Question

Bivariate normal random variable

Answer 1

Is it what we have to evaluate for part (i)
\[\int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^{{x_2}} { - \frac{1}{{57.6\pi }}} } \exp \left\{ {\frac{{25}}{{1152}}[{{({x_1} - 40)}^2} + {{({x_2} - 40)}^2} - 1.2({x_1} - 40)({x_2} - 40)]} \right\}d{x_1}d{x_2}\]

Answer 2

oh no, don't do that!!

Answer 3

prob(x2>x1)= prob(x2-x1>0)
if x1and x2 are bivariate then x2-x1 will be normal distribution...
E(x2-x1)= 40-40 = 0
var(x2-x1) = var(x2) + var(x1) - 2Cov(x2,x1) = 6^2 + 6^2 -2(0.6)(6)(6) = 28.8
stddev(x2-x1) = sqrt(28.8) = 5.367 so...
Z = (x2-x1 - 0) / 5.367 is std normal
prob(z>0) = 0.5 (which we could have said intuitively from the beginning due to the equal means and deviations of the respective distributions!)

Answer 4

Wow, that's a clever approach, thanks!