4.18 Moment Generating Function

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$\begin{array}{l} μ_{4} & = {μ'}_{4} - 4 {μ'}_{3} {μ'}_{1} + 6 {μ'}_{2} {({μ'}_{1})}^{2} - 3 {({μ'}_{1})}^{4} \\ = 26 - 4 (5.5) (1) + 6 (2.5) (1) + 3 {(1)}^{4} \\ = 26 - 22 + 15 - 3 = 16 \\ μ_{1} & = 0, μ_{2} = 1.5, μ_{3} = 0, μ_{4} = 16 \end{array}$ $\begin{array}{l} μ_{4} & = {μ'}_{4} - 4 {μ'}_{3} {μ'}_{1} + 6 {μ'}_{2} {({μ'}_{1})}^{2} - 3 {({μ'}_{1})}^{4} \\ = 26 - 4 (5.5) (1) + 6 (2.5) (1) + 3 {(1)}^{4} \\ = 26 - 22 + 15 - 3 = 16 \\ μ_{1} & = 0, μ_{2} = 1.5, μ_{3} = 0, μ_{4} = 16 \end{array}$

4.17.2 Moments About Origin

Definition 4.25

The rth moment about origin of a random variable X, denoted by V_r is the expected value of X^r, symbolically

$V_{r} = E [X^{r}] = \sum_{i = 1}^{n} x_{i}^{r} p_{i}$ $V_{r} = E [X^{r}] = \sum_{i = 1}^{n} x_{i}^{r} p_{i}$

For r=0, 1, 2, … when X is discrete, and

$V_{r} = E [X^{r}] = \int_{- \infty}^{\infty} x^{r} f (x) d x$ $V_{r} = E [X^{r}] = \int_{- \infty}^{\infty} x^{r} f (x) d x$

when X is continuous.

Relation Between μ_r and V_r:

If μ is the Arithmetic mean of X, the distribution then

$V_{r} = μ_{r} + {}^{r}C_{1} μ_{r - 1} μ + {}^{r}C_{2} μ_{r - 2} μ^{2} + \dots + μ^{r}$ $V_{r} = μ_{r} + {}^{r}C_{1} μ_{r - 1} μ + {}^{r}C_{2} μ_{r - 2} μ^{2} + \dots + μ^{r}$

Putting r=1, 2, 3,…. we get

$V_{1} = μ, V_{2} = μ_{2} + μ^{2}, V_{3} = μ_{3} + 3 μ_{2} μ + μ^{3}$ $V_{1} = μ, V_{2} = μ_{2} + μ^{2}, V_{3} = μ_{3} + 3 μ_{2} μ + μ^{3}$

Remark

From the definition of moment about arbitrary number A, we have

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

Taking A=0, we obtain

${μ'}_{r} = E [X^{r}] = \sum_{i = 1}^{n} {(x_{i})}^{r} p_{i}$ ${μ'}_{r} = E [X^{r}] = \sum_{i = 1}^{n} {(x_{i})}^{r} p_{i}$

i.e.,

${μ'}_{r} = V_{r}$ ${μ'}_{r} = V_{r}$

4.17.3 Skewness and Kurtosis

A measure of skewness is defined by

$S k_{m} = \frac{E [{(x - μ)}^{3}]}{σ^{3}} = \frac{μ_{3}}{\sqrt{μ_{2}^{3}}}$ $S k_{m} = \frac{E [{(x - μ)}^{3}]}{σ^{3}} = \frac{μ_{3}}{\sqrt{μ_{2}^{3}}}$

If we define

$β_{1} = \frac{μ_{3}^{2}}{μ_{2}^{3}}$ $β_{1} = \frac{μ_{3}^{2}}{μ_{2}^{3}}$

The moment coefficient of skewness= $\frac{μ_{3}}{\sqrt{μ_{2}^{3}}} = \pm β_{1}$ $\frac{μ_{3}}{\sqrt{μ_{2}^{3}}} = \pm β_{1}$

where the sign of $\sqrt{β_{1}}$ $\sqrt{β_{1}}$ is taken as that of $μ_{3}$ $μ_{3}$ .

The moment of coefficient of skewness is also denoted by r₁. Thus

$r_{1} - \frac{μ_{3}}{\sqrt{μ_{3}^{3}}} = \pm \sqrt{β_{1}}$ $r_{1} - \frac{μ_{3}}{\sqrt{μ_{3}^{3}}} = \pm \sqrt{β_{1}}$

If measure of skewness is positive, we say that the distribution is positively skewed or right tailed. And if the measure of skewness is negative, we say that the distribution is negatively skewed or left tailed.

If Arithmetic mean of the distribution is greater than the mode, the distribution is positively skewed (i.e., right tailed), and if the Arithmetic mean is less than the value of the mode, the distribution is negatively skewed (i.e., left tailed). In a symmetrical distribution the quartiles are equidistant from the median.

Measure of kurtosis tells us the extent to which a distribution is more peaked or more flat topped than the normal curve. If the curve of a frequency distribution is more peaked than the normal curve, it is said to be “Leptokurtic.” The curve of a frequency distribution is called “Platykurtic” if it is more flat topped than the normal curve. If the curve of a frequency is neither flat nor sharply peaked, then the curve of the distribution is called “Mesokurtic.” The measure of kurtosis denoted by β₂ and is defined as

$β_{2} = \frac{μ_{4}}{μ_{2}^{2}}$ $β_{2} = \frac{μ_{4}}{μ_{2}^{2}}$

where μ₂ and μ₄ are respectively the second and fourth central moments about mean. Kurtosis is also known as Convexity of a curve (or bulginess).

If β₂<3, the distribution is said to be “Platykurtic”

If β₂=3, the distribution is said to be “Mesokurtic”

If β₂>3, the distribution is said to be “Leptokurtic.”

The kurtosis of distribution is also measured by the considering the value of β₂−3, which is denoted by r₂.

Thus

$r_{2} = β_{2} - 3$ $r_{2} = β_{2} - 3$

If r₂=0, the distribution is “Mesokurtic”

If r₂<0, the distribution is “Platykurtic” and

If r₂>0, the distribution is “Leptokurtic”.

Example 4.34: In a certain distribution the first four moments about x=5 are 2, 20, 40, and 50. Calculate and state whether the distribution is Leptokurtic or Platykurtic?

Solution: We have A=5

${μ'}_{1} = 2, {μ'}_{2} = 20, {μ'}_{3} = 40, {μ'}_{4} = 50$ ${μ'}_{1} = 2, {μ'}_{2} = 20, {μ'}_{3} = 40, {μ'}_{4} = 50$

