Moments of distribution

[luccacerf]

Hi, im a little bit cofunsed about the how the first moment (k=1) equals to the mean.

i.e. k = 1 .: E[(y-m)^1] isnt E(y) - E(y) = 0 <> m ?
if you do k =2, goes perfectly to the variance formula.

k=2 .: E[(y-m)^2] = E(y^2) - [E(y)]^2

Could you guys help me please?

Thanks.

 

[ShaktiRathore]

Hi,
mean is first moment around 0(zero),thus mean=E[(y-0)^1] =E[y]
variance is the second moment around the mean= E[(y-m)^2] = E(y^2) - [E(y)]^2
thanks

 

[luccacerf]

Thanks!

 

[David Harper CFA FRM]

@luccacerf I think Miller is a bit unclear on this but the central moment (aka, moment about the mean, as in centered on the mean) is given by k-th central moment = E[(X - µ)^k], such that the first central moment (k = 1) is given by E[(X - µ)^1] = E[X-µ]. Now an unbiased estimator, by definition, is when E(X) = µ, so if the estimator is unbiased then E[X-µ] = 0. In this way, I hadn't actually thought it is this way, but while the variance is the second central moment, the mean is actually the first raw moment (because the first central moment is zero by definition; see https://en.wikipedia.org/wiki/Moment_(mathematics). I guess while the central moment is about the mean, the raw moment is "about the zero" so to speak

Because as k-th central moment = E[(X - µ)^k], the k-th raw moment = E[(X - 0)^k] = E[X^k], so the 1st raw moment = E[X] = mean. Thanks!