Sufficient parameters for a probability distribution We know that a Gaussian distribution can be constructed if its first two moments i.e. its mean and covariance are known. Is there any other standard distribution whose construction requires the knowledge of first three or four moments?
I know that given a set of moments, one can construct a probability distribution using the principal of maximum entropy. However, I am looking for any standard distribution having the aforementioned property.
 A: The title (original and edited versions) of your question implies you are trying to use data to say from which member
of a particular distribution family the data were sampled. But the use of the word constructed in the question seem to imply you already know some parameter values and have a specific distribution family in mind and are wondering whether they are enough to identify the member of the family.
Finally, your question seems to imply that parameters are always population moments (or estimates are always sample moments?), and they need not be. 
I don't think there is any simple rule of the kind you seem to be seeking. So I am going to write a bit about
a few distributions (all univariate, for simplicity). Maybe these examples will answer your real question, and
maybe they will help you formulate it.
Normal: mean $\mu$ and standard deviation $\sigma$ are the parameters, needed to specify a particular member of the family.
Sample mean and sample standard deviation are often used as estimates of these parameters. Both are based on sufficient statistics.
(You can use variance $\sigma^2$ instead of SD.)
Exponential: either mean $\mu$ or $\lambda = 1/\mu$ are
enough to identify which member of the family. The sample mean
estimates $\mu$ and is a sufficient statistic. For this distribution it happens that $\mu = \sigma,$ but the sample SD is not based
on a sufficient statistic.
Beta: there are two shape parameters usually called $\alpha$ and $\beta.$ Both are required to identify a specific member of the
family. Neither is a moment. Sufficient statistics are not moments.
Binomial: parameters are $n$ and $\theta$ (sometimes notation $p$ is used). Usually $n$ is known. The success probability $\theta$ is estimated as $X/n,$ where $X$ is the number of observed successes
and also the sufficient statistic. The population mean is $\mu = n\theta,$ and the population variance is $\sigma^2 = n\theta(1-\theta);$ these are the moments.
Some distribution families require three or even four parameters to
specify the family member. Perhaps look at Wikipedia for Weibull and Pareto (including the 'shifted' and more-advanced expanded families near the ends of the articles.)
A: 
Is there any other standard distribution whose construction requires the knowledge of first three or four moments?

This isn't about "any standard distribution" but rather about any standard family of distributions.
You can single out a normal distribution with only the mean if the family of distributions happens to be all normal distributions with one particular number as the variance.
You cannot single out a normal distribution with only the mean and the variance if you're working with a broader class of distributions that includes normal distributions but also some other distributions; for some such families perhaps the mean, variance, and skewness would suffice.
For the uniform distirbution on the interval $[a,b]$ we have
\begin{align}
\text{mean} & = \frac{a+b} 2, \\[10pt]
\text{variance} & = \frac{(a+b)^2}{12}.
\end{align}
Given the mean and variance one can find $a$ and $b$ by a bit of algebra, so this would be such an example. I can image parametrizing this family of distributions by the mean and the range, those being a location parameter and a scale parameter.
The standard Gamma distribution is
$$
\frac 1 {\Gamma(\alpha)} \left( \frac x \beta\right)^{\alpha-1} e^{-x/\beta}\  \frac{dx}\beta\text{ on }(0,\infty). 
$$
The parameter $\beta$ is a scale parameter so the variance must be proportional to $\beta^2$ but it also depends on $\alpha$.  We have
\begin{align}
\text{mean} & = \beta\alpha, \\[10pt]
\text{variance} & = \beta^2\alpha.
\end{align}
Given the mean and the variance one can easily find $\alpha$ and $\beta$ and thus single out one distribution.  In particular
$$
\beta = \frac{\text{variance}}{\text{mean}} \text{ and }\alpha = \frac{\text{mean}^2}{\text{variance}}.
$$
(For the family of gamma distributions, all of the cumulants are directly proportional to $\alpha$.)
The family of Pareto distributions is characterized by
$$
\Pr(X>x) = \left(\frac{\kappa}{x}\right)^\alpha \text{ for }x>\kappa.
$$
The two parameters are positive numbers, the "Pareto index" $\alpha$ and the minimum $\kappa$. Here are some exercises:


*

*Show that $\kappa$ is a scale parameter, so that the mean and standard deviation must be proportional to $\kappa$.

*Find the mean and variance as functions of $\kappa$ and $\alpha$.

*For which values of $\alpha$ does the mean exist?

*For which values of $\alpha$ does the variance exist?

*How can one find $\kappa$ and $\alpha$ given the mean and the variance (when those exist)?

