10
$\begingroup$

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

$\endgroup$
12
  • 5
    $\begingroup$ 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
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 8
    $\begingroup$ @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
3
$\begingroup$

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 $t=0$.

$\endgroup$
2
  • 4
    $\begingroup$ 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. $\endgroup$ Jan 10 '17 at 2:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.