Bivariate Normal Distribution: Joint distribution of functions of random variables

Hi, I need your help with this problem: Suppose (X, Y)' follows a Bivariate Normal Distribution with parameters μ1 ,μ2, σ1^2, σ2^2, and ρ. Let U = X + Y and V = X - Y. Considering that X and Y are not independent random variables, how will I get the joint distribution of U and V? Thanks in advance!

Re: Bivariate Normal Distribution: Joint distribution of functions of random variable

Hey Mach.

You will have to use a change of variables in your integral. Have you come across the substitution theorem for multivariable integration?

Re: Bivariate Normal Distribution: Joint distribution of functions of random variable

Well, I believe you can use their linear combination theorem since they are assumed bivariate normal, that is when we have the scalar product

U=AX+BY=(A B)'(X Y). In this case, we need A=1, B=1 to have U=X+Y

The mean of U is thus given by

$\displaystyle \mu_U=$(1 1)$\displaystyle (\mu_x \mu_y)'=\mu_x+\mu_y$

The variance of U is given by

$\displaystyle \sigma^2_U$=(1 1)$\displaystyle \Sigma_{X,Y}$(1 1)'

where $\displaystyle \Sigma_{X,Y}$ is the covariance matrix of X and Y.

Note that the variance matrix Sigma is not just a diagonal matrix like in the independence case. Here, the covariance elements $\displaystyle \sigma_{x,y}$ are assumed to be non zero.

Therefore, we obtain $\displaystyle \sigma^2_U=\sigma^2_X+\sigma^2_Y+2 \sigma_{X,Y}$ (as expected)

And since they are bivariate normal by assumption, we obtain

$\displaystyle U \sim N(\mu_x+\mu_y, \sigma^2_X+\sigma^2_Y+2 \sigma_{X,Y})$

Similar for V. In order to get V, replace (1 1) by (1 -1) as then you have V=X-Y. Then redo the matrix calculations for the joint mean and variance.