# Recent content by nowoman

1. ### Double integration with normal distribution

Thanks, jambaugh. I'll have a look at these methods. Thanks for the long discussion with you. Take care!
2. ### Double integration with normal distribution

Hi jambaugh, Thanks for much for your suggestions. The problem is actually a subroutine of a larger problem. The larger problem frequently refer to this problem. Hence, Monte Carlo method may be too expensive to implement it. Thanks though.
3. ### Double integration with normal distribution

yes. Do you have ideas how to get an approximate solution that is computable and accurate enough for general purpose? Thanks
4. ### Double integration with normal distribution

Thanks. Yes, I think if U and V are standard normal distribution, the problem is easy to solve. If U and V are general normal distributions, it is still very hard to solve. Any ideas?
5. ### Double integration with normal distribution

If we let U and V are standard normal distributions, we have P(U+V<0, U<0)=\int_{A}f(u,v)dA=\int_{3\pi /4}^{3\pi /2}\int_{0}^{\infty} f(r cos(\theta),rsin(\theta))rdrd\theta=\int_{3\pi /4}^{3\pi /2}\int_{0}^{\infty} f(r cos(\theta)) f(rsin(\theta))rdrd\theta=3/8 Does anyone verify me?
6. ### Double integration with normal distribution

Thanks for your information, tiny-tim. I have a bit more thoughts bout the problem. P(X+Y<b, X<a) Let U=X-a, V=Y-b+a, then, we have P(U+V<0, U<0), In this form, the integral area is a sector, we can use polar coordinates to solve it., Such as, P(U+V<0, U<0)=\int_{\theta}^{3\pi...
7. ### Normal Distribution Integral

I think in this case, x should be greater than zero.
8. ### Double integration with normal distribution

Thanks to both of you. I am glad to see your replies. But I still doubt the methods are feasible to solve the problem. We can see, P(X+Y<b,X<a)=\int_{-\infty}^{a}f(x)\int_{-\infty}^{b-x}f(y)dxdy =\int_{-\infty}^{a}f(x)erf(b-x)dx However, erf(b-x) does not have a close-form...
9. ### Double integration with normal distribution

Homework Statement Given X and Y are independent, normal distribution variable. a and b are constants. Homework Equations The probability of P(X+Y<b,X<a) The Attempt at a Solution P(X+Y<b,X<a)=\int_{-\infty}^{a}f(x)\int_{-\infty}^{b-x}f(y)dxdy Is there a close-form solution...