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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 $$E(X^n)=\,?$$

Scortchi - Reinstate Monica
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damla
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    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. – whuber Mar 26 '13 at 22:43
  • I do not have any restrictions. The independent and more general version of this problem is solved in this link: http://stats.stackexchange.com/questions/52646/variance-of-product-of-multiple-random-variables 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 $X_1=X_2=⋯=X_n=X$ – damla Mar 26 '13 at 23:01
  • As I wrote, the answer to the question you ask is no. The "independent and more general version" is a different question altogether, because it does not ask to compute higher moments in terms of lower ones. Please note that *all* the solutions and comments there assume independence of the variables; the case $X_1=\cdots=X_n$ does not satisfy that restriction. – whuber Mar 26 '13 at 23:16
  • Thank you whuber. So, we basically do not know what happens to variance if we take a square of the data? – damla Mar 27 '13 at 21:24
  • There are some inequalities that have to hold, but otherwise you are correct. This generalizes the fact that you cannot predict the second moment even when you know the expectation (the first moment), but you are guaranteed that the second moment exceeds the square of the first moment. – whuber Mar 27 '13 at 21:29
  • Oh, thanks so much. I can see it when I run numerical simulations on it. Is there a proof of what you said, the second moment exceeds the square of first moment? – damla Mar 27 '13 at 22:46
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    Yes: their difference is the variance and the variance, as a sum of squares, cannot be negative. – whuber Mar 27 '13 at 22:47
  • I have been misreading the question - the original question on products seems more interesting. – wolfies May 04 '13 at 18:12
  • @whuber In your first comment you seem to claim that distributions with arbitrary high moments are characterized by their moments, this is not true (considering integer powers only). – Stéphane Laurent May 04 '13 at 18:31
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    Stéphane Laurent I don't see where such a claim is made, but just to be clear, I didn't maintain any such thing. – whuber May 05 '13 at 17:27
  • @whuber Suppose we know it is a normally distributed random variable, is there then a general formula for Var(X^n)? – Bastiaan Mar 04 '16 at 06:51
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    @Bastiaan $\mathrm{Var}(X^n) = \mathbb{E}[X^{2n}] - \mathbb{E}[X^n]^2$; $\mathbb{E}[X^n]$ for normal distributions is available [here](https://en.wikipedia.org/wiki/Normal_distribution#Moments). – Danica Aug 29 '16 at 22:56

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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 $t=0$.

Nick Lim
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