Show $\frac14\leqslant P(\min(X,Y)>10)\leqslant\frac12$ for this bivariate normal $(X,Y).$

Question

I have a question that I've been struggling over. Let $(X,Y)$ have a bivariate normal distribution with $E(X)=E(Y)=10$ and $Var(X)=Var(Y)=16$ and $Corr(X,Y)=1/2.$ The question (and solution) is as follows:

I understand why $P(\min(X^*,Y^*)>0 = P(X^* > 0,\, Y^*>0).$ This is because if the minimum value is greater than $0,$ then both $X$ and $Y$ will be greater than $0.$ However I don't understand the steps after this on proving the lower/upper bounds. Any help would be greatly appreciated!

Answer 1

Notice that $Z \sim Norm(0,1)$ because $$0 = E(Y^*) = E[(1/2)X^* + (\sqrt{3}/2)Z] = 0 + (\sqrt{3}/2)E(Z)$$ and $$1 = V(Y^*) = V[(1/2)X^* + (\sqrt{3}/2)Z] = (1/4) + (3/4)V(Z).$$

Also, $X^* > 0$ and $Z > 0\,$ imply $(1/2)X^* + (\sqrt{3}/2)Z > 0.$ which takes care of the $\ge$ in the long display ending with $1/4.$

The following simulation in R may help you understand some of the relationships and the exact value of $P(\min(X,Y) > 10).$ It is based on a million realizations of each random variable. I have used 1s instead of $*$'s. Approximations should be accurate to two or three places. The first few lines of code merely verify that the random variables have the correct moments.

m = 10^6
z = rnorm(m);  x1 = rnorm(m)  # standard normal
y1 = x1/2 + sqrt(3)*z/2
x = 4*x1 + 10;  y = 4*y1 + 10
mean(x);  sd(x)
## 10.00258   # aprx E(X) = 10
## 3.999907   # aprx SD(X) = 4
mean(y);  sd(y)
## 9.998794   # aprx E(Y) = 10
## 3.994897   # aprx SD(Y) = 4
cor(x,y)
## 0.4991412  # aprx Cor(X,Y) = 1/2
w = pmin(x,y) # a million realizations of min(X,Y)
mean(w > 10)
## 0.333126   # aprx P(W > 10) btw 1/4 and 1/2

The scatterplot below shows 30,000 simulated $(X, Y)$ points, suggesting the density function of their joint multivariate normal distribution. The (roughly 10,000) points for which $\min(X,Y) > 10$ are plotted in orange.