4.17 Moments

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But

$\int_{| x - μ | \geq k} f (x) d x = P (| x - μ | \geq k)$ $\int_{| x - μ | \geq k} f (x) d x = P (| x - μ | \geq k)$

Hence we get

$σ^{2} \geq k^{2} (| x - μ | \geq k)$ $σ^{2} \geq k^{2} (| x - μ | \geq k)$

$P (| x - μ | \geq k) \leq \frac{σ^{2}}{k^{2}}$ $P (| x - μ | \geq k) \leq \frac{σ^{2}}{k^{2}}$ (4.7)

(4.7)

Also we have

$P (| x - μ | \geq k) + P (| x - μ | \geq k) = 1$ $P (| x - μ | \geq k) + P (| x - μ | \geq k) = 1$

Therefore we get

$P (| x - μ | \leq k) = 1 = P (| x - μ | \geq k)$ $P (| x - μ | \leq k) = 1 = P (| x - μ | \geq k)$

From Eq. (4.7), we get

$P (| x - μ | \leq k) \geq 1 - \frac{σ^{2}}{k^{2}}$ $P (| x - μ | \leq k) \geq 1 - \frac{σ^{2}}{k^{2}}$

Hence proved.

Remark

1. $P (| x - μ | \geq k) \geq 1 - \frac{σ^{2}}{k^{2}}$ $P (| x - μ | \geq k) \geq 1 - \frac{σ^{2}}{k^{2}}$ can be written as $P (μ - k < x < μ + k) \geq 1 - \frac{σ^{2}}{k^{2}} a$ $P (μ - k < x < μ + k) \geq 1 - \frac{σ^{2}}{k^{2}} a$ .

2. $\frac{σ^{2}}{k^{2}}$ $\frac{σ^{2}}{k^{2}}$ is called an upper bound of $P (| x - μ | \geq k)$ $P (| x - μ | \geq k)$ .

Example 4.30: If the Chebyshev’s inequality for the random variable X is given by $P (- 2 < X < 8) \geq \frac{21}{25}$ $P (- 2 < X < 8) \geq \frac{21}{25}$ . Find E(X) and V[X]?

Solution: Comparing $P (- 2 < X < 8) \geq \frac{21}{25}$ $P (- 2 < X < 8) \geq \frac{21}{25}$

with

$P (μ - k < x < μ + k) \geq 1 - \frac{σ^{2}}{k^{2}}$ $P (μ - k < x < μ + k) \geq 1 - \frac{σ^{2}}{k^{2}}$

We get

$μ - k = - 2$ $μ - k = - 2$ (4.8)

(4.8)

$μ + k = 8$ $μ + k = 8$ (4.9)

(4.9)

and

$1 - \frac{σ^{2}}{k^{2}} = \frac{21}{25}$ $1 - \frac{σ^{2}}{k^{2}} = \frac{21}{25}$ (4.10)

(4.10)

Adding Eqs. (4.8) and (4.9) we get

$2 μ = 6 or μ = 3$ $2 μ = 6 or μ = 3$

Now μ+k=8 gives 3+k=8 or k=5

Substituting in Eq. (4.10) we get $1 - \frac{σ^{2}}{k^{2}} = \frac{21}{25}$ $1 - \frac{σ^{2}}{k^{2}} = \frac{21}{25}$

i.e., ∴ $μ = {μ'}_{1} + 4$ $μ = {μ'}_{1} + 4$

$1 - \frac{21}{25} = \frac{σ^{2}}{5^{2}}$ $1 - \frac{21}{25} = \frac{σ^{2}}{5^{2}}$

$\frac{4}{25} = \frac{σ^{2}}{5^{2}}$ $\frac{4}{25} = \frac{σ^{2}}{5^{2}}$

$σ^{2} = 4$ $σ^{2} = 4$

Hence

$E (X) = 3, V [X] = 4$ $E (X) = 3, V [X] = 4$

Example 4.31: The number of components manufactured in a factory during one month period is a random variable with mean 600 and variance 100. What is the probability that the production will be between 500 and 700 over a month?

