Moments - moment generating functions and their properties

MOMENTS - MOMENT GENERATING FUNCTIONS AND THEIR PROPERTIES :

 Moments [Discrete case]

Let X be discrete R.V. taking the values x₁, x₂,....,x_n with probability mass function P₁, P₂, ... P_n respectively then the r_th moment about the origin is

Moments [Continuous case]

If X is a continuous R.V. with probability density function f(x) defined in the interval (a, b) then

Moments Generating Function: (M.G.F)

An important device that can be used to calculate the higher moments is the moment generating function.

Moment generating function of a random variable X about the origin is defined as

where t being a real parameter assuming that the integration or summation is absolutely convergent for some positive number h such that |t| < h

 

where μ_r' = r^th  moment about the origin.

E(X^f) = ∫ x^r f(x) dx or

= ∑x^rp(x) depending upon X is continuous or discrete

The coefficient of 

M_X(t) generates moments about the origin and hence we call it as moment generating function.

Note1: µ_r' = d^r/dt^r [ M_X(t)]_t=0

2: M_X(t) = E[e^t(X-a)]

 (about X = a)

Where µ_r'  = E [ (X- a)^r], r^th moment about the point X = a

Note 3 : Mean =

= r^th central moment

Limitations of m.g.f

1. A random variable X may have no moment although its m.g.f exists.

2. A random variable X can have its moment generating function and some (or all) moments, yet the moment generating function does not generate the moments.

3. A random variable X can have all or some moments, but moment generating function do not exist except perhaps at one point.

Properties of moment Generating function [A.U Tvli. A/M 2009]

1. Let Y = aX + b, where X is a R.V with moment generating function M_X (t). Then

2. M_cX(t) = E [e^cXt] = E [e^X(ct)] = M_X(ct) where c is a constant.

3. If X and Y are two independent random variables, then

M_{X + Y} (t) = M_X (t). M_Y (t)

Proof :

 

Example 1.6.1

Find the moment generating function of the RV X whose probability function

P(X = x) 1/2^x , x = 1,2,... ... Hence find its mean. [A.U Tvli A/M 2009] [A.U CBT A/M 2011] [A.U N/D 2018 PQT R-13]

Solution :

Example 1.6.2

If X represents the outcome, when a fair die is tossed, find the moment generating function (MGF) of X and hence find E(X) and Var(X). [A.U A/M 2018 R-13]

Solution: The probability distribution of X is given by

P_i = P (X = i) = 1/6, i = 1, 2, ... 6

= 1/6 [ 1 + 4 + 9 + 16 + 25 + 36 ] = 1/6 [91]

Var(X) = E (X²) - [E(X)]² = 91/6 – 49/4 = 35/12

 

Example 1.6.3

Find the probability distribution of the total number of heads obtained in four tosses of a balanced coin. Hence obtain the MGF of X, mean of X and of a variance of X. [AU A/M 2008]

Solution :

Example 1.6.4

For a discrete random variable. X with probability function

 

Show that E(X) does not exist eventhough m.g.f exists. [A.U N/D 2012]

Solution: 

Hence, E(X) does not exist.

Now, we have, by definition the m.g.f as

by using L'Hospital rule for indetermine form (0 × ∞) and M_X(t) does not exist for t > 0

 

Example 1.6.5

For the triangular distribution

f(x) =   [A.U. M/J 2006, N/D 2013]

find the mean, variance and the moment generating function (MGF) also find cdf of F(x). [A.U CBT M/J 2010, CBT N/D 2011) [A.U N/D 2013] [A.U A/M 2018 R-08] [A.U N/D 2018 R13 RP]

Solution : 

=(1/4 - 0) + (16/3 – 16/4 ) – (2/3 – 1/4 ) = 1/4 + 16/3 – 16/4 – 2/3 + 1/4

= - 14/4 + 14/3 = -42 + 56 /12 = 14/12 = 7/6

(2) Var (X) = E [X²] - [E (X)]²

= 7/6 – (1)² = 7/6 – 1 = 1/6

The moment generating function of the Random variable X is

To find the cdf of F (x)

 

Example 1.6.6

Let the random variable X have the p.d.f

Find the moment generating function, mean and variance of X.

[A.U. A/M. 2005, N/D 2012] [A.U A/M 2019 (R8) RP]

Solution: The m.g.f is given by

Mean = E [X] =Co-efficient of = t /1! = 2

E[X²] = Co-efficient of t²/2! = 8

Var [X] = E [X²] - [E [X]]²

= 8-2² = 4

 

Example 1.6.7

The density function of a random variable x is given by f(x) = Kx (2 - x), 0 ≤ x ≤ 2. Find K, mean, variance and r^th moment. [A.U. N/D 2006] [A.U. M/J 2007] [A.U Trichy A/M 2010]

Given: f(x) = Kx (2 - x), 0 ≤  x  ≤2 is a p.d.f.

