Graduate Calculating Bivariate Normal Probabilities

[showzen]

Hello good people of PF, I came across this problem today.

Problem Statement
Given bivariate normal distribution $X,Y \sim N(\mu_x=\mu_y=0, \sigma_x=\sigma_y=1, \rho=0.5)$,

determine $P(0 < X+Y < 6)$.

My Approach
I reason that
$$ P(0 < X+Y < 6) = P(-X < Y < 6-X)$$
$$ = \int_{-\infty}^{\infty} \int_{-x}^{6-x} f(x,y) dy dx$$
where $f(x,y)$ is the bivariate normal density with parameters above.
I could not solve this problem analytically, but numerically I get an answer of 0.499734.

Discussion
First, I would like to know if my reasoning is correct?
Second, is there a better method for this type of calculation? I am especially interested in any analytic solutions.

 

[StoneTemplePython]

I didn't see you explicitly use $\rho$ in here or what $f(x,y)$ is or how you did this numerically, though the answer looks about right...

Second, is there a better method for this type of calculation? I am especially interested in any analytic solutions.

Yes. In terms of streamlining the problem, a single 1-D random variable is easier to work with than a 2-D joint random variable, so consider $Z:=(X+Y)$

The fact is you should be able to easily calculate the mean, and variance of $Z$ and (not so easily) confirm that this is a normal random variable. $Z$ has zero mean but is normal so $Pr(Z \leq 0) = \frac{1}{2}$, and you now want to (i) estimate or bound the tail -- in particular find probability that $Pr(Z \geq 6)$ then (ii) add it to $\frac{1}{2}$ and (iii) then find the complement.

For (i), as a hint I'd suggest dividing both $Z$ and $6$ by the standard deviation of $Z$ as standard normal random variables are easiest to work with... you could look the result up in a table but I'll flag that this is a rare event probability i.e. it is pretty far out on the distribution of $Z$ so there are numerous analytical upper and lower bounds that may be used here to show that the associated probability is quite small. You're generally not going to find analytically useful integrals of Gaussians, so look to estimate and bound if you don't want to go a numeric route.

edit: using these analytic bounds, I can get

$0.499715 \lt P(0 < X+Y < 6) \lt 0.499736$

the bound on the left side is easy to derive, the one on the right is unfortunately rather difficult, though an internet search will of course find numerous bounds to choose from.

second edit:
The bound on the left is given by the 'slip-in trick' for integrals. While the other bound I used is a bit too involved, a close one (plus the slip-in related bound) is nicely given here:
https://www.johndcook.com/blog/norm-dist-bounds/