independent. A random vector (X, Y ) is said to be standard bivariate Gaussian if its joint density is given by: T f(x, y) = X exp E (27) V(1 + (EXY ) 2) Here E is the covariance matrix given by EXY E EY X 1 Therefore the inside of the exponent is -1/2(x2 +y +2(EXY)xy). Prove that if (X, Y) are standard bivariate and uncorrlated then they are independent. That is show that the joint denisty f(x, y) factorizes into joint densities for two standard Gaussian ran- dom variables X and Y.7. Bivariate Gaussian distribution Let X, Y be continuous random variables with correlation coefficient p and joint pdf f (x, y) -1 2 2 - HX y - MY (x - ux) (y - MY ) 2ToxoY V1 - 2xP + 2(1 - p2) - 2p ,x, y ER, OX OY OX OY where ox, oy > 0, ux, MY E R. Show that X, Y are uncorrelated they are independent. Remark: This is an example where independence coincides with uncorrelatedness.5. Suppose that (X1, X2) is a random vector with bivariate Gaussian distribu- tion such that X1 and X2 have the same variance. Define the new vector ( Y1, Y2) Y1 = X1+ X2 Y2 = X1 - X2. One can show (via some involved calculus) that (Y1, Y2) has a bivariate Gaussian distribution. Show that Y1 and Y2 are uncorrelated and thus in- dependent. Remark: It is striking that Y1 and Y2 are independent, because they are clearly derived from the same underlying randomness. What is happening is that the "positive" coupling of Y1 and Y2 through X1 exactly cancels the 'negative" coupling through X2.(a) a: has mean 2 and standard deviation 1, y has mean 4 and standard deviation 0.5, and the corre- lation between m and y is 70.5. (b) a: has mean 1 and standard deviation 1, and y is equal to 3