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"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 $Cov(X_1+X_2,X_1-X_2)=0$)

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$Cov$ 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}

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  • $\begingroup$ Thanks, Just what I had missed! Do you think there will ever be a scenario where I will need to integrate as above (If X1 and X2 are still indpt)? $\endgroup$
    – litmus
    Commented Dec 29, 2015 at 19:05
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    $\begingroup$ It may happen. But before you integrate, apply these properties first. This will simplify things for sure. In most cases it will suffice. $\endgroup$
    – Jimmy R.
    Commented Dec 29, 2015 at 21:37
  • $\begingroup$ Ok, thanks Stef! $\endgroup$
    – litmus
    Commented Dec 30, 2015 at 13:39

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