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I found on another thread* that if $(X+Y)$ is independent $(X-Y)$, and if $X,Y$ are i.i.d., then $X,Y$ are $\mathcal{N}(0,1)$ distributed.

Is also the opposite true? Being $X,Y$ i.i.d $\mathcal{N}(0,1)$, then $X+Y$ is independent of $X-Y$?

here * $X$ and $Y$ i.i.d., $X+Y$ and $X-Y$ independent, $\mathbb{E}(X)=0 $and $\mathbb{E}(X^2)=1$. Show $X \sim N(0,1)$

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Vittorio Apicella
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\begin{align*} \operatorname{cov}(X+Y,X-Y)\equiv&\,\mathbb E[(X+Y)(X-Y)]-[\mathbb E(X+Y)][\mathbb E(X-Y)]=\mathbb E(X^2)-\mathbb E(Y^2)-0\times0\\ =&\,1-1=0.\end{align*}

Hence, $X+Y$ and $X-Y$ are uncorrelated. It is not difficult to see that $X+Y$ and $X-Y$ are jointly normal. Now recall that two jointly random normal variables are independent if and only if they are uncorrelated.

triple_sec
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  • :| I feel really stupid for not having thought to that, thanks! – Vittorio Apicella Aug 23 '14 at 08:01
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    To be uncorrelated wouldn't you need $Cov(X+Y,X-Y)=E[(X+Y)(X-Y)]-E[X+Y]E[X-Y]=0$ this is still true since $E[X-Y]=0$ but just wanted to make sure that was stated. – Kamster Aug 23 '14 at 08:02
  • Now I think of it that may have been to obvious that it didn't need to be stated lol – Kamster Aug 23 '14 at 08:03
  • yes, it is, Both are $\mathcal{N}(0,1)$ – Vittorio Apicella Aug 23 '14 at 08:03
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    @user159813 You're right, I added it for completeness. – triple_sec Aug 23 '14 at 08:05
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    Okay, yea it does nags at me sometimes when thats not stated because it looks like sometimes people want to imply if two random variables are orthogonal then they are uncorrelated (PS totally unrelated but noticed from your profile you are also economics major and wanted to say hello to fellow econ major) – Kamster Aug 23 '14 at 08:06
  • @user159813 Oh hello, fellow economist. :-) – triple_sec Aug 23 '14 at 08:18
  • I know that this is not where I should be commenting about this, but as of now I am an undergraduate economics deciding if I want to grad school and I would love if I could email you some questions I have? I don't really know if there is a way to private chat on this site – Kamster Aug 23 '14 at 08:23
  • @user159813 As far as I know, StackExchange is intended to be fully public and no private chat is available. If you leave your address on your contact page, I will write you an e-mail. – triple_sec Aug 23 '14 at 08:32