Hence we get μ₁=0 (always),

$\begin{array}{l} μ_{2} & = {μ'}_{2} - {({μ'}_{1})}^{2} = (20) - 2^{2} - 16 \\ μ_{3} & = {μ'}_{3} - 3 {μ'}_{2} {μ'}_{1} + 2 {({μ'}_{1})}^{2} \\ = 40 - 3 (2) (20) + 2 {(2)}^{3} \\ = 40 - 120 + 16 = - 64 \\ μ_{4} & = {μ'}_{4} - 4 {μ'}_{3} {μ'}_{1} + 6 {μ'}_{2} {({μ'}_{1})}^{2} - 3 {({μ'}_{1})}^{4} \\ = 50 - 4 (40) (2) + 6 (20) (2^{2}) + 3 {(2)}^{4} \\ = 50 - 320 + 480 - 48 \\ = 162 \\ β_{1} & = \frac{μ_{3}^{2}}{μ_{2}^{3}} = \frac{{(- 64)}^{2}}{16^{3}} = 1 \\ β_{2} & = \frac{μ_{4}}{μ_{2}^{2}} = \frac{162}{16^{2}} = 0.63 \end{array}$ $\begin{array}{l} μ_{2} & = {μ'}_{2} - {({μ'}_{1})}^{2} = (20) - 2^{2} - 16 \\ μ_{3} & = {μ'}_{3} - 3 {μ'}_{2} {μ'}_{1} + 2 {({μ'}_{1})}^{2} \\ = 40 - 3 (2) (20) + 2 {(2)}^{3} \\ = 40 - 120 + 16 = - 64 \\ μ_{4} & = {μ'}_{4} - 4 {μ'}_{3} {μ'}_{1} + 6 {μ'}_{2} {({μ'}_{1})}^{2} - 3 {({μ'}_{1})}^{4} \\ = 50 - 4 (40) (2) + 6 (20) (2^{2}) + 3 {(2)}^{4} \\ = 50 - 320 + 480 - 48 \\ = 162 \\ β_{1} & = \frac{μ_{3}^{2}}{μ_{2}^{3}} = \frac{{(- 64)}^{2}}{16^{3}} = 1 \\ β_{2} & = \frac{μ_{4}}{μ_{2}^{2}} = \frac{162}{16^{2}} = 0.63 \end{array}$

Since β₂<3, the curve of the distribution is Platykurtic.

Example 4.35: The density of a random variable X is given by f(x)=kx(2−x), 0≤x ≤2. Find

a. k

b. Mean

c. Variance

d. β₁ and β₂

e. rth moment.

Solution:

a. Since $\int_{- \infty}^{\infty} f (x) d x = 1$ $\int_{- \infty}^{\infty} f (x) d x = 1$ , i.e., $\int_{0}^{2} f (x) d x = 1$ $\int_{0}^{2} f (x) d x = 1$
We have $\int_{0}^{2} k x (2 - x) d x = 1$ $\int_{0}^{2} k x (2 - x) d x = 1$
or $k {[x^{2} - \frac{x^{3}}{3}]}_{0}^{2} = 1$ $k {[x^{2} - \frac{x^{3}}{3}]}_{0}^{2} = 1$
or $k [4 - \frac{8}{3}] = 1$ $k [4 - \frac{8}{3}] = 1$
or $\frac{4 k}{3} = 1$ $\frac{4 k}{3} = 1$
∴ $k = \frac{3}{4}$ $k = \frac{3}{4}$

b. $\begin{array}{l} Mean & = E (X) = \int_{- \infty}^{\infty} x f (x) d x \\ = \frac{3}{4} \int_{0}^{2} (2 x^{2} - x^{3}) d x \\ = \frac{3}{4} {[\frac{2}{3} x^{3} - \frac{x^{4}}{4}]}_{0}^{2} \\ = \frac{3}{4} [\frac{16}{3} - \frac{16}{4}] \\ = \frac{3}{4} \cdot 16 [\frac{1}{3} - \frac{1}{4}] = 3.4 \frac{(4 - 3)}{3.4} \\ = 1 \end{array}$ $\begin{array}{l} Mean & = E (X) = \int_{- \infty}^{\infty} x f (x) d x \\ = \frac{3}{4} \int_{0}^{2} (2 x^{2} - x^{3}) d x \\ = \frac{3}{4} {[\frac{2}{3} x^{3} - \frac{x^{4}}{4}]}_{0}^{2} \\ = \frac{3}{4} [\frac{16}{3} - \frac{16}{4}] \\ = \frac{3}{4} \cdot 16 [\frac{1}{3} - \frac{1}{4}] = 3.4 \frac{(4 - 3)}{3.4} \\ = 1 \end{array}$

c. We have

$\begin{array}{l} {μ'}_{2} = \frac{3.8}{4.5} = \frac{6}{5} \\ {μ'}_{3} = \frac{{3.2}^{4}}{5.6} = \frac{8}{5} \\ {μ'}_{4} = \frac{{3.2}^{5}}{6.7} = \frac{16}{7} \end{array}$ $\begin{array}{l} {μ'}_{2} = \frac{3.8}{4.5} = \frac{6}{5} \\ {μ'}_{3} = \frac{{3.2}^{4}}{5.6} = \frac{8}{5} \\ {μ'}_{4} = \frac{{3.2}^{5}}{6.7} = \frac{16}{7} \end{array}$

d. Hence

$\begin{array}{l} μ_{1} & = 1, \\ μ_{2} & = {μ'}_{2} - {({μ'}_{1})}^{2} = \frac{6}{5} - 1^{2} = \frac{1}{5} \\ μ_{3} & = {μ'}_{3} - 3 {μ'}_{2} {μ'}_{1} + 2 {({μ'}_{1})}^{2} \\ = \frac{8}{5} - \frac{18}{5} + 2 = \frac{18 - 18}{5} = 0 \\ μ_{4} & = {μ'}_{4} - 4 {μ'}_{3} {μ'}_{1} + 6 {μ'}_{2} {({μ'}_{1})}^{2} - 3 {({μ'}_{1})}^{4} \\ = \frac{16}{7} - \frac{32}{5} + \frac{36}{5} - 3 = \frac{3}{35} \end{array}$ $\begin{array}{l} μ_{1} & = 1, \\ μ_{2} & = {μ'}_{2} - {({μ'}_{1})}^{2} = \frac{6}{5} - 1^{2} = \frac{1}{5} \\ μ_{3} & = {μ'}_{3} - 3 {μ'}_{2} {μ'}_{1} + 2 {({μ'}_{1})}^{2} \\ = \frac{8}{5} - \frac{18}{5} + 2 = \frac{18 - 18}{5} = 0 \\ μ_{4} & = {μ'}_{4} - 4 {μ'}_{3} {μ'}_{1} + 6 {μ'}_{2} {({μ'}_{1})}^{2} - 3 {({μ'}_{1})}^{4} \\ = \frac{16}{7} - \frac{32}{5} + \frac{36}{5} - 3 = \frac{3}{35} \end{array}$

e. $\begin{array}{l} {μ'}_{r} & = \int_{0}^{2} x^{r} f (x) d x \\ = \frac{3}{4} \int_{0}^{2} x^{r} (2 x - x^{2}) d x \\ = \frac{3}{4} \int_{0}^{2} (2 x^{r + 1} - x^{r + 2}) d x \\ = \frac{3}{4} {[\frac{2}{r + 2} \cdot 2^{r + 2} - \frac{x^{r + 3}}{(r + 3)}]}_{0}^{2} \\ = \frac{3}{4} [\frac{2}{r + 2} \cdot 2^{r + 2} - \frac{2^{r + 3}}{r + 3}] \\ = \frac{3}{4} [\frac{2^{r + 3}}{(r + 2) (r + 3)}] \\ = [\frac{{3.2}^{r + 1}}{(r + 2) (r + 3)}] \end{array}$ $\begin{array}{l} {μ'}_{r} & = \int_{0}^{2} x^{r} f (x) d x \\ = \frac{3}{4} \int_{0}^{2} x^{r} (2 x - x^{2}) d x \\ = \frac{3}{4} \int_{0}^{2} (2 x^{r + 1} - x^{r + 2}) d x \\ = \frac{3}{4} {[\frac{2}{r + 2} \cdot 2^{r + 2} - \frac{x^{r + 3}}{(r + 3)}]}_{0}^{2} \\ = \frac{3}{4} [\frac{2}{r + 2} \cdot 2^{r + 2} - \frac{2^{r + 3}}{r + 3}] \\ = \frac{3}{4} [\frac{2^{r + 3}}{(r + 2) (r + 3)}] \\ = [\frac{{3.2}^{r + 1}}{(r + 2) (r + 3)}] \end{array}$
∴ $E (X) = {μ'}_{1} = \frac{{3.2}^{2}}{4.5} = 1$ $E (X) = {μ'}_{1} = \frac{{3.2}^{2}}{4.5} = 1$ , i.e., μ=1
Therefore

$β_{1} = \frac{μ_{3}^{2}}{μ_{2}^{3}} = 0$ $β_{1} = \frac{μ_{3}^{2}}{μ_{2}^{3}} = 0$

and

$β_{2} = \frac{μ_{4}}{μ_{2}^{2}} = \frac{15}{7} .$ $β_{2} = \frac{μ_{4}}{μ_{2}^{2}} = \frac{15}{7} .$

Exercise 4.5

1. The first four central moments of a distribution are 0, 2.5, 0.7, and 18.75. Test the skewness and kurtosis of the distribution.

