As asked in the title? Does the independence of two random variables $X$ and $Y$ imply the independence of $X$ and $X+Y$? If so, what's the easiest way to prove that?
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2$\begingroup$ If $X$ and $Y$ are independent, knowing the value of $X$ tells you nothing that you already did not know about $Y$. But does knowing the value of $X$ tell you something about the value of $X+Y$ that you did not know before? $\endgroup$– Dilip SarwateApr 7, 2015 at 16:27
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No: If $X=\pm1$ with equal probability and independently $Y=\pm2$ also with equal probability, and you tell me the value of $X+Y$, then I can tell you the value of $X$ (and of $Y$).
So in this case $X$ and $X+Y$ are not independent.
And in general they usually will not be, as $\text{cov}(X,X+Y)=\text{var}(X)+\text{cov}(Y,X)$. So if $X$ and $Y$ have finite positive variances, then their being independent makes $\text{cov}(Y,X)=0$ and so $\text{cov}(X,X+Y)=\text{var}(X)>0$, meaning $X$ and $X+Y$ cannot be independent.
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$\begingroup$ What if $X$ and $Y$ are identically distributed? $\endgroup$ Apr 7, 2015 at 16:53
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3$\begingroup$ They will still not be independent: if each is $\pm1$ then knowing $X+Y=-2$ will tell you $X$. $\endgroup$– HenryApr 7, 2015 at 17:04