We have two genes X and Y. Let $(X,Y)\sim N(\mu_x=9,\mu_y=10,\sigma^2_x=3,\sigma^2_y=5,\rho\sigma_x\sigma_y=2)$. To find $P(X+0.5<Y)$ the probability that the sample mean for the second gene exceeds the sample mean of the first gene by more than 0.5
Which can convert to $P(X+0.5<Y)$ => $P(Y-X>0.5)$
From my understanding, i probably can determine the mean and variance of the distribution Z:Y-X and then use the CDF of a univariate normal distribution?
Thanks whuber and Glen_b pointed out my mistakes.
Since (X,Y) is normally distributed then $Y-X\sim N(\mu_y-\mu_x, \sigma_y^2-2\rho \sigma_x\sigma_y+\sigma_x^2)\sim N(1, 25-4+9)$ => $\sim N(1, 30)$
Using R we get,
> 1- pnorm(0.5,mean=1,sd=sqrt(30))  0.5363678
Additionally, how do i use Monte Carlo simulation to approximate this probability, providing a 95% confidence interval for your estimation?