Solution: We have

$\begin{array}{l} m = 600, σ^{2} = 100 \\ P (500 < X < 700) = P (600 - 100 < X < 600 + 100) \end{array}$ $\begin{array}{l} m = 600, σ^{2} = 100 \\ P (500 < X < 700) = P (600 - 100 < X < 600 + 100) \end{array}$

Comparing with

$P (μ - k < x < μ + k), we get k = 100$ $P (μ - k < x < μ + k), we get k = 100$

Using Chebyshev’s inequality, we get

$P (| x - 600 | < k) \geq 1 - \frac{σ^{2}}{k^{2}} = 1 - \frac{100}{100^{2}} = 1 - \frac{1}{100}$ $P (| x - 600 | < k) \geq 1 - \frac{σ^{2}}{k^{2}} = 1 - \frac{100}{100^{2}} = 1 - \frac{1}{100}$

$P (| x - 600 | < 100) \geq 1 - 0.01 = 0.99$ $P (| x - 600 | < 100) \geq 1 - 0.01 = 0.99$

i.e.,

$P (| x - 600 | < 100) \geq 0.99$ $P (| x - 600 | < 100) \geq 0.99$

Hence the probability of production in one month between 500 and 700 is 0.99.

Definition 4.21

Let X and Y be two discrete random variables and f(x, y) be the value of the joint probability distribution of X and Y at (x, y). If (x, y) is a real-valued function of (X, Y) then the expectation of g(x, y) is given by

$E (g (X, Y)) = \sum_{x} \sum_{y} g (x, y) F (x, y) d x d y .$ $E (g (X, Y)) = \sum_{x} \sum_{y} g (x, y) F (x, y) d x d y .$

If X and Y are continuous random variables and f(x, y) is the value of their joint probability density of X and Y at (x, y).

Then

$E (g (X, Y)) = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} g (x, y) f (x, y) d x d y$ $E (g (X, Y)) = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} g (x, y) f (x, y) d x d y$

For example consider the joint density of X and Y is given by

$\begin{array}{l} F (x, y) & = \frac{2}{3} (x + 2 y), & 0 < x < 1, 1 < y < 2 \\ = 0, & elsewhere \end{array}$ $\begin{array}{l} F (x, y) & = \frac{2}{3} (x + 2 y), & 0 < x < 1, 1 < y < 2 \\ = 0, & elsewhere \end{array}$

The expected value of $g (x, y) = \frac{x}{y^{3}}$ $g (x, y) = \frac{x}{y^{3}}$ is

$\begin{array}{l} E [\frac{x}{y^{3}}] & = \frac{2}{5} \int_{1}^{2} \int_{0}^{1} \frac{x}{y^{3}} (x + 2 y) d x d y \\ = \frac{2}{5} \int_{1}^{2} [\int_{0}^{1} (\frac{x^{2}}{y^{3}} + \frac{2 x y}{y^{2}}) d x] d y \\ = \frac{2}{5} \int_{1}^{2} [\frac{x^{3}}{3 y^{3}} + \frac{x^{2}}{y^{2}}] d y \\ = \frac{2}{5} \int_{1}^{2} [\frac{1}{3 y^{3}} + \frac{1}{y^{2}}] d y \\ = \frac{2}{5} {[\frac{1}{3} (\frac{y^{- 2}}{- 2}) + \frac{y^{- 1}}{(- 1)}]}_{2}^{1} \\ = \frac{2}{5} {[\frac{- 1}{6 y^{2}} - \frac{1}{y}]}_{1}^{2} \\ = \frac{1}{4} \end{array}$ $\begin{array}{l} E [\frac{x}{y^{3}}] & = \frac{2}{5} \int_{1}^{2} \int_{0}^{1} \frac{x}{y^{3}} (x + 2 y) d x d y \\ = \frac{2}{5} \int_{1}^{2} [\int_{0}^{1} (\frac{x^{2}}{y^{3}} + \frac{2 x y}{y^{2}}) d x] d y \\ = \frac{2}{5} \int_{1}^{2} [\frac{x^{3}}{3 y^{3}} + \frac{x^{2}}{y^{2}}] d y \\ = \frac{2}{5} \int_{1}^{2} [\frac{1}{3 y^{3}} + \frac{1}{y^{2}}] d y \\ = \frac{2}{5} {[\frac{1}{3} (\frac{y^{- 2}}{- 2}) + \frac{y^{- 1}}{(- 1)}]}_{2}^{1} \\ = \frac{2}{5} {[\frac{- 1}{6 y^{2}} - \frac{1}{y}]}_{1}^{2} \\ = \frac{1}{4} \end{array}$

