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Integral transforms

We consider briefly the relationship of the moment generating function and the characteristic function with well known integral transforms (hence the name of this chapter).

Moment generating function and the Laplace transform

When we examine the integral forms of the moment generating function, we see that they represent forms of the Laplace transform, widely used in engineering and applied mathematics.Suppose F X is a probability distribution function with F X ( - ) = 0 . The bilateral Laplace transform for F X is given by

- e - s t F X ( t ) d t

The Laplace-Stieltjes transform for F X is

- e - s t F X ( d t )

Thus, if M X is the moment generating function for X , then M X ( - s ) is the Laplace-Stieltjes transform for X (or, equivalently, for F X ).

The theory of Laplace-Stieltjes transforms shows that under conditions sufficiently general to include all practical distribution functions

M X ( - s ) = - e - s t F X ( d t ) = s - e - s t F X ( t ) d t

Hence

1 s M X ( - s ) = - e - s t F X ( t ) d t

The right hand expression is the bilateral Laplace transform of F X . We may use tables of Laplace transforms to recover F X when M X is known. This is particularly useful when the random variable X is nonnegative, so that F X ( t ) = 0 for t < 0 .

If X is absolutely continuous, then

M X ( - s ) = - e - s t f X ( t ) d t

In this case, M X ( - s ) is the bilateral Laplace transform of f X . For nonnegative random variable X , we may use ordinary tables of the Laplace transform to recover f X .

Use of laplace transform

Suppose nonnegative X has moment generating function

M X ( s ) = 1 ( 1 - s )

We know that this is the moment generating function for the exponential (1) distribution. Now,

1 s M X ( - s ) = 1 s ( 1 + s ) = 1 s - 1 1 + s

From a table of Laplace transforms, we find 1 / s is the transform for the constant 1 (for t 0 ) and 1 / ( 1 + s ) is the transform for e - t , t 0 , so that F X ( t ) = 1 - e - t t 0 , as expected.

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Laplace transform and the density

Suppose the moment generating function for a nonnegative random variable is

M X ( s ) = λ λ - s α

From a table of Laplace transforms, we find that for α > 0 ,

Γ ( α ) ( s - a ) α is the Laplace transform of t α - 1 e a t t 0

If we put a = - λ , we find after some algebraic manipulations

f X ( t ) = λ α t α - 1 e - λ t Γ ( α ) , t 0

Thus, X gamma ( α , λ ) , in keeping with the determination, above, of the moment generating function for that distribution.

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The characteristic function

Since this function differs from the moment generating function by the interchange of parameter s and i u , where i is the imaginary unit, i 2 = - 1 , the integral expressions make that change of parameter. The result is that Laplace transforms become Fourier transforms.The theoretical and applied literature is even more extensive for the characteristic function.

Not only do we have the operational properties (T1) and (T2) and the result on moments as derivatives at the origin, but there is an important expansion for the characteristic function.

An expansion theorem

If E [ | X | n ] < , then

φ ( k ) ( 0 ) = i k E [ X k ] , for 0 k n and φ ( u ) = k = 0 n ( i u ) k k ! E [ X k ] + o ( u n ) as u 0

We note one limit theorem which has very important consequences.

A fundamental limit theorem

Suppose { F n : 1 n } is a sequence of probability distribution functions and { φ n : 1 n } is the corresponding sequence of characteristic functions.

  1. If F is a distribution function such that F n ( t ) F ( t ) at every point of continuity for F , and φ is the characteristic function for F , then
    φ n ( u ) φ ( u ) u
  2. If φ n ( u ) φ ( u ) for all u and φ is continuous at 0, then φ is the characteristic function for distribution function F such that
    F n ( t ) F ( t ) at each point of continuity of F

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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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