Is the mean of samples still a valid sample? Suppose I sample $n$ times from a distribution
$$
x_1, \ldots, x_n \sim p_\theta(x)
$$
is the mean of the samples always a valid sample from the target distribution? I.e. is $\overline{x}$ a valid sample from $p_\theta(x)$
$$
\overline{x} = \frac{1}{n}\sum_{i=1}^n x_i
$$
 A: No. Suppose you have $X_1, X_2 \sim N(0,1)$. Then,
$$
\bar{X} = \dfrac{X_1 + X_2}{2} \sim N\left(0, \dfrac{1}{2} \right)\,.
$$
But $N(0,1) \ne N(0, 1/2)$.
A: No, $\bar x$ has its own sampling distribution. Take, for example, the variances of $\bar x$ and $x_i$, in which the former is always lower ($\leq$) than the latter, which means $\bar x$ is not sampled from $p_\theta(x)$.
A: Good examples so far but consider $$X_i \sim Bernoulli(.5)$$
In that case the distribution of the data will only have support on 0 and 1.  But the sample mean will have an ever decreasing probability of taking a value of 0 or 1 as the sample size gets larger and larger.  That alone should show that the mean isn't being sampled from the original distribution.
A: No, it is only valid in cases as the Cauchy distribution, the means of samples of the Cauchy follow the same Cauchy dstribution.
A: As an even more pathological example, consider a sample from the distribution which is uniform on the union of $[0,1]$ and $[3,4]$. As the sample size increases, the mean will tend to 2 which isn't even in the support of the distribution. Another similar example is the uniform distribution on the boundary of the unit sphere (in any number of dimensions)
A: No. 
For the average to be a sample of the distribution, it must belong to the support of the distribution.
Below are two examples where that is not the case (which is sufficient to show that the statement is not true in general).
Discrete
The distribution p(x=1) = 0.5; p(x=-1) = 0.5 has support $$S=\{-1,1\}$$ but average $0\notin S$.
Continuous
The density function
$$p(x) = \frac{1}{2}rect(x-1) + \frac{1}{2}rect(x+1)$$
(two Rectangular functions centered at 1 and -1 respectively) has support 
$$S = ]-1.5,-0.5[\cap]0.5,1.5[$$
but average $0\notin S$.
