Recent content by nowoman

Thanks, jambaugh. I'll have a look at these methods. Thanks for the long discussion with you. Take care!

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.

yes. Do you have ideas how to get an approximate solution that is computable and accurate enough for general purpose? Thanks

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?

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?

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...

I think in this case, x should be greater than zero.

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...

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...