4.16.4 Covariance

Definition 4.22

Let (X, Y) be a two-dimensional (bivariate) random variable. Then covariance between X and Y denoted by Cov (X, Y) is defined as follows:

$Cov (X, Y) = E [(X - E (X)) (Y - E (Y))]$ $Cov (X, Y) = E [(X - E (X)) (Y - E (Y))]$

where E(X) is the expectation of X and E(Y) is the expectation of Y.

From the definition of covariance, we have

$\begin{array}{l} Cov (X, Y) & = E [(X - E (X)) (Y - E (Y))] \\ = E [X Y - X E (Y) - Y E (X) + E (X) E (Y)] \\ = E (X Y) - E (X) E (Y) - E (Y) E (X) + E (X) E (Y) \\ = E (X Y) - E (X) E (Y) \end{array}$ $\begin{array}{l} Cov (X, Y) & = E [(X - E (X)) (Y - E (Y))] \\ = E [X Y - X E (Y) - Y E (X) + E (X) E (Y)] \\ = E (X Y) - E (X) E (Y) - E (Y) E (X) + E (X) E (Y) \\ = E (X Y) - E (X) E (Y) \end{array}$

Remarks

1. If X and Y are independent, then

$\begin{array}{l} Cov (X, Y) & = E (X Y) - E (X) E (Y) \\ = E (X) E (Y) - E (X) E (Y) = 0 \end{array}$ $\begin{array}{l} Cov (X, Y) & = E (X Y) - E (X) E (Y) \\ = E (X) E (Y) - E (X) E (Y) = 0 \end{array}$

But the Cov (X, Y)=0 does not imply that X and Y are independent.

2. If a, b are real constants then

$Cov (a X, b Y) = a b Cov (X, Y)$ $Cov (a X, b Y) = a b Cov (X, Y)$

and

$Cov (X + a, Y + b) = Cov (X, Y)$ $Cov (X + a, Y + b) = Cov (X, Y)$

Example 4.32: For the following Joint probability distribution. Find Cov (X, Y)?

XY	1	2	3	p_i
0	$\frac{2}{12}$ $\frac{2}{12}$	$\frac{1}{12}$ $\frac{1}{12}$	$\frac{1}{12}$ $\frac{1}{12}$	$\frac{4}{12}$ $\frac{4}{12}$
1	$\frac{1}{12}$ $\frac{1}{12}$	$\frac{1}{12}$ $\frac{1}{12}$	$\frac{2}{12}$ $\frac{2}{12}$	$\frac{4}{12}$ $\frac{4}{12}$
2	$\frac{3}{12}$ $\frac{3}{12}$	$\frac{1}{12}$ $\frac{1}{12}$	0	$\frac{4}{12}$ $\frac{4}{12}$
p_ij	$\frac{6}{12}$ $\frac{6}{12}$	$\frac{3}{12}$ $\frac{3}{12}$	$\frac{3}{12}$ $\frac{3}{12}$	1

Solution: From the given table, we get

$\begin{array}{l} E (X) & = 0 \cdot \frac{4}{12} + 1 \cdot \frac{4}{12} + 2 \cdot \frac{4}{12} = \frac{12}{12} = 1 \\ E (Y) & = 1 \cdot \frac{6}{12} + 2 \cdot \frac{3}{12} + 3 \cdot \frac{3}{12} \\ = \frac{6}{12} + \frac{6}{12} + \frac{9}{12} = \frac{21}{12} = \frac{7}{4} \end{array}$ $\begin{array}{l} E (X) & = 0 \cdot \frac{4}{12} + 1 \cdot \frac{4}{12} + 2 \cdot \frac{4}{12} = \frac{12}{12} = 1 \\ E (Y) & = 1 \cdot \frac{6}{12} + 2 \cdot \frac{3}{12} + 3 \cdot \frac{3}{12} \\ = \frac{6}{12} + \frac{6}{12} + \frac{9}{12} = \frac{21}{12} = \frac{7}{4} \end{array}$

