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Are marginal distributions of the random variables comprising a jointly gaussian random vector always Gaussian?

This stems from my confusion with the Central Limit Theorem which loosely states that sum of a sufficiently large number of independent random variables tends to be normal under mild constraints irrespective of the distributions of each random variable.

The second statement is that of a Gaussia Random process in time, which states that "for any number n of samples, any sampling times $t_1,t_2\ldots ,t_n$, and any scalar constants $a_1,a_2\ldots a_n$, the linear combination $a_1X(t_1)+a_2X(t_2)+\ldots +a_nX(t_n)$ is a jointly gaussian random variable."

Now, for Jointly gaussian random variables $X_1,X_2,\ldots,X_n$, any linear combination of these random variables is a gaussian random variable. Then, for all but one scalar coefficients $a_i$ set to zero, the resulting random variable would still be a Gaussian and hence, the marginal $X_i$ would be gaussian.

But, CLT states that it could be of any distribution!

I am confused with these two lines of thought. Please remedy my confusion.

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    $\begingroup$ The CLT says absolutely nothing about finite linear combinations of random variables. It asserts a limiting distribution is Gaussian, not that it "could be any distribution"! Why don't you review our posts on the CLT? $\endgroup$ – whuber Sep 30 '16 at 19:10
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    $\begingroup$ It is worth adding that the converse statement need not be true: R.V.s can have marginal Normal distributions but their joint distribution might not Multivariate Normal. A simple example for Bivariate Normal: let $X$ be a standard Normal R.V. and $S$ be a random sign R.V. which is independent of $X$. $S$ takes on values $-1$ or $1$ with equal probabilities of $\frac{1}{2}$. Now define $Y=SX$, which is also Normally distributed (by symmetry of the Normal distribution). Now, $P(X+Y=0)=P(S=-1)=\frac{1}{2}$. ... ctd $\endgroup$ – slazien Sep 30 '16 at 19:40
  • $\begingroup$ ctd ... Since $X+Y$ is a linear combination of two standard Normal R.V.s which is not Normal, $(X,Y)$ is not Bivariate Normal. $\:\:\:$ This neat example was taken from Blitzstein's Introduction to Probability book $\endgroup$ – slazien Sep 30 '16 at 19:40
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The assertion "$a_1X(t_1)+a_2X(t_2)+\ldots +a_nX(t_n)$ is a jointly gaussian random variable" in your second statement is almost completely correct: $a_1X(t_1)+a_2X(t_2)+\ldots +a_nX(t_n)$ is indeed a gaussian random variable but since it is just one variable all by itself, the adjective "jointly" is not needed.

Definition: A random process $\{X(t) \colon t \in \mathbb T\}$is called a Gaussian random process if all the finite-dimensional distributions of the process are (multivariate) Gaussian distributions.

A more prolix description is that each random variable $X(t), t \in \mathbb T$ is a Gaussian random variable, and for any integer $n \geq 2$ time instants $t_1, t_2, \ldots, t_n \in \mathbb T$, the $n$ random variables $X(t_1)$, $X(t_2)$, $\ldots X(t_n)$, are jointly Gaussian random variables.

Definition: The random variables $X_1, X_2, \ldots, X_n$ are said have a multivariate Gaussian distribution (or they are called jointly Gaussian random variables) if for all choices of real numbers $a_1, a_2, \ldots, a_n$, the random variable $a_1X_1+a_2X_2+\cdots+a_nX_n$ is a Gaussian random variable.

As the OP has noted, this implies that the marginal distributions of the $X_i$ are Gaussian. Note that also that when each $a_i$ is $0$, the sum is also $0$ and we are accepting this constant as a degenerate Gaussian random variable (cf. the extended discussion in the comments following this answer).

Inserting the latter definition into the former, we get the description of Gaussian random process used by the OP. Note that each random variable from a Gaussian random process is indeed a Gaussian random variable. That is, the answer to the OP's question

Are marginal distributions of the random variables comprising a jointly gaussian random vector always Gaussian?

is an unequivocal Yes.

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