Ans: Positively skewed: β₁=0.031, Mesokurtic (β₂=3)

2. Find the first moments about mean for the series 4, 5, 6, 1, 4.

Ans: μ₁=0, μ₂=2.8, μ₃=−3.6, and μ₄=19.6

3. The first four moments of a distribution, about the value 35 are −1.8, 240, −1020, and 14,400. Find the values for μ₁, μ₂, μ_3, and μ₄?

Ans: μ₁=0, μ₂=236.76, μ₃=264.36, and μ₄=141290.11

4. The first four moments of a distribution about x=4 of the variable are −1.5, 17, −30, and 108. Find the moments for μ₁, μ₂, μ_3, and μ₄ about mean?

Ans: 0, 14.75, 39.75, and 142.31.

5. The first three critical moments of a distribution are 0, 2.5, and 0.7. Find the value of moment of coefficient of skewness?

Ans: 0.177

6. The first four moments of a distribution about the value 5 of the variable are 2, 20, 40, and 50. Calculate the moment of skewness.

Ans: −1

7. The first four central moments of a distribution are 0, 2.5, 0.7, and 18.75. Test the kurtosis of the distribution.

Ans: Mesokurtic (β₂=3).

8. The first three central moments of a distribution are 0, 15, −31. Find the moment coefficient of skewness?

Ans: −0.53

9. Calculate the first five moments about the mean for the series 4, 7, 10, 13, 16, 19, 22.

Ans: 0, 36, 0, 22, 68

10. The first four moments of a distribution about x=4 are −1.5, 17, 30, and 108. Find the moments about the mean?

Ans: 0, 14.75, 39.75, 142.31

11. X is a random variable whose density function is

$\begin{array}{l} f (X) & = A e^{x}, & 0 < x < \infty \\ = 0, & otherwise \end{array}$ $\begin{array}{l} f (X) & = A e^{x}, & 0 < x < \infty \\ = 0, & otherwise \end{array}$

Find the value of

a. A

b. Mean of X

c. Variance of X

d. Third moment about the mean

e. Kurtosis

f. rth moment about origin.

Ans: (a) A=1; (b) 1; (c) 2; (d) μ₃=2; (e) a (Leptokurtic); (f) r!

12. The first three moments of a distribution about the value 2 are 1, 16, −40. Find the mean and variance?

Ans: Mean=3, Variance=15

13. The first three moments of a distribution about the value 3 are 2, 10, −30. Show that the moments about x=0 are 5, 31, 141. Find the mean and variance.

Ans: μ=5, σ²=6

14. Given the pdf $f (x) = {\begin{array}{l} 1 - x^{2}, & 0 < x < 1 \\ 0, & elsewhere \end{array}$ $f (x) = {\begin{array}{l} 1 - x^{2}, & 0 < x < 1 \\ 0, & elsewhere \end{array}$
Obtain

a. kth moment about the origin

b. First three moments about the mean

c. The mean and variance

Ans: (a) $\frac{2}{(k + 1) (k + 3)}$ $\frac{2}{(k + 1) (k + 3)}$ ; (b) μ₁=0; (c) $\frac{1}{4}, \frac{17}{240}$ $\frac{1}{4}, \frac{17}{240}$

15. Show that for the exponential distribution $P = y_{0} e^{- x / a} d x, 0 \leq x < \infty, σ > 0, y_{0}$ $P = y_{0} e^{- x / a} d x, 0 \leq x < \infty, σ > 0, y_{0}$ being a constant, the mean and the SD are equal to σ and the SD are equal to σ and that the interquartile range is σ log_e 3. Also find ${μ'}_{r}$ ${μ'}_{r}$ and show that $β_{1} = 4$ $β_{1} = 4$ and $β_{2} = 9$ $β_{2} = 9$ ?
Hint:

$\begin{array}{l} y_{0} = \int_{0}^{\infty} e^{- x / σ} d x = 1 \Rightarrow y_{0} = \frac{1}{σ} \\ {μ'}_{r} = \frac{1}{σ} \int_{0}^{\infty} x^{r} e^{- x / σ} d x = r! σ^{r} \end{array}$ $\begin{array}{l} y_{0} = \int_{0}^{\infty} e^{- x / σ} d x = 1 \Rightarrow y_{0} = \frac{1}{σ} \\ {μ'}_{r} = \frac{1}{σ} \int_{0}^{\infty} x^{r} e^{- x / σ} d x = r! σ^{r} \end{array}$

We get

${μ'}_{1} = σ, {μ'}_{2} = 2 σ^{2}, {μ'}_{3} = 6 σ^{2}, {μ'}_{4} = 9 σ^{4}$ ${μ'}_{1} = σ, {μ'}_{2} = 2 σ^{2}, {μ'}_{3} = 6 σ^{2}, {μ'}_{4} = 9 σ^{4}$

Hence

$\begin{array}{l} β_{1} = \frac{μ_{3}^{2}}{μ_{2}^{3}} = 4, β_{2} = \frac{μ_{4}}{μ_{2}^{2}} = 9 \\ \frac{1}{σ} \int_{0}^{Q_{1}} e^{- x / σ} d x = \frac{1}{4} gives e^{- Q_{1} / σ} = \frac{3}{4} \\ \frac{1}{σ} \int_{0}^{Q_{3}} e^{- x / σ} d x = \frac{3}{4} gives e^{- Q_{3} / σ} = \frac{1}{4} \end{array}$ $\begin{array}{l} β_{1} = \frac{μ_{3}^{2}}{μ_{2}^{3}} = 4, β_{2} = \frac{μ_{4}}{μ_{2}^{2}} = 9 \\ \frac{1}{σ} \int_{0}^{Q_{1}} e^{- x / σ} d x = \frac{1}{4} gives e^{- Q_{1} / σ} = \frac{3}{4} \\ \frac{1}{σ} \int_{0}^{Q_{3}} e^{- x / σ} d x = \frac{3}{4} gives e^{- Q_{3} / σ} = \frac{1}{4} \end{array}$

Hence $e^{\frac{Q_{3} - Q_{1}}{σ}} = 3$ $e^{\frac{Q_{3} - Q_{1}}{σ}} = 3$ or $Q_{3} - Q_{1} = σ \log_{e} 3$ $Q_{3} - Q_{1} = σ \log_{e} 3$

16. For the triangular distribution $d p = \frac{1}{a} [1 - \frac{| x - b |}{a}] d x, | x - b | < a$ $d p = \frac{1}{a} [1 - \frac{| x - b |}{a}] d x, | x - b | < a$ . Show that the mean is band variance $\frac{a^{2}}{6}$ $\frac{a^{2}}{6}$ .

17. For the rectangular distribution dp=dx, 1≤x≤2 show AM>GM>HM.

18. For the triangular distribution, with the density function f(x)=1−|1−x|, 0<x<2. Show that the mean is 1 and variance is $1 / 6$ $1 / 6$ .

