# Check if $Cov(X_1+X_2,X_1-X_2)=0$, i.e. if independent?

"Let $X_1$ and $X_2$ be independent, $N(0,1)$-distributed random variables. Show that $X_1+X_2$ and $X_1-X_2$ are independent."

I know that for multivariate normal distributions independence can be proven if the covariance equals zero. So if I let $Y_1=X_1+X_2$ and $Y_2=X_1-X_2$, then:

$$Cov(Y_1,Y_2)=E[(Y_1-\mu_1)(Y_2-\mu_2)],$$ where $$\mu_1=E[Y_1]=E[X_1+X_2]=EX_1+EX_2=0+0=0$$ and $$\mu_2=E[Y_1]=E[X_1-X_2]=EX_1-EX_2=0-0=0.$$ Thus we get that $$Cov(Y_1,Y_2)=E[(Y_1-0)(Y_2-0)]=E[Y_1 \cdot Y_2]=$$ $$=\int \int y_1 y_2 \frac 1 {\sqrt{2\pi}} e^{\frac {-y_1^2} 2} \frac 1 {\sqrt{2\pi}} e^{\frac {-y_2^2} 2} dy_1dy_2$$ Now this is where I get lost. I am not sure about what limits I am supposed to be interating over? Can anyone please explain how I am supposed to think here? Thanks

(I already know that the answer is going to be $Cov(X_1+X_2,X_1-X_2)=0$)

$Cov$ has all the properties of an inner product. So, instead of trying to calculate this expected value explicitly, use these properties \begin{align}Cov(X_1+X_2,X_1-X_2)&=Cov(X_1,X_1)-Cov(X_1,X_2)+Cov(X_2,X_1)-Cov(X_2,X_2)\\[0.2cm]&=σ^2_{Χ_1}-Cov(X_1,X_2)+Cov(X_1,X_2)-σ^2_{X_2}\\[0.2cm]&=1-1=0\end{align}