Would the joint distribution of Normal Random Variable and the distribution of a X bar from the same sample be bivariate Normal? I know this question is somewhat redundant... but here goes:
My text asserts that the joint distribution of $$X_1=N(\theta, 1)\text{ and } \bar X = N(\theta, \frac 1n)$$ 
is Bivariate normal with $$\text{mean vector} = (\theta, \theta), \, \sigma_1^2 = 1,\ \sigma_2^2 = \frac 1n,\text{ and } \rho = 1/\sqrt n$$
Given the numerous reasons to not assume a Bivariate normal joint distribution  as a result of two marginal normals, many summed up nicely here: https://stats.stackexchange.com/questions/30159/is-it-possible-to-have-a-pair-of-gaussian-random-variables-for-which-the-joint-d/30205#30205,
why is it okay (to make assumptions) in this case? Also, how could the covariance be computed in this situation? The text offers a problem to show this result later in problems, thus prompting this question.
Thanks
 A: Nobody is making assumptions in this matter; rather, someone is drawing conclusions. And those conclusions are not based simply on the marginal distributions.  Conventionally, one has
$$
X_1,\ldots,X_n \overset{\mathrm{i.i.d.}}\sim N(\theta,1),
$$
$$
\bar X = \frac{X_1+\cdots+X_n} n \sim N\left(\theta,\frac 1 n \right)
$$
This gives use the mean vector and the two variances.  Then we have
\begin{align}
& \operatorname{cov}\left( X_1, \bar X \right) = \operatorname{cov}\left( X_1, \frac{X_1+\cdots+X_n} n \right) \\[6pt]
= {} & \frac 1 n \Big( \operatorname{cov}(X_1,X_1)+\operatorname{cov}(X_1,X_2) + \operatorname{cov}(X_1,X_3)+\cdots +\operatorname{cov}(X_1,X_n) \Big) \\[6pt]
= {} & \frac 1 n \Big( 1 + 0 + 0 + 0 + \cdots + 0 \Big) = \frac 1 n.
\end{align}
Finally, we have the question of why this has a bivariate normal distribution. The definition of the statement that $(U,V)$ has a bivariate normal distribution is that for every pair of constants $a$, $b$ (and "constant" means not random) the random variable $aU+bV$ has a univariate normal distribution.  So we need the distribution of $aX_1+b\bar X$.  Observe that
$$
aX_1+b\bar X = aX_1+b\frac{X_1+X_2+\cdots+X_n} n = \underbrace{\left(a+\frac b n\right)X_1 + b\frac{X_2}n+\cdots+b\frac{X_n}n}.
$$
Notice that the part over the underbrace is a sum of independent random variables each of which is normally distributed.  Consequently it is itself normally distributed.  This conclusion is not based merely on the marginal distributions because it relies on independence.
