Moment Generating Function

 Moment Generating Function

The moment generating. function (m.g.f.). of a, random variable X (about·origin) having the probability function f(x) is given by

M_x (t) = E(e^tx) = ∫e^txf(x)dx 
                                             (for continuous  probability function)
= ∑e^txf(x)

                                             (for discrete random variable )
the integration or summation being extended to the entire range of x, t being the real parameter and it is being assumed that the right-hand side of above equation is absolutely  convergent for some positive number h such that  -h <t < h Thus

M_x (t) = E(e^tx) = E[1 + tx + t²X² /2! + ....... + t^'X^' /r! ]
1+tE(X) + t² /2!E(X²) + t^' /r!E(X^') + ........

is the rth moment of X about origin. Thus we see that the coefficient of  t^' /r! in  M_x (t) gives μ_r'  (above origin). since M_x (t) generates moments, it is known as moment generties function. Differentiating  w.r.t. t and then' putting 1 = 0,    we get

[ d^r/dt^'{M_x (t) } ]_r=0  =[{μ_r' /r!}r! + μ_r+1't + μ_r+2' t² /2! +......]_r=0

μ_r' =[ d^r/dt^'{M_x (t) } ]_r=0 

In general, the moment generating function of X about the point X. a is defined as
 M_x (t) (about X = a) = E [e^t(x+1) ]
                         =E [1 + t(X-a) + t² /2!(X-a)² + .......... + t² /r!(X-a)^r ]
                        = 1+ tμ'_r+ t² /2!μ'₂+.................+ t^r /r!μ'_r +............

where μ'_r.= E {(X - a)²}, is the r^th moment-about the point X - a.

A Discrete Example

Suppose a discrete PDF is given by the following table.

X=x$X=x$	P(X=x)$P(X=x)$
X=0$X=0$
	0.25$0.25$

We obtain the moment generating function  from the expected value of the exponential function.

MX(t)=E(etx)=∑all xetxP(x)=0.4e0t+0.35e1t+0.25e2t=0.4+0.35et+0.25e2t

We can then compute derivatives and obtain the moments about zero.

E(X)Var(X)Skew(X)Kurt(X)=M′(0)=0.85=E(X2)−E(X)2=1.35−0.852=0.6275=E(X3)−3E(X)E(X2)+2E(X)3=2.35−3(0.85)(1.35)+2(0.853)=0.13575=E(X4)−4E(X)E(X3)+6E(X)2E(X2)−3E(X)4=4.35−4(0.85)(2.35)+6(0.852)(1.35)−3(0.854)≈0.6462

Uniqueness theorem of m.g.f

 The moment generating function of a distribution if it exists uniquelly determines the distribution. This implies that corresponding to a given probability distribution there is only one m.g.f provided it exists and corresponding to a given m.g.f, there is only one probability distribution. 

 Hence 

$M_{X}(t)$ = $M_{Y}(t)$ = X And Y are identically distributed.

Limitations 

Some of the important limitations on m.g.f are given below:

1) A random variable X may have no moments although its m.g.f exits. 
2) A random variable X can have m.g.f and some moments yet the m.g.f does not generate the moments. 
3) A random variable X can have all are some moments but m.g.f does exist except perhaps at one point.

Mean and Variance of m.g.f

The mean and variance are therefore
$\mu$    = x
= $M^{'}(0)$  

$\sigma^{2}$ = $x^{2}$ - $x^2$   

= $M^{''}(0)$ - $[M^{'}(0)]^{2}$.   

It is also true that

$\mu_n$ = $\sum_{(j=0)}^{n}$ $(n; j)$ (-1)$^{n-j}$ $\mu_{j}^{'}$($\mu_{1}^{'}$)$^{(n-j)}$,    

where $\mu_{0}^{'}$ = 1 and $\mu_{j}^{'}$ is the jth raw moment. 

Examples

Some problems are solved below on moment generating function:

Example 1: f the moments of a variate 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 $\geq$ 2) = 0.

Solution: The m.g.f of variate X is

$M_{X}$(t) = $\sum_{r = 0}^{\infty}$ $\frac{t^{r}}{r!}$ $\mu_{r}$'

= 1 + $\sum_{r = 1}^{\infty}$(0.6)

= 0.4 + 0.6 $\sum_{r = 0}^{\infty}$ $\frac{t^{r}}{r!}$

= 0.4 + 0.6 e$^{t}$

but $M_{x}(t)$ = $E(e^{tX})$ = $\sum_{x = 0}^{\infty}e^{tx}$ $P(X = x)$

= $P(X = 0)$ + $e^{t}$.$P(X = 1)$ + $\sum_{x = 2}^{\infty}e^{tx}$ $P(X = x)$

P(X = 0) = 0.4, P(X = 1) = 0.6, P(X $\geq$ 2) = 0.

Hence proved

Example 2: Find the m.g.f of the random variable whose moments are
$M_{r}$' = $(r + 1)!2^{r}$

Solution:
The m.g.f is given by

$M_{x}$(t) = $\sum_{r = 0}^{\infty}$ $\frac{t^{r}}{r!}$ $\mu_{r}'$

= $\sum_{r  = 0}^{\infty}$ $\frac{t^{r}}{r!}$ $(r + 1)!2^{r}$

= $\sum_{r = 0}^{\infty}(r + 1)(2t)^{r}$

= 1 +  2 (2t) + 3(2t)$^{2}$ + 4 (2t)$^{3}$ + ............................

= ( 1 - 2t)$^{-2}$