Normal distribution and independence I was reading about white noise and it stated:

Although $\varepsilon_t$ & $y_t$ are serially uncorrelated, they are not necessarily serially independent, because they are not necessarily normally distributed. If in addition to being serially uncorrelated, $y$ is serially independent, then we say $y$ is independent white noise.

I cannot understand the link between the normal distribution and being serially independent. Can anyone help me with this?
 A: My interpretation of the (slightly paraphrased) statement that the OP is reading, viz.

"If $\{Y_n\}$ is a sequence of (serially) uncorrelated random variables, then $\{Y_n\}$ is not necessarily a sequence of independent random variables, because (emphasis added) they are not necessarily normally distributed"

is that the author is asserting (quite correctly) that uncorrelated random variables need not be independent but the reason for this
failure of uncorrelatedness to imply independence is that we cannot assert that the uncorrelated random variables are normally distributed.
If the author is not implying that uncorrelated random variables that are normally distributed are independent random variables, then he is
certainly begging the reader to jump to that (false) conclusion,
by musing in the the subjunctive mood: "Gol dang it, if only those pesky
$Y_n$'s were normally distributed in addition to being
uncorrelated, then we could take those
uncorrelated (and normal) $Y_n$'s to be independent random variables
and avoid a lot of headaches."   However, Moderator @whuber has stated
(in a comment on the main question) that he does not interpret that
sentence that way, and that the
statement quoted by the OP is perfectly accurate.
In my opinion, the second sentence quoted by the OP.

If in addition to being serially uncorrelated, the $\{Y_n\}$ are serially independent, then we say $\{Y_n\}$ is independent white noise.

also incorrect. Independent
random variables are always uncorrelated and it is unnecessary
to start with uncorrelated random variables and then impose the
additional constraint that they are independent random variables.
Furthermore, if by serially independent it is meant that
for all $n\neq m$, $Y_m$ and $Y_n$ are independent random variables
(that is, only pairwise independence is required), then I disagree
vehemently with the assertion that $\{Y_n\}$ is a white noise process.
For a random process to be called a white noise process, the
random variables need to be mutually independent, not just
pairwise independent, and most people, upon encountering the
phrase white noise process, are likely to
assume that the the random variables
constituting the white noise process also are zero-mean random variables with common finite variance $\sigma^2$. This property of the $Y_n$'s
is nowhere mentioned in the paragraph fragment quoted by the OP.

Finally, turning to the OP's complaint

I cannot understand the link between the normal distribution and being serially independent

I say that it is a red herring. Uncorrelated (marginally)
normal random variables are not necessarily independent
random variables while uncorrelated jointly normal random
random variables are always independent random variables.

It is not true that if $\{Y_n\}$ is a sequence of normally distributed random variables that happen to be uncorrelated, then the random variables are independent.

A standard counterexample begins with $X \sim N(0,1)$ and an independent random
variable $Z$ that takes on values $+1$ and $-1$ with equal probability.
Then $Y = XZ$ is also a standard normal random variable since
\begin{align}P\{Y \leq x\} &= P\{X \leq x\mid Z=+1\}P\{Z=+1\}
+ P\{X \geq -x\mid Z=-1\}P\{Z=-1\}\\
&= \frac 12 \Phi(x) + \frac 12 (1 - \Phi(-x))\\
&= \Phi(x).
\end{align}
Also,
$\quad\operatorname{cov}(X,Y) = E[XY]-E[X]E[Y]= E[X^2Z]=E[X^2]E[Z]=0$
showing that $X$ and $Y$ are
are uncorrelated random variables. But $X$ and $Y$, although they are
uncorrelated normal
random variables, are not independent random variables but instead
very much dependent random variables since given that $X=x$, $Y$ takes on values $\pm x$ with equal probability $\frac 12$.
Now, with $\{Z_n\}$ being a sequence
of independent random variables with the same distribution as $Z$ (and all independent of $X$ also), set
$Y_n = XZ_n$. It follows from our construction that $Y_n \sim N(0,1)$.
But,
$$E[Y_nY_{m}] = E[X^2Z_nZ_m] = E[X^2]E[Z_n]E[Z_m] = 0~\text{provided that} ~ n \neq m.$$
Thus, the $\{Y_n\}$ are uncorrelated and normal.
But they are not independent because if we know that $Y_m = y$, then we know that
$Y_n$ is necessarily either $y$ or $-y$.
Normal random variables need not  be jointly normal
random variables and it is only in the case of
joint normality that one can assert that uncorrelated (jointly)
normal random variables are independent.
