Statistics Standard Deviation

[Pekis]

The proof:
Knowing that:

\(\sigma^{2}x=\frac{M}{n^{2}}\)

\(s^{2}x=\frac{M}{n(n-1)}\)

where \(M=nb-a^{2}\)
\(a=\sum_{}^{}x\)
\(b=\sum_{}^{}x^{2}\)

If we add \(\bar{x}\) to the dataset, then:

\(n^{'}=n+1\)

\(a'=a+\bar{x}=a+\frac{a}{n}=a\frac{n+1}{n}\)

\(b'=b+\bar{x}^{2}=b+\frac{a^{2}}{n^{2}}\)

\(s^{2^{'}}x=\frac{M'}{n'(n'-1)}\)

As for M':

\(M'=n'b'-a'^{2}=(n+1)(b+\frac{a^{2}}{n^{2}})-a^{2}\frac{(n+1)^{2}}{n^{2}}\)

\(=\frac{n+1}{n^{2}}(bn^{2}+a^{2}-a^{2}(n+1))=\frac{n+1}{n^{2}}(bn^{2}-a^{2}n)=n\frac{n+1}{n^{2}}(bn-a^{2})=\frac{n+1}{n}M\)

Then

\(s^{2^{'}}x=\frac{\frac{n+1}{n}M}{n'(n'-1)}=\frac{\frac{n+1}{n}M}{(n+1)n}=\frac{M}{n^{2}}=\sigma^{2}x\) QED