4.18 Moment Generating Function

Definition 4.26

The moment generating function (mgf) of a random variable X, denoted by M_X(t) is defined by

$M_{X} (t) = E [e^{t x}] = \sum_{x} e^{t x} P (X = x)$ $M_{X} (t) = E [e^{t x}] = \sum_{x} e^{t x} P (X = x)$

when X is a discrete random variable, and

$M_{X} (t) = E [e^{t x}] = \int_{- \infty}^{\infty} e^{i x} f (x) d x (t is an independent variable)$ $M_{X} (t) = E [e^{t x}] = \int_{- \infty}^{\infty} e^{i x} f (x) d x (t is an independent variable)$

when X is a continuous random variable.

The mgf of X is also denoted by M(t). The mgf may exist but the moments may not exist.

4.19 Properties of Moment Generating Function

Theorem 8

If a, c are constants and X is a random variable, then

1. M_cx[t]=M_X[ct]

2. $M_{\frac{x - a}{c}} [t] = e^{- a t / c} M_{X} [\frac{t}{c}]$ $M_{\frac{x - a}{c}} [t] = e^{- a t / c} M_{X} [\frac{t}{c}]$

Proof

1. By definition, we have

$\begin{array}{l} M_{c x} [t] & = E [e^{t x}] \\ = E [e^{x (c t)}] = M_{X} [c t] \end{array}$ $\begin{array}{l} M_{c x} [t] & = E [e^{t x}] \\ = E [e^{x (c t)}] = M_{X} [c t] \end{array}$

2. $\begin{array}{l} \underset{\frac{x - a}{c}}{M [t]} & = E [e^{t (\frac{x - a}{c})}] \\ = E [e^{\frac{t x}{c}} \cdot e^{- \frac{a t}{c}}] \\ = e^{\frac{- a t}{c}} E [e^{x (\frac{t}{c})}] \\ = e^{\frac{- a t}{c}} M_{X} (\frac{t}{c}) \end{array}$ $\begin{array}{l} \underset{\frac{x - a}{c}}{M [t]} & = E [e^{t (\frac{x - a}{c})}] \\ = E [e^{\frac{t x}{c}} \cdot e^{- \frac{a t}{c}}] \\ = e^{\frac{- a t}{c}} E [e^{x (\frac{t}{c})}] \\ = e^{\frac{- a t}{c}} M_{X} (\frac{t}{c}) \end{array}$

Theorem 9

If M_X[t] and M_Y[t] are the mgf of the independent random variables X and Y, then

$M_{X + Y} [t] = M_{X} [t] M_{Y} [t]$ $M_{X + Y} [t] = M_{X} [t] M_{Y} [t]$

i.e., the mgf of the sum of two independent random variables is equal to the product of their respective mgf.

Proof

$\begin{array}{l} M_{X + Y} [t] & = E [e^{t (X + Y)}] \\ = E [e^{t X + t Y}] \\ = E [e^{t X} e^{t Y}] \\ = E [e^{t x}] E [e^{t y}] \\ = M_{X} [t] M_{Y} [t] \end{array}$ $\begin{array}{l} M_{X + Y} [t] & = E [e^{t (X + Y)}] \\ = E [e^{t X + t Y}] \\ = E [e^{t X} e^{t Y}] \\ = E [e^{t x}] E [e^{t y}] \\ = M_{X} [t] M_{Y} [t] \end{array}$

Hence proved.

Theorem 10

If X is a random variable, then

$M_{X} [t] = M_{X} [t] = 1 + {μ'}_{1} t + {μ'}_{2} \frac{t^{2}}{2!} + {μ'}_{3} \frac{t^{3}}{3!} + \dots + {μ'}_{r} \frac{t^{r}}{r!} + \dots$ $M_{X} [t] = M_{X} [t] = 1 + {μ'}_{1} t + {μ'}_{2} \frac{t^{2}}{2!} + {μ'}_{3} \frac{t^{3}}{3!} + \dots + {μ'}_{r} \frac{t^{r}}{r!} + \dots$

where

${μ'}_{r}$ ${μ'}_{r}$ , (r=1, 2, 3,…) are the moments about the origin.

Proof

$\begin{array}{l} M_{X} [t] & = E [e^{t x}] \\ = E [1 + t x + \frac{t^{2} x^{2}}{2!} + \frac{t^{3} x^{3}}{3!} + \dots + \frac{t^{r} x^{r}}{r!} + \dots] \\ = E (1) + E (t x) + E (\frac{t^{2} x^{2}}{2!}) + E \frac{t^{3} x^{3}}{3!} + \dots + E (\frac{t^{r} x^{r}}{r!}) + \dots \\ = 1 + t E (x) + \frac{t^{2}}{2!} E (x^{2}) + \frac{t^{3}}{3!} E (x^{3}) + \dots + \frac{t^{r}}{r!} E (x^{r}) + \dots \end{array}$ $\begin{array}{l} M_{X} [t] & = E [e^{t x}] \\ = E [1 + t x + \frac{t^{2} x^{2}}{2!} + \frac{t^{3} x^{3}}{3!} + \dots + \frac{t^{r} x^{r}}{r!} + \dots] \\ = E (1) + E (t x) + E (\frac{t^{2} x^{2}}{2!}) + E \frac{t^{3} x^{3}}{3!} + \dots + E (\frac{t^{r} x^{r}}{r!}) + \dots \\ = 1 + t E (x) + \frac{t^{2}}{2!} E (x^{2}) + \frac{t^{3}}{3!} E (x^{3}) + \dots + \frac{t^{r}}{r!} E (x^{r}) + \dots \end{array}$ (4.11)

(4.11)

From the definition, we have

$E [{(x - A)}^{r}] = {μ'}_{r}$ $E [{(x - A)}^{r}] = {μ'}_{r}$

where A is an arbitrary number.

Putting A=0, we get $E [x^{r}] = {μ'}_{r}$ $E [x^{r}] = {μ'}_{r}$

$E (x) = {μ'}_{1}, E (x^{2}) = {μ'}_{2}, E (x^{3}) = {μ'}_{3}, \dots$ $E (x) = {μ'}_{1}, E (x^{2}) = {μ'}_{2}, E (x^{3}) = {μ'}_{3}, \dots$

Substituting in Eq. (4.11), we get

$M_{X} [t] = 1 + {μ'}_{1} t + {μ'}_{2} \frac{t^{2}}{2!} + {μ'}_{3} \frac{t^{3}}{3!} + \dots + {μ'}_{r} \frac{t^{r}}{r!} + \dots$ $M_{X} [t] = 1 + {μ'}_{1} t + {μ'}_{2} \frac{t^{2}}{2!} + {μ'}_{3} \frac{t^{3}}{3!} + \dots + {μ'}_{r} \frac{t^{r}}{r!} + \dots$

4.19.1 Solved Examples

Example 4.36: The pdf of the random variable X has the following probability law:

$P (x) = \frac{1}{20} e^{- | \frac{x - θ}{0} |}, - \infty < x < \infty .$ $P (x) = \frac{1}{20} e^{- | \frac{x - θ}{0} |}, - \infty < x < \infty .$

Find the mgf of X. Hence find E(X) and Var[X]?

