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.
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...
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...