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I have tried to separate the case into $\bar{X} \geq 0$ and $\bar{X} < 0$.

But I can't work out the pattern of $\mathrm{cov} (x,x^2)$

Any ideas?

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3  
$$\mathrm{cov}(X, X^{2}) = \mathbb{E}[(X-\mathbb{E}(X))(X^{2}-\mathbb{E}(X^{2}))] = \mathbb{E}(X^{3}) - \mathbb{E}(X)\mathbb{E}(X^{2}).$$ It seems that it cannot be simplified further without any additinal condition. –  sos440 Dec 14 '11 at 7:02
1  
@sos440: That looks like an answer –  Henry Dec 14 '11 at 7:58
    
As a comment to sos440's answer, you can rewrite this as $$\mathrm{cov}(X, X^{2}) = (\mu+ \sigma\gamma_1 )\sigma^2 $$ where $\mu$ is the mean of $X$, $\sigma$ the standard deviation and $\gamma_1$ the skewness –  Henry Dec 14 '11 at 8:13
    
As an additional comment on sos440's answer, $E[X^2] = \sigma^2 + \mu^2$ and so $\text{cov}(X,X^2)$ can also be expressed as $$\text{cov}(X,X^2) = E[X^3] - \mu(\sigma^2 + \mu^2) = (E[X^3] - (E[X])^3) - \sigma^2E[X]$$ where $\mu$ is the mean of $X$ and $\sigma$ the standard deviation. –  Dilip Sarwate Dec 14 '11 at 13:56
    
Thanks for the solution, I think that is the farthest point we can reach without any additional information. –  Rein Dec 15 '11 at 3:17

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