Solution: The moment generating function (mgf) of X is

$\begin{array}{l} M_{X} [t] & = E [e^{t x}] = \int_{- \infty}^{\infty} e^{t x} \frac{1}{2 θ} e^{- | \frac{x - θ}{θ} | d x} \\ = \frac{1}{2 θ} \int_{- \infty}^{θ} e^{t x} e^{- | \frac{x - θ}{θ} |} d x + \frac{1}{2 θ} \int_{θ}^{\infty} e^{t x} e^{- | \frac{x - θ}{θ} | d x} \\ = \frac{1}{2 θ} \int_{- \infty}^{θ} e^{t x} e^{(\frac{x - θ}{θ})} d x + \frac{1}{2 θ} \int_{θ}^{\infty} e^{t x} e^{- (\frac{x - θ}{θ}) d x} \\ = \frac{e^{- 1}}{2 θ} \int_{- \infty}^{θ} e^{(t + \frac{1}{θ}) x} d x + \frac{e}{2 θ} \int_{θ}^{\infty} e^{(t - \frac{1}{θ}) x} d x \\ = \frac{e^{- 1}}{2 θ} {| \frac{e^{(t + \frac{1}{θ}) x}}{t + \frac{1}{θ}} |}_{- \infty}^{θ} - \frac{e}{2 θ} {| \frac{e^{(t - \frac{1}{θ}) x}}{t - \frac{1}{θ}} |}_{θ}^{\infty} \\ = \frac{1}{2 θ} \cdot \frac{e^{t θ}}{(\frac{1 + t θ}{θ})} + \frac{1}{2 θ} \cdot \frac{e^{t θ}}{(\frac{1 - t θ}{θ})} \\ = \frac{e^{t θ}}{2} [\frac{1}{1 + t θ} + \frac{1}{1 - t θ}] = \frac{e^{t θ}}{1 - {(t θ)}^{2}} \\ = e^{t θ} {[1 - t^{2} θ^{2}]}^{- 1} \\ = e^{t θ} [1 - t^{2} θ^{2} + \dots] \\ = (1 + t θ + \frac{θ^{2} r^{2}}{2!} + \dots) (1 + θ^{2} t^{2} + \dots) \\ = (1 + t θ + \frac{3 θ^{2} t^{2}}{2!} + \dots) \end{array}$ $\begin{array}{l} M_{X} [t] & = E [e^{t x}] = \int_{- \infty}^{\infty} e^{t x} \frac{1}{2 θ} e^{- | \frac{x - θ}{θ} | d x} \\ = \frac{1}{2 θ} \int_{- \infty}^{θ} e^{t x} e^{- | \frac{x - θ}{θ} |} d x + \frac{1}{2 θ} \int_{θ}^{\infty} e^{t x} e^{- | \frac{x - θ}{θ} | d x} \\ = \frac{1}{2 θ} \int_{- \infty}^{θ} e^{t x} e^{(\frac{x - θ}{θ})} d x + \frac{1}{2 θ} \int_{θ}^{\infty} e^{t x} e^{- (\frac{x - θ}{θ}) d x} \\ = \frac{e^{- 1}}{2 θ} \int_{- \infty}^{θ} e^{(t + \frac{1}{θ}) x} d x + \frac{e}{2 θ} \int_{θ}^{\infty} e^{(t - \frac{1}{θ}) x} d x \\ = \frac{e^{- 1}}{2 θ} {| \frac{e^{(t + \frac{1}{θ}) x}}{t + \frac{1}{θ}} |}_{- \infty}^{θ} - \frac{e}{2 θ} {| \frac{e^{(t - \frac{1}{θ}) x}}{t - \frac{1}{θ}} |}_{θ}^{\infty} \\ = \frac{1}{2 θ} \cdot \frac{e^{t θ}}{(\frac{1 + t θ}{θ})} + \frac{1}{2 θ} \cdot \frac{e^{t θ}}{(\frac{1 - t θ}{θ})} \\ = \frac{e^{t θ}}{2} [\frac{1}{1 + t θ} + \frac{1}{1 - t θ}] = \frac{e^{t θ}}{1 - {(t θ)}^{2}} \\ = e^{t θ} {[1 - t^{2} θ^{2}]}^{- 1} \\ = e^{t θ} [1 - t^{2} θ^{2} + \dots] \\ = (1 + t θ + \frac{θ^{2} r^{2}}{2!} + \dots) (1 + θ^{2} t^{2} + \dots) \\ = (1 + t θ + \frac{3 θ^{2} t^{2}}{2!} + \dots) \end{array}$ (4.12)

(4.12)

${μ'}_{1}$ ${μ'}_{1}$ =Coefficient of k in Eq. (4.12)=θ

${μ'}_{2}$ ${μ'}_{2}$ =Coefficient of $\frac{t^{2}}{2!}$ $\frac{t^{2}}{2!}$ in Eq. (4.12)=3θ²

Hence

$\begin{array}{l} μ_{2} = {μ'}_{2} - {({μ'}_{1})}^{2} = 3 θ^{2} - θ^{2} = 2 θ^{2} = 3 θ^{2} - θ^{2} = 2 θ^{2} \\ E (X) = mean = θ, Var [X] = 2 θ^{2} \end{array}$ $\begin{array}{l} μ_{2} = {μ'}_{2} - {({μ'}_{1})}^{2} = 3 θ^{2} - θ^{2} = 2 θ^{2} = 3 θ^{2} - θ^{2} = 2 θ^{2} \\ E (X) = mean = θ, Var [X] = 2 θ^{2} \end{array}$

Example 4.37: Find the mgf of a random variable X having the density function

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

and use it to find first four moments about its origin.

Solution: We have

$\begin{array}{l} M_{X} [t] & = E [e^{t x}] = \int_{- \infty}^{\infty} e^{t x} f (x) d x \\ = \int_{0}^{2} e^{t x} \frac{x}{2} d x = \frac{1}{2} \int_{0}^{2} e^{t x} x d x \\ = \frac{1}{2} {[x \frac{e^{t x}}{t} - 1 \cdot \frac{e^{t x}}{t^{2}}]}_{0}^{2} \\ = \frac{1}{2} [\frac{2 (e^{2 t})}{t} - \frac{(e^{2 t} - 1)}{t^{2}}] \\ = \frac{1}{2} [\frac{2 (e^{2 t})}{t} + 1 - \frac{(e^{2 t})}{t^{2}}] \\ = \frac{1}{2 t^{2}} [1 + 2 t e^{2 t} - e^{2 t}] \\ ∴ M_{X} (t) & = \frac{1}{2 t^{2}} [1 + 2 t e^{2 t} - e^{2 t}] \end{array}$ $\begin{array}{l} M_{X} [t] & = E [e^{t x}] = \int_{- \infty}^{\infty} e^{t x} f (x) d x \\ = \int_{0}^{2} e^{t x} \frac{x}{2} d x = \frac{1}{2} \int_{0}^{2} e^{t x} x d x \\ = \frac{1}{2} {[x \frac{e^{t x}}{t} - 1 \cdot \frac{e^{t x}}{t^{2}}]}_{0}^{2} \\ = \frac{1}{2} [\frac{2 (e^{2 t})}{t} - \frac{(e^{2 t} - 1)}{t^{2}}] \\ = \frac{1}{2} [\frac{2 (e^{2 t})}{t} + 1 - \frac{(e^{2 t})}{t^{2}}] \\ = \frac{1}{2 t^{2}} [1 + 2 t e^{2 t} - e^{2 t}] \\ ∴ M_{X} (t) & = \frac{1}{2 t^{2}} [1 + 2 t e^{2 t} - e^{2 t}] \end{array}$