$\begin{array}{l} E (X Y) & = 0.1 \cdot \frac{1}{12} + 0.2 \frac{1}{12} + 0.3 \cdot \frac{1}{12} + 1.1 \frac{1}{12} \\ + 1.2 \cdot \frac{1}{12} + 1.3 \frac{2}{12} + 2.1 \cdot \frac{3}{12} + 2.2 \cdot \frac{1}{12} + 2.3.0 \\ = 0 + 0 + 0 + \frac{1}{12} + \frac{2}{12} + \frac{6}{12} + \frac{6}{12} + \frac{4}{12} + 0 = \frac{19}{12} \end{array}$ $\begin{array}{l} E (X Y) & = 0.1 \cdot \frac{1}{12} + 0.2 \frac{1}{12} + 0.3 \cdot \frac{1}{12} + 1.1 \frac{1}{12} \\ + 1.2 \cdot \frac{1}{12} + 1.3 \frac{2}{12} + 2.1 \cdot \frac{3}{12} + 2.2 \cdot \frac{1}{12} + 2.3.0 \\ = 0 + 0 + 0 + \frac{1}{12} + \frac{2}{12} + \frac{6}{12} + \frac{6}{12} + \frac{4}{12} + 0 = \frac{19}{12} \end{array}$

Hence

$\begin{array}{l} Cov (X, Y) & = E (X Y) - E (X) E (Y) \\ = \frac{19}{12} - 1 \cdot \frac{7}{4} \\ = \frac{19}{12} - \frac{7}{4} = \frac{- 2}{12} = \frac{- 1}{6} \end{array}$ $\begin{array}{l} Cov (X, Y) & = E (X Y) - E (X) E (Y) \\ = \frac{19}{12} - 1 \cdot \frac{7}{4} \\ = \frac{19}{12} - \frac{7}{4} = \frac{- 2}{12} = \frac{- 1}{6} \end{array}$

Exercise 4.4

1. Let X be random variable with the following probability distribution. Find E(k)?

X	3	6	9
P(X=x)	$\frac{1}{6}$ $\frac{1}{6}$	$\frac{1}{2}$ $\frac{1}{2}$	$\frac{1}{3}$ $\frac{1}{3}$

Ans: $\frac{13}{2}$ $\frac{13}{2}$

2. If X is a random variable and the density function of X is

$\begin{array}{l} f (x) & = e^{- x}, & x \geq 0 \\ = 0, & 0 \end{array}$ $\begin{array}{l} f (x) & = e^{- x}, & x \geq 0 \\ = 0, & 0 \end{array}$

Find

a. E(X)

b. E(X²)

c. E(X−1)²

d. V(X)

Ans: (a) 1; (b) 2; (c) 1; (d) 1

3. Define expectation of random variable.

4. A random variable X is given by

X	−2	3	1
P(X=x)	$\frac{1}{3}$ $\frac{1}{3}$	$\frac{1}{2}$ $\frac{1}{2}$	$\frac{1}{6}$ $\frac{1}{6}$

Find

a. E(X)

b. V[X]

Ans: (a) 1; (b) 5

5. Find the mean and variance for the distribution:

X	−1	0	1	otherwise
P(X=x)	$\frac{1}{8}$ $\frac{1}{8}$	$\frac{3}{8}$ $\frac{3}{8}$	$\frac{1}{2}$ $\frac{1}{2}$	0

Ans: Mean= $\frac{3}{8}$ $\frac{3}{8}$ ; Variance= $\frac{5}{8}$ $\frac{5}{8}$

6. The distribution of the number of raisins in a cookie is given in the following table. Find the mean and variance of raisins in a cookie?

