Changing
We are going to employ the new method in four types of distribution.
Let us revisit the traditiona method with the help of an example.
Problem: Let us suppose first three moments about 2 are -1, 7 & 39. We have to find them about 5.
Traditional Solution:
Given
Now we have to find \(\mu_1'(5), \mu_3'(5), \text{ and } \mu_3'(5)\)
We have: \[ \mu_1'(2) = \frac{1}{n} \sum_{i=1}^n (x_i - 2) = -1 \]
To find the first moment about 5, we rewrite the expression as: \[ \begin{aligned} \mu_1'(5) &= \frac{1}{n} \sum_{i=1}^n (x_i - 5) \\ &= \frac{1}{n} \sum_{i=1}^n \left[(x_i - 2) - 3\right] \\ &= \frac{1}{n} \left( \sum_{i=1}^n (x_i - 2) - 3n \right) \\ &= \mu_1'(2) - 3 = -1 - 3 = -4 \end{aligned} \]
Thus, the first moment about 5 is \(\mu_1'(5) = -4\).
We are given: \[ \mu_2'(2) = \frac{1}{n} \sum_{i=1}^n (x_i - 2)^2 = 7 \]
To find the second moment about 5, we proceed as follows: \[\begin{equation} \begin{aligned} \mu_2'(5) &= \frac{1}{n} \sum_{i=1}^n (x_i - 5)^2 \\ &= \frac{1}{n} \sum_{i=1}^n \left[(x_i - 2) - 3\right]^2 \\ &= \frac{1}{n} \sum_{i=1}^n \left[(x_i - 2)^2 - 6(x_i - 2) + 9\right] \\ &= \frac{1}{n} \sum_{i=1}^n (x_i - 2)^2 - 6 \cdot \frac{1}{n} \sum_{i=1}^n (x_i - 2) + 9 \\ &= \mu_2'(2) - 6\mu_1'(2) + 9 \\ &= 7 - 6(-1) + 9 = 22 \end{aligned} \end{equation}\]
Thus, the second moment about 5 is \(\mu_2'(5) = 22\).
We are given: \[ \mu_3'(2) = \frac{1}{n} \sum_{i=1}^n (x_i - 2)^3 = 39 \]
To compute the third moment about 5, we use the binomial expansion: \[\begin{equation} \begin{aligned} \mu_3'(5) &= \frac{1}{n} \sum_{i=1}^n (x_i - 5)^3 \\ &= \frac{1}{n} \sum_{i=1}^n \left[(x_i - 2) - 3\right]^3 \\ &= \frac{1}{n} \sum_{i=1}^n \left[(x_i - 2)^3 - 9(x_i - 2)^2 + 27(x_i - 2) - 27\right] \\ &= \frac{1}{n} \sum_{i=1}^n (x_i - 2)^3 - 9 \cdot \frac{1}{n} \sum_{i=1}^n (x_i - 2)^2 + 27 \cdot \frac{1}{n} \sum_{i=1}^n (x_i - 2) - 27 \\ &= \mu_3'(2) - 9\mu_2'(2) + 27\mu_1'(2) - 27 \\ &= 39 - 9 \cdot 7 + 27 \cdot (-1) - 27 \\ &= -78 \end{aligned} \end{equation}\]
Therefore, the third moment about 5 is \(\mu_3'(5) = -78\).
Other higher order moments are converted in the same fashion, and the calculation keeps getting more difficult.
The following formulas are used:
\[ \begin{aligned} \mu_1 &= 0 \\[6pt] \mu_2 &= \mu_2'(a) - \left[\mu_1'(a)\right]^2 \\[6pt] \mu_3 &= \mu_3'(a) - 3\mu_2'(a)\mu_1'(a) + 2\left[\mu_1'(a)\right]^3 \\[6pt] \mu_4 &= \mu_4'(a) - 4\mu_3'(a)\mu_1'(a) + 6\mu_2'(a)\left[\mu_1'(a)\right]^2 - 3\left[\mu_1'(a)\right]^4 \\[6pt] \mu_5 &= \mu_5'(a) - 5\mu_4'(a)\mu_1'(a) + 10\mu_3'(a)\left[\mu_1'(a)\right]^2 - 10\mu_2'(a)\left[\mu_1'(a)\right]^3 + 4\left[\mu_1'(a)\right]^5 \end{aligned} \]
Here, \(\mu_r'(a)\) denotes the \(r^{\text{th}}\) raw moment about point \(a\) (an arbitrary number), and \(\mu_r\) denotes the \(r^{\text{th}}\) central moment.
The traditional methods for transforming moments—either from one arbitrary origin to another, or from raw to central moments—are often tedious and procedurally distinct. While central moments can, in principle, be derived by treating the arithmetic mean as a new origin, this approach is computationally intensive, while computing central moments using these formulas also poses a significant challenge, as it requires careful handling of various combinations of raw moments of differing orders.