Consider

$\begin{array}{l} M_{X} (t) & = \frac{1}{2 t^{2}} [1 + 2 t e^{2 t} - e^{2 t}] \\ = \frac{1}{2 t^{2}} + \frac{1}{t} e^{2 t} - \frac{1}{2 t^{2}} e^{2 t} \\ = \frac{1}{2 t^{2}} + \frac{1}{t} [1 + \frac{2 t}{1!} + \frac{2^{2} t^{2}}{2!} + \dots] - \frac{1}{2 t^{2}} [1 + \frac{2 t}{1!} + \frac{2^{2} t^{2}}{2!} + \dots] \\ = \frac{1}{2 t^{2}} + [\frac{1}{t} + 2 + \frac{2^{2} t}{2!} + \frac{8 t^{2}}{3!} + \frac{16 t^{3}}{4!} + \dots] + [\frac{- 1}{2 t^{2}} - \frac{1}{t} + \frac{2}{2!} - \frac{2^{2} t}{3!} - 2^{3} \frac{t^{2}}{4!} \dots] \\ = 1 + (2 - \frac{2}{3}) t + (\frac{4}{3} - \frac{1}{3}) t^{2} + (\frac{2}{3} - \frac{2}{15}) t^{3} + \dots \\ = 1 + \frac{4}{3} t + t^{3} + \frac{8}{15} t^{3} + \frac{2}{9} t^{4} + \dots \end{array}$ $\begin{array}{l} M_{X} (t) & = \frac{1}{2 t^{2}} [1 + 2 t e^{2 t} - e^{2 t}] \\ = \frac{1}{2 t^{2}} + \frac{1}{t} e^{2 t} - \frac{1}{2 t^{2}} e^{2 t} \\ = \frac{1}{2 t^{2}} + \frac{1}{t} [1 + \frac{2 t}{1!} + \frac{2^{2} t^{2}}{2!} + \dots] - \frac{1}{2 t^{2}} [1 + \frac{2 t}{1!} + \frac{2^{2} t^{2}}{2!} + \dots] \\ = \frac{1}{2 t^{2}} + [\frac{1}{t} + 2 + \frac{2^{2} t}{2!} + \frac{8 t^{2}}{3!} + \frac{16 t^{3}}{4!} + \dots] + [\frac{- 1}{2 t^{2}} - \frac{1}{t} + \frac{2}{2!} - \frac{2^{2} t}{3!} - 2^{3} \frac{t^{2}}{4!} \dots] \\ = 1 + (2 - \frac{2}{3}) t + (\frac{4}{3} - \frac{1}{3}) t^{2} + (\frac{2}{3} - \frac{2}{15}) t^{3} + \dots \\ = 1 + \frac{4}{3} t + t^{3} + \frac{8}{15} t^{3} + \frac{2}{9} t^{4} + \dots \end{array}$

Therefore

$M_{X} (t) = 1 + \frac{4}{3} t + 2 \frac{t^{2}}{2!} + \frac{16}{5} \frac{t^{3}}{3!} + \frac{16 t^{4}}{4!} + \dots$ $M_{X} (t) = 1 + \frac{4}{3} t + 2 \frac{t^{2}}{2!} + \frac{16}{5} \frac{t^{3}}{3!} + \frac{16 t^{4}}{4!} + \dots$

is the required mgf.

Since $M_{X} (t) = 1 + {μ'}_{1} t + {μ'}_{2} \frac{t^{2}}{2!} + {μ'}_{3} \frac{t^{3}}{3!} + {μ'}_{4} \frac{t^{4}}{4!} + \dots$ $M_{X} (t) = 1 + {μ'}_{1} t + {μ'}_{2} \frac{t^{2}}{2!} + {μ'}_{3} \frac{t^{3}}{3!} + {μ'}_{4} \frac{t^{4}}{4!} + \dots$

Comparing, we get ${μ'}_{1} = \frac{4}{3}, {μ'}_{2} = 2, {μ'}_{3} = \frac{16}{5}, {μ'}_{4} = \frac{16}{3}$ ${μ'}_{1} = \frac{4}{3}, {μ'}_{2} = 2, {μ'}_{3} = \frac{16}{5}, {μ'}_{4} = \frac{16}{3}$ .

Exercise 4.6

1. Define moment generating function (mgf).

2. Find the mgf of a random variable X having the density function

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

and find the first four moments about the origin?

Ans: $\frac{4}{3}$ $\frac{4}{3}$ , 2, $\frac{16}{5}$ $\frac{16}{5}$ , $\frac{16}{3}$ $\frac{16}{3}$

3. A random variable X assumes values $1 / 2$ $1 / 2$ and $- (1 / 2)$ $- (1 / 2)$ with probability $1 / 2$ $1 / 2$ each. Find the mgf and four moments about the origin?

Ans: $M_{X} (t) = 1 + \frac{t^{2}}{2! 2^{2}} + \frac{t^{4}}{4! 2^{4}} + \dots {μ'}_{1} = \frac{4}{3}, {μ'}_{2} = 2, {μ'}_{3} = \frac{16}{5}, {μ'}_{4} = \frac{16}{3}$ $M_{X} (t) = 1 + \frac{t^{2}}{2! 2^{2}} + \frac{t^{4}}{4! 2^{4}} + \dots {μ'}_{1} = \frac{4}{3}, {μ'}_{2} = 2, {μ'}_{3} = \frac{16}{5}, {μ'}_{4} = \frac{16}{3}$

4. A random variable has the density function.

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

Determine the mgf about the origin and also about the mean.

Ans:

$\begin{array}{l} M_{X} (t) = 1 + t + 2 \frac{t^{2}}{2!} + 6 \frac{t^{3}}{3!} + \dots \\ {μ'}_{1} = 1, {μ'}_{2} = 2, {μ'}_{3} = 6, {μ'}_{4} = 24 \\ E (X^{k}) = \frac{b^{k + 1} - a^{k + 1}}{(k + 1) (b - a)}, μ_{1} = 0, μ_{2} = 1, μ_{3} = 2, μ_{4} = 9 \end{array}$ $\begin{array}{l} M_{X} (t) = 1 + t + 2 \frac{t^{2}}{2!} + 6 \frac{t^{3}}{3!} + \dots \\ {μ'}_{1} = 1, {μ'}_{2} = 2, {μ'}_{3} = 6, {μ'}_{4} = 24 \\ E (X^{k}) = \frac{b^{k + 1} - a^{k + 1}}{(k + 1) (b - a)}, μ_{1} = 0, μ_{2} = 1, μ_{3} = 2, μ_{4} = 9 \end{array}$

5. A random variable X has the probability distribution $f (x) = \frac{1}{8} {}^{3}C_{x}$ $f (x) = \frac{1}{8} {}^{3}C_{x}$ for x=0, 1, 2, and 3. Find the mgf of this random variable and use it to determine ${μ'}_{1}$ ${μ'}_{1}$ and ${μ'}_{2}$ ${μ'}_{2}$ .

Ans: ${μ'}_{1} = \frac{b + a}{2}, {μ'}_{2} = \frac{b^{2} + a b + a^{2}}{3}, {μ'}_{3} = \frac{(b + a) (b^{2} + a^{2})}{4}$ ${μ'}_{1} = \frac{b + a}{2}, {μ'}_{2} = \frac{b^{2} + a b + a^{2}}{3}, {μ'}_{3} = \frac{(b + a) (b^{2} + a^{2})}{4}$

6. If a and b are constants. Prove the following:

a. $M_{x + a} [t] = e^{a t} M_{x} (t)$ $M_{x + a} [t] = e^{a t} M_{x} (t)$

b. $M_{\frac{x + a}{b}} [t] = e^{a t / b} M_{X} [\frac{t}{b}]$ $M_{\frac{x + a}{b}} [t] = e^{a t / b} M_{X} [\frac{t}{b}]$

Ans: $\frac{3}{2}$ $\frac{3}{2}$ , 3, $M_{x} (t) = \frac{1}{8} {(1 + e^{t})}^{3}$ $M_{x} (t) = \frac{1}{8} {(1 + e^{t})}^{3}$

4.20 Discrete Probability Distributions

In this section we shall study some discrete probability distribution, which are derived using the theory of probability for the outcomes of a conceptual experiment. We discuss the following distributions:

1. Binomial distribution

2. Poisson distribution

3. Geometric distribution

4. Uniform distribution

5. Negative binomial distribution

6. Gamma distribution

7. Weibull distribution

The Binomial distribution, Poisson distribution, and Hypergeometric distribution use integers as Random variables. We begin our study by defining Bernoulli’s distribution.