No. of raisins (x)	0	1	2	3	4	5
Probability P(x)	0.05	0.1	0.2	0.4	0.15	0.1

Ans: Mean=2.8; Variance=1.56

7. If X is a random variable defined by the density function

$\begin{array}{l} f (x) & = 3 x^{2}, & 0 \leq x \leq 1 \\ = 0, & otherwise \end{array}$ $\begin{array}{l} f (x) & = 3 x^{2}, & 0 \leq x \leq 1 \\ = 0, & otherwise \end{array}$

Find

a. E(X)

b. E(3X−2)

c. E(X²)

d. Var[X]

Ans: (a) 0.75; (b) 1; (c) 0.6; (d) 0.0375

8. If X and Y are random variables each having the density functions

$\begin{array}{l} f (x) & = e^{- x}, & x \geq 0 \\ = 0, & Otherwise \end{array}$ $\begin{array}{l} f (x) & = e^{- x}, & x \geq 0 \\ = 0, & Otherwise \end{array}$

Find

a. E(X+Y)

b. E(X²+Y²)

c. E(XY)

d. V[X] and V[Y]

Ans: (a) $\frac{1}{2}$ $\frac{1}{2}$ ; (b) 1; (c) $\frac{1}{4}$ $\frac{1}{4}$ ; (d) $\frac{1}{4}, \frac{1}{4}$ $\frac{1}{4}, \frac{1}{4}$

9. Two unbiased dice are thrown. Find the expected value of the sum of the numbers obtained on the top faces of them?

Ans: 7

10. Find the expectation of the number of failures preceding the first successes in an infinite series of independent trials with constant probability p of successes in each trial?

Ans: $\frac{q}{p}$ $\frac{q}{p}$

11. The density function of a continuous random variable X is given by

$\begin{array}{l} f (x) & = \frac{1}{2} x, & 0 < x < 2 \\ = 0, & Otherwise \end{array}$ $\begin{array}{l} f (x) & = \frac{1}{2} x, & 0 < x < 2 \\ = 0, & Otherwise \end{array}$

Find E(X), Var[X], and E(3x²−2x)?

Ans: $\frac{4}{3}, \frac{2}{9}, \frac{10}{3}$ $\frac{4}{3}, \frac{2}{9}, \frac{10}{3}$

12. For the density function

$\begin{array}{l} f (x) & = \frac{1}{4} & for x = - 1 \\ = \frac{1}{4} & for x = 0 \\ = \frac{1}{2} & for x = 2 \end{array}$ $\begin{array}{l} f (x) & = \frac{1}{4} & for x = - 1 \\ = \frac{1}{4} & for x = 0 \\ = \frac{1}{2} & for x = 2 \end{array}$

Find E(X) and V[X]?

Ans: $\frac{3}{4}, \frac{27}{16}$ $\frac{3}{4}, \frac{27}{16}$

13. If a>0 and n>0, find E(X) and V[X] when

$f (x) = [\frac{a^{n + 1} \cdot x^{n}}{n!}] e^{- a x}$ $f (x) = [\frac{a^{n + 1} \cdot x^{n}}{n!}] e^{- a x}$

Ans: $\frac{n + 1}{a}, \frac{n + 1}{a^{2}}$ $\frac{n + 1}{a}, \frac{n + 1}{a^{2}}$

14. From the following table, find (a) k; (b) E(2X+3); (c) V(x)?

X	−3	−2	−1	0	1	2	3
P(X=x)	0.05	0.10	2k	0	0.30	k	0.10

Ans: (a) 0.15; (b) 3.5; (c) 2.887

15. The function $f (x) = \frac{1}{π} \cdot \frac{1}{1 + x^{2}} - \infty < x < \infty$ $f (x) = \frac{1}{π} \cdot \frac{1}{1 + x^{2}} - \infty < x < \infty$
Defines the pdf of X. Find E(X)?

Ans: E(X) does not exist.

16. If X is a random variable $Y = \frac{(x - μ)}{σ}$ $Y = \frac{(x - μ)}{σ}$ , is also a random variable, show that

a. E(Y)=0

b. V(Y)=1

17. For a random variable with $f (x) = \frac{1}{2} e^{- | x |}$ $f (x) = \frac{1}{2} e^{- | x |}$ , find its SD?

Ans: $σ = \sqrt{2}$ $σ = \sqrt{2}$

18. A random variable X with unknown probability distribution has an average of 14 and variance 9. Find the probability that X will be between 7 and 21?

