# 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))


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 $$XX$$ and $$YY$$ which have mean 0 and the same variance, skew, and kurtosis, but for which $$E(|X|)≠E(|Y|)\mathbb{E}(|X|) \ne \mathbb{E}(|Y|)$$.

 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:

• $$XX$$ and $$YY$$ would both be symmetric about $$00$$, so that the mean and skew would automatically be $$00$$.
• $$XX$$ and $$YY$$ would both be discrete taking values on $${−n,..,+n}\{-n, .., +n\}$$ for some $$nn$$.

For a given distribution $$XX$$, we want to find another distribution $$YY$$ satisfying the simultaneous equations $$E(Y2)=E(X2)\mathbb{E}(Y^2) = \mathbb{E}(X^2)$$ and $$E(Y4)=E(X4)\mathbb{E}(Y^4) = \mathbb{E}(X^4)$$. We find $$n=2n = 2$$ isn’t enough to provide multiple solutions, because subject to the above constraints we only have 2 degrees of freedom: once we pick $$f(2)f(2)$$ and $$f(1)f(1)$$, the rest of the distribution is fixed, and our two simultaneous equations in two variables have a unique solution, so $$YY$$ must have the same distribution as $$XX$$. But $$n=3n = 3$$ gives us 3 degrees of freedom, so should lead to infinite solutions.

Given $$XX$$, our 3 degrees of freedom in picking $$YY$$ are:

$$fY(1)=fX(1)+pfY(2)=fX(2)+qfY(3)=fX(3)+rf_Y(1) = f_X(1)+p \\ f_Y(2) = f_X(2)+q \\ f_Y(3) = f_X(3)+r$$

Then our simultaneous equations become:

p+4q+9r=0p+16q+81r=0 \begin{align} p + 4q + 9r& = 0 \\ p + 16q + 81r& = 0 \end{align}

The general solution is:

$$p=15rq=−6r p = 15r \\ q = -6r \\$$

Finally I arbitrarily picked
fX(1)=0.1fX(2)=0.1fX(3)=0.1r=−0.001 \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.