Finding the P(2x1>x2) for a bivariate normal distribution

ahuds001

1. The problem statement, all variables and given/known data

Given a bivariate normal distribution with E(x1)=4 and E(x2) = 6 and Var(X) = [3 2.5]
[2.5 7]
Find P(2*x1>x2)

2. Relevant equations

The cdf of this bivariate normal distribution is given by:

f(x1,x2)=1/(2*pi*var(x1)*var(x2)*sqrt(1-rho^2)) * e^(-0.5*(z/(2*1-rho^2)))

where var(x1) = 3, var(x2) = 7, E(x1)=4 and E(x2) = 6, and rho = 2.5/(sqrt(3)*sqrt(7))

and z = ((x1-E(x1))^2)/var(x1) + ((x2-E(x2))^2)/var(x2) - (2*rho*(x1-E(x1))*(x2-E(x2)))/sqrt(var(x1)*var(x2))

3. The attempt at a solution

The approximate cdf of this bivariate normal distribution is given (in terms of x and y) by:

.041440417*%e^(-0.71186*((((x-4)^2)/3)+(((y-6)^2)/7)-(1.09109*(x-4)*(y-6))/4.582576))

Taking the integral from (-infinity to 2*x) dy and then from (-infinity to + infinity) dx should do the trick, but I have been unable to do so even using approximations.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

Ray Vickson

Can you figure out the distribution of the single random variable Y = 2*X1 - X2?

RGV

ahuds001

I can't, not sure if I am missing something or just being thick.
Working only on z=((x1-4)^2)/3 + ((x2-6)^2)/7 - 5*(x1-4)(x2-6)/21

Distributing I get z=(7x1^2+3x2^2-26x1-16x2-5x1x2+100)/21

My best attempt has z = (2y^2-4y-x1^2+x2^2+3x1x2-20x2-18x1+100)/21

ahuds001

Figured it out, thanks for the idea, it was extremely helpful :-)