Does a random sample $X_1, \dots, X_n$ imply $n > 1$? I am calculating the variance for an estimator which is a function of a random sample $X_1, \dots, X_n$. However, if $n=1$, then $E(\theta^2) = \infty$ and the variance is infinite. Can I assume $n > 1$ due to the random sample definition $X_1, \dots, X_n$ or does this also include the random sample of {$X_1$}?
 A: 
Can I assume $n > 1$ due to the random sample definition $X_1, \dots, X_n$ or does this also include the random sample of {$X_1$}?

Technically a single observation is a sample as well.
The notation $X_1, \dots, X_n$, which implies at least two elements, is for convenience. It is used because often, in practice, a sample has more than a single observation (and the implied $n > 2$ or $n > 1$ does not matter).
The notation implies more than one observation (and it even also implies more than two observations because the dots make no sense, $X_1, \dots, X_2$, when $n=2$). But often this is not intentional. The notation is not used to restrict the size of the sample. The notation is used for convenience.


However, if $n=1$, then $E(\theta^2) = \infty$ and the variance is infinite.

In your particular case, a population or sample variance, it would make sense to use a sample of at least size 2. But that is a different story than the use of the notation ("due to the random sample definition ") that you mention.
A: It is totally reasonable to get an infinite variance of an estimator, such as the variance of $\bar X$ for a distribution like $t_2$.
Further, estimator variances definitely make sense for samples of just $1$, such as $\text{var}\left(\bar X\right)=\sigma^2$ for $X\sim N\left(\mu,\sigma^2\right)$.
Therefore, this does not seem like a problem. You just have infinite variance when you observe one point.
