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Relation Between Moments About Mean and Arbitrary Point

The \(r\)th moment about the mean is given by:

\[
u_r = \frac{1}{N} \sum_{i=1}^n f_i (x_i – \overline{x})^r, \quad r = 0, 1, \dots
\]

Using substitutions, we can express this as:

\[
u_r = \frac{1}{N} \sum_{i=1}^n f_i ((x_i – A) – (\overline{x} – A))^r
\]

If \(d_i = x_i – A\), then:

\[
\mu’_1 = \bar{x} – A
\]

By substituting \(d_i\) and \(\mu’_1\), we get:

\[
\mu_r = \frac{1}{N} \sum_{i=1}^n f_i (d_i – \mu’_1)^r
\]

Using the Binomial Theorem

Using the binomial expansion \((a-b)^n = a^n – C^n_1 a^{n-1} b + \dots\), the relationship becomes:

\[
\mu_r = \mu’_r – C^r_1 \mu’_1 \mu’_{r-1} + C^r_2 (\mu’_1)^2 \mu’_{r-2} – \dots + (-1)^r (\mu’_1)^r
\]

For specific values of \(r\):

• For \(r = 2\): \[
  \mu_2 = \mu’_2 – (\mu’_1)^2
  \]
• For \(r = 3\): \[
  \mu_3 = \mu’_3 – 3\mu’_1 \mu’_2 + 2 (\mu’_1)^3
  \]
• For \(r = 4\): \[
  \mu_4 = \mu’_4 – 4\mu’_1 \mu’_3 + 6 (\mu’_1)^2 \mu’_2 – 3 (\mu’_1)^4
  \]

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