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Recall the 'law of the unconscious statistician' (wikipedia), namely that one can calculate the expectation of the transform $g(X)$ of a random variable $X$, given the transformation and the probability density function $f$ of the original random variable. If one is being fancy, then this is a statement about pushforward measures.

But I would like to know if one has a similarly simple result about calculating the higher central moments. If there is no closed form, then I should say I would be happy with the first four as a compromise.

To be concrete, I have a smooth function $g:\mathbb{R}^2 \to [0,1]$, and $X$ is bivariate normal (so no worries about moments existing).

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Take a look at this link books.google.co.in/… The authors use Taylor's series expansion to estimate expected value of $g(x)$, and say that the closed form of $g(x)$ instead of Taylor series expansion can be used to estimate higher order moments. – NikBels Dec 19 '11 at 6:45
    
@NikhilBellarykar - google won't let me see that page. – David Roberts Dec 19 '11 at 9:34
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Can't you just calculate $\int (g(t) - \mu)^n f_X (t) dt $ where $\mu = \int g(t) f_X (t) dt $? or is this not regarded as "simple"? – yaakov Dec 19 '11 at 10:03
    
@DavidRoberts I have no clue why. It opens fine on my system. Anyways, the expressions are no different there. The authors used Taylor series expansion for calculating only $E(g(x))$ but said that the closed form $g(x)$ can be used for higher order moments-thats it. I don't think that's much of help, but that's what I could find out with a quick google search. – NikBels Dec 19 '11 at 10:23
    
As a slight emendation to yaakov's comment, since $Z=g(X,Y)$ is a function of the bivariate normal random variable $(X,Y)$, $$E[Z] = E[g(X,Y)] = \int\int g(x,y) f(x,y) dx dy = \mu$$ and $Z$ has $n$-th central moment $$E[(Z-\mu)^n] = \int\int (g(x,y) - \mu)^n f(x,y) dx dy$$ where $f(x,y)$ is the joint density of $X$ and $Y$, a bivariate normal in this case. Is it that the integrals are hard to compute analytically? If so, can you evaluate them numerically? – Dilip Sarwate Dec 19 '11 at 18:25

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