- #1
showzen
- 34
- 0
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.
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.