x>a, y>b)$$ but I'm not sure how to move forward with $\Pr(x+y > c | x>a, y>b)$. Perhaps use a truncated normal? But if so, how?

Answer 1

This is an extended comment rather than an answer. Using Mathematica the end result (assuming $c \geq a+b$) is

$$\int_a^{c-b} \frac{e^{-\frac{x^2}{2}} \Phi (x-c)}{\sqrt{2 \pi }} \, dx+\frac{1}{2} \Phi (-b) (2-2 \Phi (c-b))$$

I don't see it simplifying much more than that except in a few special cases such as $c=0$:

$$-\frac{1}{2} \Phi (-a)^2+\Phi (-a)-\frac{1}{2} \Phi (-b)^2+\Phi (-b)-\frac{1}{2}$$