Relationship between mean and variance of samples I am thinking about the relationship between sample mean and variance in an example. If we want to look at the average goals per month for a soccer team. And we have mean and variance of goals for each month. Now we find average goals per match is higher if goals per match in a month variates a lot. Is the high correlation between mean and variance defined by math, or it deserved to explore. 
It may indicate that some part of the team is impacting the number of goals. For example the strategy, the team may have a rotation on players. So substitutes have no goals in first 2 matches, and starting lineup have extremely high goals after well rest. In this case, mean would be greater than having starting 11 players on every match and getting really tired.
Further, if we add samples variance to the nonlinear model to predict mean, dose the variance provides a unique contribution to the model, or it is more of a self-learning?
 A: As goals scored in soccer matches are typically rare events, you might want to consider your data as coming from Poisson distributions. If so, then the variance in terms of goals per match will be equal to the average number of goals per match. Any changes in strategies, player health, opponent strength, and so on that affect the average number of goals per match from month to month would then necessarily also affect the variance.
If your variances are higher than corresponding means then a negative binomial distribution might be a better fit. But even then the variance will tend to increase with the mean values.
So your observation of a high correlation between mean and variance in soccer scores has a solid mathematical basis.
A: For data distributed normally, the mean and variance are independent. The PDF is parameterized by both values, which you can tweak to your heart's content.
For other distributions, the mean and variance are related. Let's consider the exponential distribution, which has PDF $f(x\vert\lambda) = \lambda e^{-\lambda x}$.
$$\mu_{f} = \dfrac{1}{\lambda}$$
$$\sigma^2_f = \dfrac{1}{\lambda^2}$$
There are mean-variance combinations that simply are not possible, even if the mean is possible and the variance is possible. For instance, we can have a mean of 1 when $\lambda=1$, but then $\lambda = 1$ and the variance cannot be $1/4$. Likewise, we can have a variance of $1/4$, but then $\lambda = 2$ and the mean cannot be one.
As Glen_b mentioned, something with counts may be more helpful for you. The Poisson distribution has an interesting property where the mean and variance are equal. There are mean-variance combinations that are impossible for Poisson-distributed data, and they're easy to predict (anything that isn't $(\lambda,\lambda), \lambda>0)$.
So the mean and variance can be independent (normal) but don't have to be (exponential and Poisson).
