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If we have three random variables (not independent in general) $X$, $Y$, $Z$ can we conclude that $(X+Y, X+Z)\approx (Y,Z)$ about their joint distributions? If not, what is the minimum requirement on $X$, $Y$, $Z$ to hold this? $(X+Y, X+Z)$ is a two dimensional random variable and so $(Y,Z)$.

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What is $(X,Y)$ in this context? –  Ross Millikan Mar 9 '11 at 14:58
    
It is two dimensional random variable. –  user8020 Mar 9 '11 at 15:05

1 Answer 1

You would have to show that $M_{X+Y,X+Z}(s,t) = M_{Y,Z}(s,t)$ (i.e. the joint moment generating functions are the same).

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ok, if $X$ is independent of $Y$ and $Z$, $(X+Y,X+Z)\approx (Y,Z)$, but is there any other result for which they are dependent? –  user8020 Mar 9 '11 at 15:22
    
@black It is not true that if $X$ is independent of $Y$ and $Z$, then $(X+Y,X+Z)\approx (Y,Z)$. I assume that you use $\approx$ to mean "identically distributed". –  Byron Schmuland Mar 9 '11 at 16:02
    
@ Byron Schmuland, in my context $X \approx Y$ means distribution function of random variable $X$ approximately equals $Y$. Then if $X$ is independent of $Y$ and $Z$, then it actually means $M_{X+Y,X+Z}\approx M_{Y,Z}$. Am I wrong? or is there any precise notion that I can use? –  user8020 Mar 9 '11 at 16:33
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@black If you want a good answer, I think you ought to add a precise definition of $\approx$ to your question. –  Byron Schmuland Mar 9 '11 at 17:58
    
even if X is independent of Y and Z (and even if you shift the coordinates to the resulting mean, as you seem to implicitly do ; what you be equivalent of taken X with mean 0), even then, the nuew variable would have larger variance. –  leonbloy Jul 7 '11 at 17:37

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