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$X_1$ and $X_2$ , $...$ be variables that have standard normal distribution , How Can We prove the $X_1+X_2$ And $X_1-X_2$ Are Independent ?

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    $\begingroup$ Are $X_1$ and $X_2$ jointly normal? Maybe independent as well? $\endgroup$
    – gunes
    Jan 13, 2021 at 11:58
  • $\begingroup$ Yes They are independent and normal. $\endgroup$
    – ghostDs
    Jan 13, 2021 at 12:07
  • $\begingroup$ stats.stackexchange.com/questions/431329 and there's probably more $\endgroup$ Jan 14, 2021 at 20:42

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Since $A=X_1+X_2, B=X_1-X_2$ are linear transformations of jointly normal RVs, you'll just need to show $\operatorname{cov}(X_1+X_2,X_1-X_2)=0$ in order to show their independence.

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Consider that if two random variables $X$ and $Y$ are independent then $E[XY]=E[X]E[Y]$. Now, if $X$ and $Y$ are jointly normal also the inverse is true.

If $X_1$ and $X_2$ are jointly normal so is their sum and difference. Therefore, check that $E[(X_1+X_2)(X_1-X_2)]-E[X_1+X_2]E[X_1-X_2]$ = 0 holds.

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Since$$\left(\begin{matrix}X_1\\X_2\end{matrix}\right)\sim\mathcal N_2\left(\left(\begin{matrix}0\\0\end{matrix}\right),\mathbf I_2\right)$$ where $\mathbf I_2$ denote the identity matrix \begin{align*}\left(\begin{matrix}X_1+X_2\\X_1-X_2\end{matrix}\right)&= \left(\begin{matrix}1 &1\\1 &-1\end{matrix}\right)\left(\begin{matrix}X_1\\X_2\end{matrix}\right)\\ &\sim\mathcal N_2\left(\left(\begin{matrix}1 &1\\1 &-1\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right), \left(\begin{matrix}1 &1\\1 &-1\end{matrix}\right)\mathbf I_2 \left(\begin{matrix}1 &1\\1 &-1\end{matrix}\right)\right) \end{align*} and $$\left(\begin{matrix}1 &1\\1 &-1\end{matrix}\right)\mathbf I_2 \left(\begin{matrix}1 &1\\1 &-1\end{matrix}\right)= \left(\begin{matrix}2 &0\\0 &2\end{matrix}\right)$$ shows the absence of correlation.

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  • $\begingroup$ Why do the sum and difference have to be jointly normal? Only if they’re jointly normal does the lack of correlation imply independence. $\endgroup$
    – Dave
    Jan 13, 2021 at 13:06
  • $\begingroup$ @Dave: this is an assumption made by the OP. $\endgroup$
    – Xi'an
    Jan 13, 2021 at 13:12

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