I need your help with this problem: Suppose $(X, Y)'$ follows a Bivariate Normal Distribution with parameters $μ_1 ,μ_2, σ_1^2, σ_2^2$, 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!
[Math] Bivariate Normal Distribution: Finding the joint distribution of functions of random variables
statistics
Related Solutions
Nobody is making assumptions in this matter; rather, someone is drawing conclusions. And those conclusions are not based simply on the marginal distributions. Conventionally, one has $$ X_1,\ldots,X_n \overset{\mathrm{i.i.d.}}\sim N(\theta,1), $$ $$ \bar X = \frac{X_1+\cdots+X_n} n \sim N\left(\theta,\frac 1 n \right) $$ This gives use the mean vector and the two variances. Then we have \begin{align} & \operatorname{cov}\left( X_1, \bar X \right) = \operatorname{cov}\left( X_1, \frac{X_1+\cdots+X_n} n \right) \\[6pt] = {} & \frac 1 n \Big( \operatorname{cov}(X_1,X_1)+\operatorname{cov}(X_1,X_2) + \operatorname{cov}(X_1,X_3)+\cdots +\operatorname{cov}(X_1,X_n) \Big) \\[6pt] = {} & \frac 1 n \Big( 1 + 0 + 0 + 0 + \cdots + 0 \Big) = \frac 1 n. \end{align}
Finally, we have the question of why this has a bivariate normal distribution. The definition of the statement that $(U,V)$ has a bivariate normal distribution is that for every pair of constants $a$, $b$ (and "constant" means not random) the random variable $aU+bV$ has a univariate normal distribution. So we need the distribution of $aX_1+b\bar X$. Observe that $$ aX_1+b\bar X = aX_1+b\frac{X_1+X_2+\cdots+X_n} n = \underbrace{\left(a+\frac b n\right)X_1 + b\frac{X_2}n+\cdots+b\frac{X_n}n}. $$ Notice that the part over the underbrace is a sum of independent random variables each of which is normally distributed. Consequently it is itself normally distributed. This conclusion is not based merely on the marginal distributions because it relies on independence.
That's the way to goal.
Let $X= \mu_1+\sigma_1~U~,~ Y= \mu_2+\sigma_2~(\rho~U+\sqrt{1-\rho^2}~V)$ where $U,V\mathop{\sim}\limits^\textrm{iid}\mathcal N(0,1)$ and $\rho=c/\sigma_1\sigma_2$
$$\begin{align}\mathsf {Cov}(X^2, Y^2) ~=~& \mathsf {Cov}\Big(\big(\mu_1+\sigma_1U\big)^2, \big(\mu_2+\sigma_2(\rho~U+\sqrt{1-\rho^2}~V)\big)^2\Big) \\[1ex] ~=~& \mathsf{Cov}\Big(\mu_1^2+2\mu_1\sigma_1U+U^2~,~ \mu_2^2+2\mu_2\sigma_2(\rho~U+\sqrt{1-\rho^2}~V)+\sigma_2^2\big(\rho^2U^2+2\rho\sqrt{1-\rho^2}~UV+(1-\rho^2)V^2\big)\Big) \end{align}$$
Simplify with the bilinearity of covariance, making use of: $\mathsf {Cov}(U,U) = 1, \mathsf {Cov}(U,V)=0, \mathsf{Cov}(U, U^2)=0,\mathsf {Cov}(U,UV)=0$ etc. (...why?)
Best Answer
The vector $(X,Y)$ is gaussian and $(U,V)=(X+Y,X-Y)$ is a linear function of $(X,Y)$ hence $(U,V)$ is gaussian as well. In particular, the distribution of $(U,V)$ is characterized by its mean vector $M$ and covariance matrix $C$. By definition, $$ M=\begin{pmatrix}\mathbb E(U) \\ \mathbb E(V)\end{pmatrix},\qquad C=\begin{pmatrix}\mathrm{var}(U) & \mathrm{cov}(U,V)\\ \mathrm{cov}(U,V)&\mathrm{var}(V) \end{pmatrix}, $$ with
Except when $\varrho^2=\sigma_X^2\sigma_Y^2$, the distribution of $(U,V)$ has a density $f_{U,V}$, defined by $$ f_{U,V}(u,v)=\frac1{2\pi\sqrt{\det C}}\exp\left(-\frac12\left(u-\mathbb E(U),v-\mathbb E(V)\right)^*C^{-1}\left(u-\mathbb E(U),v-\mathbb E(V)\right)\right). $$