Let,
Initial origin = \(a\)
Final origin = \(k\)
First moment around \(a = \mu_1'(a)\)
First moment around \(k = \mu_1'(k)\)
\[\begin{equation} \begin{split} \mu_1'(a) & =\frac{\sum(x_i-a)}{n} \\ \mu_1'(k) & = \frac{\sum(x_i-k)}{n}\\ & = \frac{\sum(x_i-a-k+a)}{n}\\ & = \frac{x_i-a}{n} + a - k\\ & = \mu_1'(a) + (a - k) \end{split} \end{equation}\]
Second moment around \(a = \mu_2'(a)\)
Second moment around \(k = \mu_2'(k)\)
\[\begin{equation} \begin{split} \mu_2'(a) & =\frac{\sum(x_i-a)^2}{n} \\ \mu_2'(k) & = \frac{\sum(x_i-k)^n}{n}\\ & = \frac{\{\sum(x_i-a-k+a)^2+2(a-k)(x_I-a)+(a-k)^2\}}{n}\\ & = \frac{(x_i-a)^2}{n} + 2(a - k)\frac{\sum(x_i-a)}{n}+(a-k)^2\\ & = \mu_2'(a) + 2(a - k)\mu_1'(a)+(a-k)^2 \end{split} \end{equation}\]
Other moments are dealt with in the same fashion.
To generalize the process, let
So we have:
The formulae look a lot like binomial expansions (which is natural due to the formula of the moments themselves). One modification will make them look (also serving our purpose) perfectly like binomial expansions.
We let \(\mu_r'(a) = a^r\)
Which simply means whatever the power of \(a\), we are dealing with that particular moment.
Thus letting the terms of binomial expansion \(a\) and \(b\), then for converting the first moment, we use the formula of \((a+b)^1 = a + b\). The table below shows the formulae required to convert the first four moments:
| Moment | Formula |
|---|---|
| First | \(a+b\) |
| Second | \(a^2+2ab+b^2\) |
| Third | \(a^3+3a^2b+3ab^2+b^3\) |
| Forth | \(a^4+4a^3b+6a^2b^2+4ab^3+b^4\) |
Observe the coefficients. They are essentially the binomial coefficients. The coefficients can be obtained from the Pascal’s triangle, which is a triangular array of the binomial coefficients.
\[ {\begin{array}{c} 1 \\ 1 \quad 1 \\ 1 \quad 2 \quad 1 \\ 1 \quad 3 \quad 3 \quad 1 \\ 1 \quad 4 \quad 6 \quad 4 \quad 1 \\ 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1 \\ 1 \quad 6 \quad 15 \quad 20 \quad 15 \quad 6 \quad 1 \\ 1 \quad 7 \quad 21 \quad 35 \quad 35 \quad 21 \quad 7 \quad 1 \\ \end{array}} \]
The first row is 1. Each entry of each subsequent row is constructed by adding the number above and to the left with the number above and to the right, treating blank entries as 0. For example, the initial number of row 1 (or any other row) is 1 (the sum of 0 and 1), whereas the numbers 1 and 3 in row 3 are added to produce the number 4 in row 4.
From the previous problem,
\(b = a-k = 2-5=-3\)
\(\mu_3'(5) = 39 + 3(-3)\times 7 + 3 (-3)^2(-1) + (-3)^3 = -78\)
We know, the central moments are nothing but moments around arithmetic mean (\(\bar X\))
The rth central moment is \(\mu_r = \frac{\sum (x_i-\bar x)^r}{n}\)
The central moments (\(\mu_r\)) can be derived from the raw moments. The conventional, formulae, however, look esoteric and difficult to remember.
But we can derive them quickly and easily by means of binomial theorem, and we can get the coefficient from Pascal’s triangle.
Let \(X \sim \text{Bin}(n, p)\). The moment generating function (MGF) is defined as:
\[ M_X(t) = \mathbb{E}[e^{tX}] = \sum_{x=0}^{n} e^{tx} \binom{n}{x} p^x (1-p)^{n-x} \]
Factor out \(e^{tx}\):
\[ M_X(t) = \sum_{x=0}^{n} \binom{n}{x} (p e^t)^x (1-p)^{n-x} \]
This is a binomial expansion of:
\[ M_X(t) = \left( (1 - p) + p e^t \right)^n \]
Differentiate the MGF with respect to \(t\):
\[ M'_X(t) = \frac{d}{dt} \left[ (1 - p + p e^t)^n \right] = n (1 - p + p e^t)^{n-1} \cdot p e^t \]
Evaluate at \(t = 0\) to get the first moment (mean):
\[ M'_X(0) = n (1 - p + p)^{n - 1} \cdot p = n p \]
Therefore, the first moment (mean) is:
\[ \boxed{\mathbb{E}[X] = n p} \]
\[ M_{X - \mu}(t) = e^{- \mu t} \cdot M_X(t) \]
Substituting \(\mu = np\):
\[ M_{X - \mu}(t) = e^{-np t} \cdot \left( (1 - p) + p e^t \right)^n \]
To find the first central moment, we differentiate this and evaluate at \(t = 0\):
\[ \mu_1 = \left. \frac{d}{dt} M_{X - \mu}(t) \right|_{t = 0} \]
Let:
\[ A(t) = e^{-np t}, \quad B(t) = \left( (1 - p) + p e^t \right)^n \]
Then:
\[ M_{X - \mu}(t) = A(t) \cdot B(t) \]
Differentiate using the product rule:
\[ \frac{d}{dt}(A(t)B(t)) = A'(t)B(t) + A(t)B'(t) \]
Now evaluate each part at \(t = 0\):
So:
\[ \left. \frac{d}{dt}(A(t)B(t)) \right|_{t = 0} = A'(0) B(0) + A(0) B'(0) = (-np)(1) + (1)(np) = 0 \]
\[ \boxed{\mu_1 = 0} \]
The first central moment of the Binomial distribution is zero, as expected.