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Given a R.V $X$ and that $\ E(X) = 0 $ and $\ E(X^2) = \sigma^2 $. Is there anyway to compute $\ E(X^3) $ without knowing the density function of $X$?

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A similar question came up recently. See this comment and the related question. –  cardinal Nov 9 '12 at 12:43
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Look into Pearson distribution. This family is reach enough so that every 4-tuple of moments $m_1$, $m_2$, $m_3$ and $m_4$ corresponds to a unique (up to equivalence) member of the family.

Here is the parametrization on the case at hand in terms of skewness $\mathcal{s}$ and kurtosis $\kappa$:

$$ \operatorname{Pearson}\left(\underbrace{2 \left(5 \kappa -6 \mathcal{s}^2-9\right)}_{a_1}, \underbrace{(\kappa +3) \mathcal{s} \sigma^2}_{a_0}, \underbrace{2 \kappa -3 \left(\mathcal{s}^2+2\right)}_{b_2},\underbrace{(\kappa +3) \mathcal{s} \sigma^2}_{b_1}, \underbrace{\sigma^4 \left(4 \kappa -3 \mathcal{s}^2\right)}_{b_0} \right) $$ The pdf of this distribution $f(x)$ satisfies differential equation: $$ f^\prime(x) + \frac{a_1 x+a_0}{b_2 x^2+b_1 x+b_0} f(x) = 0 $$

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I am afraid if you don't have additional information about $X$, you can't compute $\mathbb E(X^3)$.

As a reminder

$\mathbb E(X^3) = \int x^3dF_X = \int x^3 f_X(x)dx$

so if you don't know $f_X$ you are in trouble, I think.

More intuitively, the third moment gives information about the skewness of the distribution. It is easy to imagine two different distributions with same mean (can be zero) and same variance but differently skewed

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