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Given $x$ and $y$ independent and $N(0,1)$, show that $x+y$ and $x-y$ are independent.

I started like this $x+y=u$ and $x-y=v$ :

I want to prove this :

$$P[u \leq t,v \leq s]=P[u \leq t][v \leq s]$$

However, I have some trouble starting this calulation since i don't know what is $f_{u,v}$ the density of $u$ and $v$. Anyone can help? thank you

Dave ddd
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1 Answers1

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Hint:

If $u,v$ have a joint normal distribution then they are independent if their covariance equals $0$.

Apply this on $u=x+y$ and $v=x-y$.

edit (here I use capitals for random variables)

In general if $\vec{Z}\sim N\left(\vec{\mu},\Sigma\right)$ then $$A\vec{Z}\sim N\left(A\vec{\mu},A\Sigma A^{T}\right)$$ for matrix $A$.

Note that $$\left(\begin{array}{c} X\\ Y\end{array}\right)\sim N\left(\left(\begin{array}{c} 0\\ 0\end{array}\right),\left(\begin{array}{cc} 1 & 0\\ 0 & 1\end{array}\right)\right)$$ and $$\left(\begin{array}{c} U\\ V\end{array}\right)=\left(\begin{array}{cc} 1 & 1\\ 1 & -1\end{array}\right)\left(\begin{array}{c} X\\ Y\end{array}\right)$$ leading to $$\left(\begin{array}{c} U\\ V\end{array}\right)\sim N\left(\left(\begin{array}{c} 0\\ 0\end{array}\right),\left(\begin{array}{cc} 2 & 0\\ 0 & 2\end{array}\right)\right)$$

Actually this shows that $\left(\begin{array}{c} U\\ V\end{array}\right)$ has the same distribution as $\sqrt{2}\left(\begin{array}{c} X\\ Y\end{array}\right)$ so an immediate conclusion about independence is at hand.

If you insist on proving the line mentioned in your question then based on this you can also find $f_{U,V}$.

drhab
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  • Thank you. My purpose is doing it using the definition of indepedence as given – Dave ddd Mar 22 '16 at 09:57
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    @Cm7F7Bb In principle if $\vec{X}\sim N\left(\vec{\mu},\Sigma\right)$ then $A\vec{X}\sim N\left(A\vec{\mu},A\Sigma A^{T}\right)$ for matrix $A$. Easy? It might be that the OP is aware of it allready. – drhab Mar 22 '16 at 10:02
  • Have you seen what me and @Cm7F7Bb discussed here? In your question you wonder how to find the density of $(u,v)$. For that you can use it. – drhab Mar 22 '16 at 10:16
  • @ Cm7F7Bb yes i know that the sum of two independet normal r.v. is normal as well that's why i put $P[u \leq t]P[v \leq s]$ as a target in the calculation. but now what about the joint distribution of $u$ and $v$ ? i mean how do you write $P[u \leq t,v \leq s]$ as an integral of the density? (and what is the explicit density here) – Dave ddd Mar 22 '16 at 10:26
  • Can you find expectation and covariance-matrix of $(u,v)^T$? They completely determine the distribution of the vector. – drhab Mar 22 '16 at 10:36