# Help with bivariate normal dist question

• Nov 23rd 2009, 06:20 PM
Danneedshelp
Help with bivariate normal dist question
Q: Consider $U_{1}=Y_{1}+Y_{2}$ and $U_{2}=Y_{1}-Y_{2}$. Assume $U_{1}$ and $U_{2}$ have a bivariate normal distribution. Show that $U_{1}$ and $U_{2}$ are independent.

My work: Suppose $U_{1}=Y_{1}+Y_{2}$ and $U_{2}=Y_{1}-Y_{2}$ have a bivariate normal distribution. To show $U_{1}$ and $U_{2}$ are independent we must show $Cov(U_{1},U_{2})=\sum_{i}\sum_{j}\\a_{i}b_{j}Cov(U _{1},U_{2})=0$.

I am having trouble with the latter computation:

$Cov(U_{1},U_{2})=\sum_{i}\sum_{j}\\a_{i}b_{j}Cov(U _{1},U_{2})
=a_{1}b_{1}Cov(Y_{1},Y_{1})+$

$a_{1}b_{2}Cov(Y_{1},Y_{2})+
a_{2}b_{2}Cov(Y_{2},Y_{2})$

How do I treat $Cov(Y_{1},Y_{1})$ given $U_{1}=Y_{1}+Y_{2}$ and $U_{2}=Y_{1}-Y_{2}$?

All in all, I'm pretty stuck. Any help would be greatly appreciated.

Thanks
• Nov 23rd 2009, 08:34 PM
matheagle
$Cov(Y_{1},Y_{1})=V(Y_1)$

you can prove that via the definition or the short cut formula

Use $Cov(X,Y)=E(XY)-E(X)E(Y)$ and replace X with Y.

$Cov(Y,Y)=E(YY)-E(Y)E(Y)=E(Y^2)-(E(Y))^2=V(Y)$