Can we calculate mean of absolute value of a random variable analytically? Let's assume that we have a distribution with known statistical properties (mean, variance, skewness, kurtosis). Let's also assume that mean is equal to zero. Is there an analytical expressions for the average value of the absolute values of the considered random variable?
In other words can we say that:
avg(abs(x)) = F(var(x), skew(x), kurt(x))

 A: In general knowing these 4 properties is not enough to tell you the expectation of the absolute value of a random variable. As proof, here are two discrete distributions $X$ and $Y$ which have mean 0 and the same variance, skew, and kurtosis, but for which $\mathbb{E}(|X|) \ne \mathbb{E}(|Y|)$.
 t   P(X=t)  P(Y=t)
-3   0.100   0.099
-2   0.100   0.106
-1   0.100   0.085
 0   0.400   0.420
 1   0.100   0.085
 2   0.100   0.106
 3   0.100   0.099

You can verify that the 1st, 2nd, 3rd, and 4th central moments of these distributions are the same, and that the expectation of the absolute value is different.

Edit: explanation of how I found this example.
For ease of calculation I decided that:

*

*$X$ and $Y$ would both be symmetric about $0$, so that the mean and skew would automatically be $0$.

*$X$ and $Y$ would both be discrete taking values on $\{-n, .., +n\}$ for some $n$.

For a given distribution $X$, we want to find another distribution $Y$ satisfying the simultaneous equations $\mathbb{E}(Y^2) = \mathbb{E}(X^2)$ and $\mathbb{E}(Y^4) = \mathbb{E}(X^4)$. We find $n = 2$ isn't enough to provide multiple solutions, because subject to the above constraints we only have 2 degrees of freedom: once we pick $f(2)$ and $f(1)$, the rest of the distribution is fixed, and our two simultaneous equations in two variables have a unique solution, so $Y$ must have the same distribution as $X$. But $n = 3$ gives us 3 degrees of freedom, so should lead to infinite solutions.
Given $X$, our 3 degrees of freedom in picking $Y$ are:
$$f_Y(1) = f_X(1)+p \\
f_Y(2) = f_X(2)+q \\
f_Y(3) = f_X(3)+r$$
Then our simultaneous equations become:
$$
\begin{align}
p + 4q + 9r& = 0 \\
p + 16q + 81r& = 0
\end{align}
$$
The general solution is:
$$
p = 15r \\
q = -6r \\
$$
Finally I arbitrarily picked
$$
\begin{align}
f_X(1) & = 0.1 \\
f_X(2) & = 0.1 \\
f_X(3) & = 0.1 \\
r & = -0.001
\end{align}
$$
giving me the above counterexample.
A: It depends on what you mean by an "analytical" calculation. In general, this is just
$$ E(|X|) = \int |x| f(x)\,dx, $$
so you do have a formula. But I assume that "evaluating a (possibly improper) integral" is not what you had in mind.
Then again, probably the simplest non-trivial example would be that of the absolute value of a normal distribution, which is the folded normal distribution. And even here, the expression given by Wikipedia for the expectation involves evaluating $\Phi$, which is the CDF of the standard normal - and here again, you need to evaluate an improper integral.
So if you don't let integral evaluations count, the answer is no in general, even for simple cases like the folded normal.
A: $$
\DeclareMathOperator{\Var}{Var}
\DeclareMathOperator{\Skew}{Skew}
\DeclareMathOperator{\Kurt}{Kurt}
\newcommand{\E}{\mathbb{E}}
$$
Intuitive answer:
Shifting the distribution for the random variable $X$ to the left or the right changes the mean $\mu = \E[X]$ ("center") by the exact same amount. However, the variance $\sigma^2$ ("width"), skewness, and kurtosis ("tailedness") do not change because they are calculated based on the distances from the center $\mu$ of the distribution. Thus, they cannot possibly be used to express a function of $\mu$. For similar reasons, they cannot be used to express $\E[X]$, which for a positive-valued random variable $X$ behaves in exactly the same way under right-shifts.

Rigorous answer:
To simplify the problem, let's consider why you can't express $\E[X]$ as a function of $\Var[X]$, $\Skew[X]$, and $\Kurt[X]$.
By definition,
\begin{align}
  \mu      &= \E[X] \\[1em]
  \sigma^2 = \Var[X]  &= \E[(X - \mu)^2] \\[1em]
  \Skew[X] &= \E\left[\left(\frac{X - \mu}{\sigma}\right)^3\right] \\[1em]
  \Kurt[X] &= \E\left[\left(\frac{X - \mu}{\sigma}\right)^4\right].
\end{align}
Notice that if you add a constant shift $\gamma$ to $X$,
$$X' = X + \gamma,$$
then the $\mu'$ associated with $X'$ also shifts by the same amount:
$$\mu' = \E[X'] = \E[X + \gamma] = \E[X] + \E[\gamma] = \E[X] + \gamma = \mu + \gamma.$$
However, variance, skewness, and kurtosis don't change at all:
\begin{align}
  \Var[X']
    &= \E[(X + \gamma - \mu')^2]
    &&= \E[(X - \mu)^2]
    &&= \Var[X]
    \\[1em]
  \Skew[X']
    &= \E\left[\left(\frac{X + \gamma - \mu'}{\sigma'}\right)^3\right]
    &&= \E\left[\left(\frac{X - \mu}{\sigma}\right)^3\right]
    &&= \Skew[X]
    \\[1em]
  \Kurt[X']
    &= \E\left[\left(\frac{X + \gamma - \mu'}{\sigma'}\right)^4\right]
    &&= \E\left[\left(\frac{X - \mu}{\sigma}\right)^4\right]
    &&= \Kurt[X].
\end{align}
Thus, for any value of $\gamma$, these quantities are invariant. Though the value of $\E[X]$ may change, these values clearly do not! Any function of these variables is thus constant under $\gamma$-shift, and so it cannot possibly express $\E[X]$.
The same proof holds for $\E[|X|]$.