Ans: $\frac{40}{49}$ $\frac{40}{49}$

19. A random variable X has the density function e^−x for x≥0. Show that Chebyshev’s inequality gives $P (| x - 1 | > 2) < \frac{1}{4}$ $P (| x - 1 | > 2) < \frac{1}{4}$ .

Ans: $\frac{5}{9}$ $\frac{5}{9}$

4.17 Moments

Moments about mean and an arbitrary number were discussed in Chapter 1, An Overview of Statistical Applications. In this section we introduce moments of a random variable, which serve to describe the shape of the distribution of a random variable.

Moments About Arithmetic Mean (Central Moments)

Definition 4.23

If X is a discrete random variable, the rth moment of X about mean μ, denoted by μ_r is defined by

$μ_{r} = E {(x - μ)}^{r} = \sum_{i = 1}^{n} {(x_{i} - μ)}^{r} p_{i} r = 0, 1, 2, \dots$ $μ_{r} = E {(x - μ)}^{r} = \sum_{i = 1}^{n} {(x_{i} - μ)}^{r} p_{i} r = 0, 1, 2, \dots$

If X is a continuous random variable, the rth moment about mean is defined by

$μ_{r} = E {(x - μ)}^{r} = \int_{- \infty}^{\infty} {(x - μ)}^{r} f (x) d x$ $μ_{r} = E {(x - μ)}^{r} = \int_{- \infty}^{\infty} {(x - μ)}^{r} f (x) d x$

Remarks

1. Since μ=E(X), we can write

$μ_{r} = E {[X - E [x]]}^{r}$ $μ_{r} = E {[X - E [x]]}^{r}$

2. If x₁, x₂, …, x_n are in dual series, i.e., values of a random variable, with mean μ then the rth moment about μ is

$μ_{r} = \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - μ)}^{r} r = 0, 1, 2, \dots$ $μ_{r} = \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - μ)}^{r} r = 0, 1, 2, \dots$

From the definition of rth moment we have,

$μ_{r} = \sum_{i = 1}^{n} {(x_{i} - μ)}^{r} p_{i}$ $μ_{r} = \sum_{i = 1}^{n} {(x_{i} - μ)}^{r} p_{i}$

Therefore we get

$\begin{array}{l} μ_{0} & = \sum_{i = 1}^{n} {(x_{i} - μ)}^{0} p_{i} \\ = \sum p_{i} = 1 \\ μ_{1} & = \sum_{i = 1}^{n} (x_{i} - μ) p_{i} \\ = \sum_{i = 1}^{n} x_{i} p_{i} - \sum_{i = 1}^{n} μ p_{i} \\ = E [X] - μ \sum p_{i} \\ = E [X] - μ \cdot 1 \\ = E [X] - E [X] = 0 (since μ = E (X)) \end{array}$ $\begin{array}{l} μ_{0} & = \sum_{i = 1}^{n} {(x_{i} - μ)}^{0} p_{i} \\ = \sum p_{i} = 1 \\ μ_{1} & = \sum_{i = 1}^{n} (x_{i} - μ) p_{i} \\ = \sum_{i = 1}^{n} x_{i} p_{i} - \sum_{i = 1}^{n} μ p_{i} \\ = E [X] - μ \sum p_{i} \\ = E [X] - μ \cdot 1 \\ = E [X] - E [X] = 0 (since μ = E (X)) \end{array}$

Therefore μ₀=1, μ₁=0 for any variable X.