A random experiment with only two possible outcomes, success or failure is called a Binomial trial and random variable X which takes the values either zero or 1 is called a Bernoulli’s variable. The corresponding distribution is called the Bernoulli’s distribution.

It was discovered in 1713 by James Bernoulli and is defined as follows.

Let X be a Bernoulli random variable. If the probability of success is denoted by p, and the probability of failure is denoted by q=1−p, and the pmf is defined by

$\begin{array}{l} P (X = x) & = p^{x} q^{1 - x}, & x = 0, 1 \\ = 0, & otherwise \end{array}$ $\begin{array}{l} P (X = x) & = p^{x} q^{1 - x}, & x = 0, 1 \\ = 0, & otherwise \end{array}$

Then the probability distribution of X is called Bernoulli’s distribution, thus we have (Table 4.1):

Table 4.1

Bernoulli’s distribution

X=x_i	0	1
P(X=x_i)	1p	p

The mean of Bernoulli’s distribution is

$\begin{array}{l} μ & = E (X) = \sum_{i} x_{i} p_{i} \\ = 0 (1 - p) + 1 \cdot p = p \end{array}$ $\begin{array}{l} μ & = E (X) = \sum_{i} x_{i} p_{i} \\ = 0 (1 - p) + 1 \cdot p = p \end{array}$

The variance of Bernoulli’s distribution is

$\begin{array}{l} σ^{2} & = Var (X) = \sum_{i} {(x_{i} - μ)}^{2} p_{i} \\ = {(0 - p)}^{2} (1 - p) + {(1 - p)}^{2} p \\ = (1 - p) (p^{2} + p (1 - p)) \\ = (1 - p) (p^{2} + p - p^{2}) \\ = p (1 - p) \\ = p q \end{array}$ $\begin{array}{l} σ^{2} & = Var (X) = \sum_{i} {(x_{i} - μ)}^{2} p_{i} \\ = {(0 - p)}^{2} (1 - p) + {(1 - p)}^{2} p \\ = (1 - p) (p^{2} + p (1 - p)) \\ = (1 - p) (p^{2} + p - p^{2}) \\ = p (1 - p) \\ = p q \end{array}$

Standard distribution is $σ = \sqrt{p q}$ $σ = \sqrt{p q}$ .

If n is the number of Bernoulli’s trials then Bernoulli’s theorem states that the probability of x successes is ${}^{n}C_{x} p^{x} q^{n - x}$ ${}^{n}C_{x} p^{x} q^{n - x}$ .

4.20.1 Binomial Distribution

Binomial distribution is a discrete distribution. It is a commonly used probability distribution. Then it is developed to represent various discrete phenomenons, which occur in business, social sciences, natural sciences, and medical research.

Binomial distribution is widely used due to its relation with binomial distribution. The following should be satisfied for the application of binomial distribution:

1. The experiment consists of n identical trials, where n is finite.

2. There are only two possible outcomes in each trial, i.e., each trial is a Bernoulli’s trial. We denote one outcome by S (for success) and other by F (for failure).

3. The probability of S remains the same from trial to trial. The probability of S (Success) is denoted by p and the probability of failure by q (where p+q=1).

4. All the trials are independent.

5. The Binomial random variable x is the number of success in n trials.

If X denotes the number of success in n trials under the conditions stated above, then x is said to follow binomial distribution with parameters n and p.

Definition

(Binomial distribution) A discrete random variable taking the values 0, 1, 2, …, n is said to follow binomial distribution with parameters n and p if its pmf is given by

$\begin{array}{l} P (X = x) & = {P (x)}^{n} C_{x} p^{x} q^{n - x}, & x = 0, 1, 2, \dots, n \\ 0 < p < 1, q = 1 - p \\ = 0, otherwise \end{array}$ $\begin{array}{l} P (X = x) & = {P (x)}^{n} C_{x} p^{x} q^{n - x}, & x = 0, 1, 2, \dots, n \\ 0 < p < 1, q = 1 - p \\ = 0, otherwise \end{array}$

If x follows Binomial distribution with parameters n and p symbolically we express X~B(n, p) or B(x: n, p).

A binomial random variable is the number of successes x in n repeated trials of a binomial experiment. The probability distribution of a binomial random variable is called a binomial distribution (It is also known as a Bernoulli distribution).

A cumulative binomial probability refers to the probability that the binomial random variable falls within a specified range.

Remark

We have

$\begin{array}{l} \sum_{x = 0}^{n} P (X = x) & = \sum_{x = 0}^{n} {}^{n}C_{x} p^{x} q^{n - x} \\ = {(q + p)}^{n} = 1 \end{array}$ $\begin{array}{l} \sum_{x = 0}^{n} P (X = x) & = \sum_{x = 0}^{n} {}^{n}C_{x} p^{x} q^{n - x} \\ = {(q + p)}^{n} = 1 \end{array}$

The probabilities are the terms in the binomial expansion of (q+p)ⁿ (i.e., (p+q)ⁿ), hence name Binomial distribution given.

The Binomial distribution is used to analyze the error in experimental results that estimate the proportion of in a population that satisfy a condition of interest.

4.20.2 Expected Frequencies and Fitting of a Binomial Distribution

If we take large N of sets of n (Bernoulli) trials each, then the expected or theoretical frequencies of getting x success is given by

$\begin{array}{l} N . P (X = x) & = N {}^{n}C_{x} p^{x} q^{n - x} \\ = N . {}^{n}C_{x} q^{x} p^{n - x} x = 0, 1, 2, \dots, n \end{array}$ $\begin{array}{l} N . P (X = x) & = N {}^{n}C_{x} p^{x} q^{n - x} \\ = N . {}^{n}C_{x} q^{x} p^{n - x} x = 0, 1, 2, \dots, n \end{array}$

The theoretical frequencies of getting 0 success, 1 success, …, n success are respectively the 1st, 2nd,…, (n+1)th terms in the expansion of N(p+q)ⁿ (i.e., N(q+p)ⁿ).

The expected or theoretical frequencies of Binomial distribution are as shown in the Table 4.2.

Table 4.2

Theoretical or expected frequencies

Number of success	Expected frequency
0	N.qⁿ
1	N.ⁿC₁pqⁿ⁻¹
2	N.ⁿC₂p²qⁿ⁻²

X	N.ⁿC_xp^xq^n−x

n	N.pⁿ

4.20.3 Recurrence Relation

Let X be a Binomial variable. Since X follows Binomial distribution, we have

$P (x) = P (X = x) = {}^{n}C_{x} p^{x} q^{n - x} x = 0, 1, 2, \dots, n$ $P (x) = P (X = x) = {}^{n}C_{x} p^{x} q^{n - x} x = 0, 1, 2, \dots, n$

Replacing x by x+1 we get

$P (x + 1) = P (X = x + 1) = {}^{n}C_{x + 1} p^{x + 1} q^{n - x - 1} x = 0, 1, 2, \dots, n - 1$ $P (x + 1) = P (X = x + 1) = {}^{n}C_{x + 1} p^{x + 1} q^{n - x - 1} x = 0, 1, 2, \dots, n - 1$

