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I have two normally distributed random variables $X$ and $Y$. Then I know that the sum $X-Y$ is also normally distributed (i).

However, I want to show (preferably by a counter example) that the opposite is not true (ii). If $X-Y$ is normally distributed, this does not mean that $X$ and $Y$ are normally distributed. \begin{align} X\sim N(\mu_x,\Sigma_x)\text{ and }Y\sim N(\mu_y,\Sigma_y) \underbrace{\overbrace{\stackrel{\Rightarrow}{\nLeftarrow}}^{(i)}}_{(ii)}Z=X-Y\sim N(\mu_z,\Sigma_z) \end{align}

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Let $W$ (for example) be a random variable which takes on values $0$ and $1$, each with probability $\frac{1}{2}$. let $Z$ be standard normal, and suppose $W$ and $Z$ are independent. Let $X=W+Z$ and let $Y=W$. Then $X-Y$ is normally distributed, but neither $X$ nor $Y$ is normal.

Remark: Note that if $X$ and $Y$ are normal and not independent, then $X-Y$ need not be normal. For example, let $X$ be standard normal, and let $Y=RX$, where $R$ and $X$ are independent, and $R$ takes on values $-1$ and $1$ each with probability $1/2$. Then $X-Y$ is not normal.

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  • $\begingroup$ I believe $X-Y$ is indeed normal, even if $X$ and $Y$ aren't independent. Non-independence just means the variances aren't additive and you need to include the correlation coefficient in the calculation. $\endgroup$ – Stanley May 22 '15 at 6:59
  • $\begingroup$ @SimonRigby: Thank you for the comment. But $X-Y$ need not be normal if $X$ and $Y$ are (non-independent) normal. I expanded the remark by giving an example. $\endgroup$ – André Nicolas May 22 '15 at 7:09
  • $\begingroup$ Interesting example. I'm guessing this $X-Y$ would have a bimodal distribution? $\endgroup$ – Stanley May 22 '15 at 7:19
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    $\begingroup$ We get that $\Pr(X-Y=0)=1/2$. So $X-Y$ does not even have continuous distribution. $\endgroup$ – André Nicolas May 22 '15 at 7:25
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Take some $Z$ following a normal distribution, and $X$ any other. Then set $Y=X-Z$.

$X-Y$ follows a normal distribution, while $X$ doesn't.

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