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Given the mean & variance of a dataset of linear measurements X ... N~(14,2), I want to find the mean of X^3.

So by taking mean as being expected value, & rearranging: -

$$Var(X)=E[X^2]−(E[X])^2$$

I get

$$ E[X^2] = Var(X) + (E[X])^2$$

$$E[X^2] = SD^2 + (E[X])^2$$

How can I go from this to using an exponent of 3 to solve the original problem?

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1 Answer 1

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Easy way is to use moment generating functions. For the normal this is

$$M(t) = e^{t\mu + \sigma^2t^2/2}$$

Then we can get the moments by using

$$\text{E}[X^n] = \left.\dfrac{d^n M(t)}{dt^n}\right|_{t=0}$$

the $n$th moment is the $n$th derivative of the moment generating function evaluated at 0. For your particular example we get

$$\begin{align*}M(t) &= e^{t(14) + (2)t^2/2}\\ \dfrac{d^3 M(t)}{dt^3} &= (8t^3 + 168t^2 + 1188t +2828)e^{t^2+14t}\\ \left.\dfrac{d^3 M(t)}{dt^3}\right|_{t=0} &= 2828 \end{align*}$$

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  • $\begingroup$ Many thanks bdeonovic - Is this approach still possible if the exponent in not an integer, e.g. 2.4? $\endgroup$
    – Michael P
    Jan 22, 2016 at 2:11
  • $\begingroup$ No... What would the 2.4th derivative be????? $\endgroup$
    – bdeonovic
    Jan 22, 2016 at 2:14
  • $\begingroup$ Non negative integers only $\endgroup$
    – bdeonovic
    Jan 22, 2016 at 2:14
  • $\begingroup$ Yes - that's what I was thinking. In addition to the original question involving X^3, I need to calculate for X^2.4. Is there any approach that may be used for this? $\endgroup$
    – Michael P
    Jan 22, 2016 at 2:18
  • $\begingroup$ Just use definition of expectation. $E(g(X)) = \int_{-infty}^{infty} g(x)f(x)dx$ where $f(x)$ is the normal pdf $\endgroup$
    – bdeonovic
    Jan 22, 2016 at 2:21

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