$\begin{array}{l} μ_{2} & = \sum_{i = 1}^{n} {(x_{i} - μ)}^{2} p_{i} \\ = \sum_{i = 1}^{n} {(x_{i}^{2} - 2 μ x_{i} + μ^{2})}^{2} p_{i} \\ = \sum_{i = 1}^{n} x_{i}^{2} p_{i} - 2 μ \sum_{i = 1}^{n} x_{i} p_{1} + μ^{2} \sum_{i = 1}^{n} p_{i} \\ = E (X^{2}) - μ^{2} \end{array}$ $\begin{array}{l} μ_{2} & = \sum_{i = 1}^{n} {(x_{i} - μ)}^{2} p_{i} \\ = \sum_{i = 1}^{n} {(x_{i}^{2} - 2 μ x_{i} + μ^{2})}^{2} p_{i} \\ = \sum_{i = 1}^{n} x_{i}^{2} p_{i} - 2 μ \sum_{i = 1}^{n} x_{i} p_{1} + μ^{2} \sum_{i = 1}^{n} p_{i} \\ = E (X^{2}) - μ^{2} \end{array}$

$μ_{2} = E (X^{2}) - {[E (x)]}^{2}$ $μ_{2} = E (X^{2}) - {[E (x)]}^{2}$

μ² is called the variance of the distribution of X. It is denoted by σ².

The positive square root of $σ^{2}$ $σ^{2}$ , i.e., variance is called the SD.

The moments about mean are also called the central moments or simply the moments of the distribution.

4.17.1 Moments About an Arbitrary Number

Definition 4.24

Let X be a discrete random variable and A be an arbitrary number. The rth moment about A of X denoted by ${μ'}_{r}$ ${μ'}_{r}$ , is defined as

${μ'}_{r} = E [{(x - A)}^{r}] = \sum_{i = 1}^{n} {(x_{i} - μ)}^{r} r = 0, 1, 2, \dots$ ${μ'}_{r} = E [{(x - A)}^{r}] = \sum_{i = 1}^{n} {(x_{i} - μ)}^{r} r = 0, 1, 2, \dots$

If X is a continuous random variable the ${μ'}_{3}$ ${μ'}_{3}$ is defined as

${μ'}_{r} = \int_{- \infty}^{\infty} {(x - A)}^{r} f (x) d x, r = 0, 1, 2, \dots$ ${μ'}_{r} = \int_{- \infty}^{\infty} {(x - A)}^{r} f (x) d x, r = 0, 1, 2, \dots$

The moments about an arbitrary number are also called raw moments.

Relation Between $μ_{r}$ $μ_{r}$ and ${μ'}_{r}$ ${μ'}_{r}$

If μ_r denotes the rth moment of a distribution on about the mean and ${μ'}_{r}$ ${μ'}_{r}$ denotes the rth moment about an arbitrary number, then

$μ_{r} = {μ'}_{r} - {}^{r}C_{1} {μ'}_{r + 1} {μ'}_{1} + {}^{r}C_{2} {μ'}_{r - 2} {μ'}_{2} + \dots + {(- 1)}^{r} {({μ'}_{1})}^{r}$ $μ_{r} = {μ'}_{r} - {}^{r}C_{1} {μ'}_{r + 1} {μ'}_{1} + {}^{r}C_{2} {μ'}_{r - 2} {μ'}_{2} + \dots + {(- 1)}^{r} {({μ'}_{1})}^{r}$

