Can Positive Values of Observations imply Population Positive Mean? Suppose that we have N-observations for a random variable $V$ from a population:
$$\{v_1,\;\ldots,\;v_N\}$$
, and all of the observations is positive: $v_i>0\quad \forall i=1,\;\ldots,\;N$.
Here, assume that the observations are "randomly sampled" (i.i.d.)
I think, from the positive values of observations, it is reasonable to guess that the population mean $\mathbb{E}[V]$ is also positive because the observations may represent the population distribution due to the random sampling assumption.
But, I am also skeptical about whether a statement for data can imply a statement for a population.
So, is it possible to derive $\mathbb{E}[V]>0$ from $v_i>0\quad \forall i=1,\;\ldots,N$?
In other words, can we mathematically prove the following implication?
$$v_i>0\quad \forall i=1,\;\ldots,N \quad \Longrightarrow \quad \mathbb{E}[V]>0$$
(Note that I am not asking an inference question. I want to know whether we can prove that under the random sampling assumption.)

First, you usually can't prove much from samples, since any sample is possible to happen, and a proof is for something that must always be true regardless of what happens.
Second, your intuition would closely apply to the median but not the mean.
Such a data set gives very strong evidence that the median is positive: If the median were negative, it would imply that each observation has at least a 50% chance of being negative, and the likelihood of having $n$ positive values would be smaller than $2^{-n}$. (However, you still could not mathematically prove that the median is positive; it's possible that it's negative but the extremely unlikely event did happen.)
On the other hand, there are distributions where the mean is negative, but getting all positive observations is overwhelmingly likely. A more extreme example than whuber's is $p(-2^n) = \frac{1}{2^n}$, $p(1) = 1 - \frac{1}{2^n}$.