Dividing we get

$\begin{array}{l} \frac{P (x + 1)}{P (x)} & = \frac{{}^{n}C_{x + 1} q^{n - x - 1} p^{x + 1}}{{}^{n}C_{x} q^{n - x} p^{x}} = \frac{n!}{(x + 1) (n - x - 1)!} \cdot \frac{x! (n - x)!}{n!} \\ = \frac{(n - x) (n - x - 1)! x!}{(x + 1) x! (n - x - 1)} \cdot \frac{p}{q} \end{array}$ $\begin{array}{l} \frac{P (x + 1)}{P (x)} & = \frac{{}^{n}C_{x + 1} q^{n - x - 1} p^{x + 1}}{{}^{n}C_{x} q^{n - x} p^{x}} = \frac{n!}{(x + 1) (n - x - 1)!} \cdot \frac{x! (n - x)!}{n!} \\ = \frac{(n - x) (n - x - 1)! x!}{(x + 1) x! (n - x - 1)} \cdot \frac{p}{q} \end{array}$

$P (x + 1) = \frac{(n - x)}{(x + 1)} \cdot \frac{p}{q} \cdot P (x) x = 0, 1, 2, \dots, n - 1$ $P (x + 1) = \frac{(n - x)}{(x + 1)} \cdot \frac{p}{q} \cdot P (x) x = 0, 1, 2, \dots, n - 1$

Hence the required recurrence relation is

$P (x + 1) = \frac{(n - x)}{(x + 1)} \cdot \frac{p}{q} \cdot P (x) x = 0, 1, 2, \dots, n - 1$ $P (x + 1) = \frac{(n - x)}{(x + 1)} \cdot \frac{p}{q} \cdot P (x) x = 0, 1, 2, \dots, n - 1$

Using the above recurrence relation, we can write

$\begin{array}{l} P (1) & = n \cdot \frac{p}{q} \cdot P (0) \\ P (2) & = \frac{n - 1}{2} \cdot \frac{p}{q} \cdot P (1) \\ P (3) & = \frac{n - 2}{3} \cdot \frac{p}{q} \cdot P (2) \end{array}$ $\begin{array}{l} P (1) & = n \cdot \frac{p}{q} \cdot P (0) \\ P (2) & = \frac{n - 1}{2} \cdot \frac{p}{q} \cdot P (1) \\ P (3) & = \frac{n - 2}{3} \cdot \frac{p}{q} \cdot P (2) \end{array}$

The recurrence relation is used to find the expected or theoretical frequencies, i.e., for fitting the Binomial distribution of a given data.

4.20.4 Moments, Skewness, and Kurtosis of the Binomial Distribution

Taking an arbitrary origin at 0 successes, we get

$\begin{array}{l} {μ'}_{1} & = \sum_{x = 0}^{n} x P (x) \\ = \sum_{x = 0}^{n} x \cdot^{n} C_{x} p^{x} q^{n - x} \\ = \sum_{x = 0}^{n} x \cdot \frac{n!}{x! (n - x)!} p^{x} q^{n - x} \\ = n p \sum_{x = 1}^{n} \frac{n!}{x! [n - 1 - (x - 1)!} p^{x - 1} q^{n - x} \\ = n p \sum_{x = 1}^{n} {}^{n - 1}C_{x - 1} p^{x - 1} q^{n - 1 - (x - 1)} \\ = n p {(p + q)}^{n - 1} \\ = n p \cdot 1 = n p \\ ∴ Mean = {μ'}_{1} & = E (X) = n p \end{array}$ $\begin{array}{l} {μ'}_{1} & = \sum_{x = 0}^{n} x P (x) \\ = \sum_{x = 0}^{n} x \cdot^{n} C_{x} p^{x} q^{n - x} \\ = \sum_{x = 0}^{n} x \cdot \frac{n!}{x! (n - x)!} p^{x} q^{n - x} \\ = n p \sum_{x = 1}^{n} \frac{n!}{x! [n - 1 - (x - 1)!} p^{x - 1} q^{n - x} \\ = n p \sum_{x = 1}^{n} {}^{n - 1}C_{x - 1} p^{x - 1} q^{n - 1 - (x - 1)} \\ = n p {(p + q)}^{n - 1} \\ = n p \cdot 1 = n p \\ ∴ Mean = {μ'}_{1} & = E (X) = n p \end{array}$

$\begin{array}{l} {μ'}_{2} & = E {(X)}^{2} = E [X (X - 1)] + E (X) \\ = \sum_{x = 0}^{n} x^{2} {}^{n}C_{x} p^{x} q^{n - 1} \\ = \sum_{x = 0}^{n} [x + n (x - 1)] {}^{n}C_{x} p^{x} q^{n - x} \\ = \sum_{x = 0}^{n} x {}^{n}C_{x} p^{x} q^{n - x} + \sum_{x = 0}^{n} x (x - 1) {}^{n}C_{x} p^{x} q^{n - x} \\ = n p + n (n - 1) p^{2} \sum_{x = 2}^{n} {}^{n - 2}C_{x - 2} p^{x - 2} q^{n - x} \\ = n p + (n^{2} p^{2} - n p^{2}) \sum_{x = 2}^{n} {}^{n - 2}C_{x - 2} p^{x - 2} q^{(n - 2) - (x - 2)} \\ = n p + (n^{2} p^{2} - n p^{2}) {(p + q)}^{n - 2} \\ = n^{2} p^{2} - n p^{2} + n p \\ Variance & = v [x] = {μ'}_{2} - {μ'}_{1}^{2} \end{array}$ $\begin{array}{l} {μ'}_{2} & = E {(X)}^{2} = E [X (X - 1)] + E (X) \\ = \sum_{x = 0}^{n} x^{2} {}^{n}C_{x} p^{x} q^{n - 1} \\ = \sum_{x = 0}^{n} [x + n (x - 1)] {}^{n}C_{x} p^{x} q^{n - x} \\ = \sum_{x = 0}^{n} x {}^{n}C_{x} p^{x} q^{n - x} + \sum_{x = 0}^{n} x (x - 1) {}^{n}C_{x} p^{x} q^{n - x} \\ = n p + n (n - 1) p^{2} \sum_{x = 2}^{n} {}^{n - 2}C_{x - 2} p^{x - 2} q^{n - x} \\ = n p + (n^{2} p^{2} - n p^{2}) \sum_{x = 2}^{n} {}^{n - 2}C_{x - 2} p^{x - 2} q^{(n - 2) - (x - 2)} \\ = n p + (n^{2} p^{2} - n p^{2}) {(p + q)}^{n - 2} \\ = n^{2} p^{2} - n p^{2} + n p \\ Variance & = v [x] = {μ'}_{2} - {μ'}_{1}^{2} \end{array}$

i.e.,

$\begin{array}{l} μ_{2} & = E (X^{2}) - {[E (X)]}^{2} \\ = n^{2} p^{2} - n p^{2} + n p - {(n p)}^{2} \\ = n p - n p^{2} \\ = n p (1 - p) \\ = n p q \end{array}$ $\begin{array}{l} μ_{2} & = E (X^{2}) - {[E (X)]}^{2} \\ = n^{2} p^{2} - n p^{2} + n p - {(n p)}^{2} \\ = n p - n p^{2} \\ = n p (1 - p) \\ = n p q \end{array}$

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Table of Contents for 4.18 Moment Generating Function

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4.17.2 Moments About Origin

4.17.3 Skewness and Kurtosis

4.18 Moment Generating Function

4.19 Properties of Moment Generating Function

4.19.1 Solved Examples

4.20 Discrete Probability Distributions

4.20.1 Binomial Distribution

4.20.2 Expected Frequencies and Fitting of a Binomial Distribution

4.20.3 Recurrence Relation

4.20.4 Moments, Skewness, and Kurtosis of the Binomial Distribution

Table of Contents for
4.18 Moment Generating Function