It is easy to show that if we have n independent normally distributed random variables, then a linear combination fo them ar normally distributed.
It is also said that if (x1,x2,..,xn) is multivariate normally distributed, but not nececarrily independent, then any linear combination is also normally distributed. This is stated here:
But does this mean that any linear combination of normally distributed random variables are normally distributed, even if they are not independent? This will follow from the definition if the joint distribution of set of normally distributed random variables(not nececarrily independent) are jointly multivarite distributed?
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$\begingroup$This will follow from the definition if the joint distribution of set of normally distributed random variables(not nececarrily independent) are jointly multivarite distributed?
Yes.
But does this mean that any linear combination of normally distributed random variables are normally distributed, even if they are not independent?
No. A frequently mentioned counterexample is based on $X$ standard normal and $Y=SX$ with $S=\pm1$ symmetric and independent of $X$. Then $X$ and $Y$ are normal but $X+Y$ is not since $P(X+Y=0)=\frac12$, a property that no normal random variable satisfies.
$\endgroup$ 5 $\begingroup$You need independence. Otherwise the simplest counterexample is $X$ and $-X$, which are both normally distributed, but their sum $X+(-X) = 0$ is not.
$\endgroup$ 5 $\begingroup$Just thinking in the univariate case, note that if two RVs are iid, distributed as $N(\mu, \sigma^2)$, then their sum is distributed as $N(2\mu, 2\sigma^2)$. If they are correlated with $\rho=1$, it is the same as $2X=X+X$, which is distributed as $N(2\mu, 4\sigma^2)$ - in other words, it is 'stretched out' twice as far in variance. This makes sense, because uncorrelated would have a standard deviation that grows by $\sqrt 2$, whereas adding to itself must double the standard deviation (which is equivalent to $4\sigma^2$).
All other cases of correlation fall in-between. Though, as people have pointed out, $pho=-1$ gives you zero as a degenerate point.
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