We know that, if f (x) is a p.d.f then,

Var X = E(X²]-[E[X]]² = 6/5 – (1)² = 6/5 – 1 = 1/5

 

Example 1.6.8

A continuous R.V. X has the p.d.f f(x) given by f(x) = ce ̄^-:x: -∞ < x < ∞. Find the value of c and moment generating function of X. [A.U. M/J 2007]

Solution :

Given: f(x) = ce^-:x:

Given f(x) is a p.d.f.

 

Example 1.6.9

If a R.V X has the mgf My (t) = 3 /3 – ť  obtain the standard deviation of X. [A.U A/M 2018 R-08]

Solution :

 

Example 1.6.10

The first four moments of a distribution about X = 4 are 1, 4, 10 and 45 respectively. Show that the mean is 5, variance is 3, µ₃ = 0 and μ₄ = 26. [A.U. N/D. 2004]

Solution : Given that μ₁'=1; μ₂'=4; μ₃' = 10 and μ₄' = 45.

The point A = 4 + μ₁' = 4 + 1 = 5

Variance = μ₂ = μ₂' - μ₁'² = 4 - 1 = 3

μ₃  = μ₃' - 3 μ₂' μ₁' + 2 μ₁'³

= 10 - 3 (4) (1) + 2(1)³ =  10 - 12 + 2 = 0

μ₄ = μ₄' - 4μ₃' μ₁'+ 6 μ₂' μ₁'² - 3 μ₁'⁴

= 45- 4 (10) (1) +6 (4) (1)² - 3 (1)⁴

= 45 - 40 + 24 – 3 = 26

 

Example 1.6.11

If a RV X has the moment generating function M_X (t) = 2/2 - t determine the

variance of X. [A.U M/J 2012]

Solution:

Given: M_X (t) = 2/2 - t = 2/ 2(1- t/2) = ( 1 – t/2)^-1

The coefficient of 1/1! is 1/2 i.e. E(X) = 1/2

The coefficient of of t²/2! is 1/2. E(X²) = 1/2

Var (X) = E (X²) - [E(X)]² = 1/2 -1/4 = 1/4

 

Example 1.6.12

Find the moment generating function of the RV whose moments are m_r = (r+ 1)! 2^r

Solution : The moment generating function is given by 

= 1 + 2 (2t) + 3 (2t)² + ....=(1- 2t)^-2 = 1/(1-2t)²

 

Example 1.6.13

A random variable X has density function given by

Find

(1) m.g.f. (2) r^th moment (3) mean (4) variance  [AU N/D 2006]

Solution: 

Variance = E (X²) - [E(X)]²

= k²/3- k²/4 = k²/12

Example 1.6.14

If the moments of a random variable 'X' are defined by E (X^r) = 0.6; r = 1, 2, 3, ... Show that P (X = 0) = 0.4, P (X = 1) = 0.6, P (X ≥2) = 0 [AU N/D 2008]

Solution: We know that,

Compare (1) & (2), we get

p (0) = p[X = 0] = 0.4

P (1) = p[X= 1] = 0.6

P[X ≥ 2] = 0

 

Example 1.6.15

Prove that the moment generating function of the sum of a number of independent random variables is equal to the product of their respective moment generating functions. [AU N/D 2006]

Solution: Given:

If X₁, X₂, ... X_n are n independent R.V's then

 

Example 1.6.16

Let X be a R.V. with p.d.f 

Find, (1) P(X > 3)

 (2) Moment generating function of X

 (3) E(X) and Var(X)

Solution: 

Mean = E [X] =Co-efficient of = t /1! = 3

E[X²] = Co-efficient of t²/2! = 18

Var [X] = E [X²] - [E [X]]²

= 18 – [3]² = 18 - 9 = 9

 

Example 1.6.17

Find the first four moments about the origin for a random variable X having

the pdf. 

[AU N/D 2008] [A.U N/D 2016 R13 PQT] [A.U N/D 2019 (R17) PS]

Solution: 

Note : m₁ = 4/81 [ 1/3 – 1/5 ]3⁵ , m₂ = 4/81 [1/4- 1/6 ]3⁶

M₃ = 4/81 [1/5 – 1/7]3⁷ , m₄ = 4/81 [1/6 – 1/8 ]3⁸

 

Example 1.6.18

Give an example to show that if pdf exists but MGF does not exist.

Solution: 

R.H.S is a divergent series  MGF does not exist.

Probability and Random Variables

 

Example 1.6.19

Find the M.G.F. of the random variable X having the probability density function

Also deduce the first four moments about the origin. [A.U N/D 2010, M/J 2012] [A.U A/M 2017 R13]

Solution: