bivariate normal integral

[Avinash Burra]

I submit to the the collective wisdom of quantnet this humble request -- I am trying to find a closed form solution (if possible) for the following integral:

( \frac{1}{2 \pi \sqrt{1-\rho^2}} \int_\epsilon^\infty {\int_{-\infty}^{\infty} {e^{\frac{-(x^2 + y^2 -2 \rho x y)}{2 (1-\rho^2)}} dx dy ) where (\rho) is a non-zero correlation between the two normal random variables.

If I could be pointed towards a technique that can get the answer in terms of error functions or something similar, I would be much obliged. I understand that lookup tables exist for this sort of thing, but I am trying to avoid that and brute force methods like simpson's rule. Is it possible to transform this somehow into a conic section?

In my naivete, I entered this into mathematica and for lack of a better word, it vomited a nonsense answer after three hours of churning.

 

[Mr Doe]

write the pdf as (const_1 \exp\{-const_2 \cdot \textbf{x}^{\tau }\Sigma^{-1} \textbf{x}\}) where (\textbf{x}=(x,y)). There's 1-1 link between the covariance matrix and the ellipsoid it spans. Now stretch-contract and rotate that ellipsoid (hint eigenvalue decomposition), it will split into two separate integrals, then you can write it with error functions.

Also, you are missing a couple of things in your pdf.

 

[Avinash Burra]

write the pdf as (const_1 \exp\{-const_2 \cdot \textbf{x}^{\tau }\Sigma^{-1} \textbf{x}\}) where (\textbf{x}=(x,y)). There's 1-1 link between the covariance matrix and the ellipsoid it spans. Now stretch-contract and rotate that ellipsoid (hint eigenvalue decomposition), it will split into two separate integrals, then you can write it with error functions.

Thanks a lot, I'm going to give that a shot. In my googling, I came across integrals for the above over an offset ellipse which used eigenvalue decomposition, but I assumed that they were dealing with horizontal sections, rather than the vertical section I need.

Also, you are missing a couple of things in your pdf.

Do you mean the Mu and Sigma terms? The standard bivariate normal is good enough for my purposes.