Substituting r=2, 3, 4, … we get

$\begin{array}{l} μ_{2} & = {μ'}_{2} - {}^{2}C_{1} {μ'}_{1} {μ'}_{1} + {}^{2}C_{2} {({μ'}_{1})}^{2} = {μ'}_{2} - {({μ'}_{1})}^{2} \\ μ_{3} & = {μ'}_{3} - {}^{3}C_{1} {μ'}_{2} {μ'}_{1} + {}^{3}C_{2} {μ'}_{1} {({μ'}_{1})}^{2} = {}^{3}C_{3} {({μ'}_{1})}^{3} \\ = {μ'}_{3} - 3 {μ'}_{2} {μ'}_{1} + 2 {({μ'}_{1})}^{2} \\ μ_{4} & = {μ'}_{4} {}^{4}C_{1} {μ'}_{3} {μ'}_{1} + {}^{4}C_{2} {μ'}_{2} {({μ'}_{1})}^{2} - {}^{4}C_{3} {μ'}_{1} {({μ'}_{1})}^{3} + {}^{4}C_{4} {({μ'}_{1})}^{4} \\ = {μ'}_{4} - 4 {μ'}_{3} {μ'}_{1} + 6 {μ'}_{2} {({μ'}_{1})}^{2} - 3 {({μ'}_{1})}^{4} \end{array}$ $\begin{array}{l} μ_{2} & = {μ'}_{2} - {}^{2}C_{1} {μ'}_{1} {μ'}_{1} + {}^{2}C_{2} {({μ'}_{1})}^{2} = {μ'}_{2} - {({μ'}_{1})}^{2} \\ μ_{3} & = {μ'}_{3} - {}^{3}C_{1} {μ'}_{2} {μ'}_{1} + {}^{3}C_{2} {μ'}_{1} {({μ'}_{1})}^{2} = {}^{3}C_{3} {({μ'}_{1})}^{3} \\ = {μ'}_{3} - 3 {μ'}_{2} {μ'}_{1} + 2 {({μ'}_{1})}^{2} \\ μ_{4} & = {μ'}_{4} {}^{4}C_{1} {μ'}_{3} {μ'}_{1} + {}^{4}C_{2} {μ'}_{2} {({μ'}_{1})}^{2} - {}^{4}C_{3} {μ'}_{1} {({μ'}_{1})}^{3} + {}^{4}C_{4} {({μ'}_{1})}^{4} \\ = {μ'}_{4} - 4 {μ'}_{3} {μ'}_{1} + 6 {μ'}_{2} {({μ'}_{1})}^{2} - 3 {({μ'}_{1})}^{4} \end{array}$

Remarks

1. $\begin{array}{l} {μ'}_{0} = \sum_{i = 1}^{n} {(x_{i} - μ)}^{0} p_{i} \\ \sum p_{i} = 1 \end{array}$ $\begin{array}{l} {μ'}_{0} = \sum_{i = 1}^{n} {(x_{i} - μ)}^{0} p_{i} \\ \sum p_{i} = 1 \end{array}$

2. $\begin{array}{l} {μ'}_{1} & = \sum_{i = 1}^{n} (x_{i} - A) p_{i} \\ = \sum_{i = 1}^{n} x_{i} p_{i} - A \sum_{i = 1}^{n} p_{i} \\ = E (X) - A \\ ∴ μ & = {μ'}_{1} + A \end{array}$ $\begin{array}{l} {μ'}_{1} & = \sum_{i = 1}^{n} (x_{i} - A) p_{i} \\ = \sum_{i = 1}^{n} x_{i} p_{i} - A \sum_{i = 1}^{n} p_{i} \\ = E (X) - A \\ ∴ μ & = {μ'}_{1} + A \end{array}$

Example 4.33: The first four moments of a distribution about x=2 are 1, 2.5, 5.5, and 26, respectively. Find the values of first four central moments?

Solution: We have $μ' = 1$ $μ' = 1$ , ${μ'}_{2} = 2.5$ ${μ'}_{2} = 2.5$ , ${μ'}_{3} = 5.5$ ${μ'}_{3} = 5.5$ , ${μ'}_{4} = 26$ ${μ'}_{4} = 26$ , and A=2.

Thus

$\begin{array}{l} μ_{1} & = 0, μ_{2} = {μ'}_{2} - {({μ'}_{1})}^{2} = (2.5) - 1^{2} = 1.5 \\ μ_{3} & = {μ'}_{3} - 3 {μ'}_{2} {μ'}_{1} + 2 {({μ'}_{1})}^{2} \\ = 5.5 - 3 (2.5) (1) + 2 {(1)}^{3} \\ = 5.5 - 7.5 + 2 = 0 \end{array}$ $\begin{array}{l} μ_{1} & = 0, μ_{2} = {μ'}_{2} - {({μ'}_{1})}^{2} = (2.5) - 1^{2} = 1.5 \\ μ_{3} & = {μ'}_{3} - 3 {μ'}_{2} {μ'}_{1} + 2 {({μ'}_{1})}^{2} \\ = 5.5 - 3 (2.5) (1) + 2 {(1)}^{3} \\ = 5.5 - 7.5 + 2 = 0 \end{array}$

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Table of Contents for 4.17 Moments

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4.16.4 Covariance

4.17 Moments

4.17.1 Moments About an Arbitrary Number

Table of Contents for
4.17 Moments