Talk:Method of moments (statistics)

Some strange things happened here. There is a portion in the last part saying:

"However, in some cases, as in the above example of the gamma distribution, the likelihood equations may be intractable without computers..."

There is no such example above and there seems to have never been. I wonder, was this article taken from a source? Is it incomplete?

Ferred — Preceding unsigned comment added by 193.254.231.85 (talk) 08:26, 3 July 2014 (UTC)

choice of using mu is confusing

mu is reserved for means m_1 is better for moments. 68.134.243.51 (talk) 13:07, 16 September 2022 (UTC)

Chebyshev proof

The argument in the Proving the central limit theorem section does not hold as stated. Consider random variables $X_{i}$ such that $\mathrm {P} [X_{i}=-{\tfrac {1}{2}}]={\tfrac {4}{5}}$ and $\mathrm {P} [X_{i}=2]={\tfrac {1}{5}}$. Then these indeed have mean $0$ and variance $1$, but their skewness is not zero, so neither will their third moments be. Indeed, $${\begin{aligned}\mathrm {E} [S_{1}^{3}]={}&\mathrm {E} [X_{1}^{3}]=(-{\tfrac {1}{2}})^{3}{\tfrac {4}{5}}+2^{3}{\tfrac {1}{5}}=-{\tfrac {1}{10}}+{\tfrac {8}{5}}={\tfrac {3}{2}}\neq 0\\\mathrm {E} [S_{2}^{3}]={}&2^{-3/2}\mathrm {E} [(X_{1}+X_{2})^{3}]=2^{-3/2}{\big (}-1\cdot {\tfrac {16}{25}}+({\tfrac {3}{2}})^{3}{\tfrac {8}{25}}+4^{3}{\tfrac {1}{25}}{\big )}={\frac {3}{\sqrt {8}}}\neq 0\end{aligned}}$$ The way that the argument is structured, it looks as though the claim was rather that a sum of normally distributed random variables with these parameters will be normally distributed, but that's a whole lot weaker than the Central limit theorem! 130.243.94.123 (talk) 13:34, 31 October 2024 (UTC)

The skewness becomes 0 at the limit of the sum thanks to the central limit theorem. The statements is inaccurate as is and needs to be modified. The editor who posted it takes the limit at the end. Limit-theorem (talk) 12:46, 23 December 2024 (UTC)