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.)
 A: 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 $p(-2^n) = \frac{1}{2^n}$, $p(1) = 1 - \frac{1}{2^n}$.
A: No. A sample generally does not allow you to "prove" anything about the underlying population with 100% certainty, it just allows you to infer it with varying levels of confidence. Suppose you have a distribution that truly has a negative expected value, but has non-zero probability for positive values. No matter how many samples you pick, it's possible that every single observed value is positive, despite the fact that the expected value is actually negative. It becomes increasingly unlikely to that the expected value is negative as you observe more and more positive samples, but you can never be 100% sure.
Suppose you have a coin with a -2 on one face, and a +1 on the other. The true expected value of the coin flip is -0.5, and yet it's possible to conduct a series of coin flips and observe the +1 face every single time. Observing only +1's does not guarantee the expected value is positive, no matter how many you observe.
A: PROVE? NO
set.seed(21)
N <- 3 # Sample size
x <- rnorm(N, -0.1, 1)

I get three positive numbers, resulting in a positive sample mean, yet the population mean is $-0.1$.
EDIT
Being more formal with mathematics, we could calculate the distribution of the first order statistic (smallest value) of a random sample drawn from this $N(-0.1, 1)$, and we could do that for any sample size. Doing so will reveal that this gets less probable as the sample size increases, but its probability never drops to zero for any finite sample size.
(You might find the order statistic calculation to be more pleasant by using something like $U(-0.2, 0.1)$.)
A: You can't prove it, but you can calculate a confidence interval for the population mean given the sample. This requires an assumption on the population distribution.
More information is required for a better answer
