Is it possible to derive a formula for variance of powers of a random variable in terms of expected value and variance of X? $$\operatorname{var}(X^n)= \,?$$ and $$E(X^n)=\,?$$

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    $\begingroup$ Do you have a particular distribution in mind? To obtain a solution, you need some such restriction. If any general formula existed, then there would be remarkably few distributions: that formula would determine all higher moments and so all distributions could be parameterized by the expectation and variance, which clearly is not the case. $\endgroup$ – whuber Mar 26 '13 at 22:43
  • $\begingroup$ I do not have any restrictions. The independent and more general version of this problem is solved in this link: stats.stackexchange.com/questions/52646/… So, I am wondering if we can derive a general equation for this one too. In other words, I am trying to find ${\rm var}(X_1X_2 \cdots X_n)= \ ?$ where $X_1=X_2=⋯=X_n=X$ $\endgroup$ – damla Mar 26 '13 at 23:01
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    $\begingroup$ Yes: their difference is the variance and the variance, as a sum of squares, cannot be negative. $\endgroup$ – whuber Mar 27 '13 at 22:47
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    $\begingroup$ Stéphane Laurent I don't see where such a claim is made, but just to be clear, I didn't maintain any such thing. $\endgroup$ – whuber May 5 '13 at 17:27
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    $\begingroup$ @Bastiaan $\mathrm{Var}(X^n) = \mathbb{E}[X^{2n}] - \mathbb{E}[X^n]^2$; $\mathbb{E}[X^n]$ for normal distributions is available here. $\endgroup$ – Danica Aug 29 '16 at 22:56

if you have the Moment generating function for the distribution X, you can calculate the expected value of $X^n$ using $\frac{d^n}{dt^n} MGF(x)$ and evaluating it at $t=0$.

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    $\begingroup$ Can you go on to explain how this may help find $\operatorname{Var}(X^n)$? $\endgroup$ – Silverfish Aug 29 '16 at 22:59
  • $\begingroup$ @Silverfish Dougal's comment on the main question, posted just three minutes before your comment above, answers your query completely. $\endgroup$ – Dilip Sarwate Jan 10 '17 at 2:42

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