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Calculate directly the generating function g X ( s ) for the geometric ( p ) distribution.

g X ( s ) = E [ s X ] = k = 0 p k s k = p k = 0 q k s k = p 1 - q s (geometric series)
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Calculate directly the generating function g X ( s ) for the Poisson ( μ ) distribution.

g X ( s ) = E [ s X ] = k = 0 p k s k = e - μ k = 0 μ k s k k ! = e - μ e μ s = e μ ( s - 1 )
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A projection bulb has life (in hours) represented by X exponential (1/50). The unit will be replaced immediately upon failure or at 60 hours, whichever comes first.Determine the moment generating function for the time Y to replacement.

Y = I [ 0 , a ] ( X ) X + I ( a , ) ( X ) a e s Y = I [ 0 , a ] ( X ) e s X + I ( a , ) ( X ) e a s
M Y ( s ) = 0 a e s t λ e - λ t d t + e s a a λ e - λ t d t
= λ λ - s 1 - e - ( λ - s ) a + e - ( λ - s ) a
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Simple random variable X has distribution

X = [ - 3 - 2 0 1 4 ] P X = [ 0 . 15 0 . 20 0 . 30 0 . 25 0 . 10 ]
  1. Determine the moment generating function for X .
  2. Show by direct calculation the M X ' ( 0 ) = E [ X ] and M X ' ' ( 0 ) = E [ X 2 ] .
M X ( s ) = 0 . 15 e - 3 s + 0 . 20 e - 2 s + 0 . 30 + 0 . 25 e s + 0 . 10 e 4 s
M X ' ( s ) = - 3 0 . 15 e - 3 s - 2 0 . 20 e - 2 s + 0 + 0 . 25 e s + 4 0 . 10 e 4 s
M X ' ' ( s ) = ( - 3 ) 2 0 . 15 e - 3 s + ( - 2 ) 2 0 . 20 e - 2 s + 0 + 0 . 25 e s + 4 2 0 . 10 e 4 s

Setting s = 0 and using e 0 = 1 give the desired results.

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Use the moment generating function to obtain the variances for the following distributions

Exponential ( λ ) Gamma ( α , λ ) Normal ( μ , σ 2 )

  1. Exponential:
    M X ( s ) = λ λ - s M X ' ( s ) = λ ( λ - s ) 2 M X ' ' ( s ) = 2 λ ( λ - s ) 3
    E [ X ] = λ λ 2 = 1 λ E [ X 2 ] = 2 λ λ 3 = 2 λ 2 Var [ X ] = 2 λ 2 - 1 λ 2 = 1 λ 2
  2. Gamma ( α , λ ) :
    M X ( s ) = λ λ - s α M X ' ( s ) = α λ λ - s α - 1 λ ( λ - s ) 2 = α λ λ - s α 1 λ - s
    M X ' ' ( s ) = α 2 λ λ - s α 1 λ - s 1 λ - s + α λ λ - s α 1 ( λ - s ) 2
    E [ X ] = α λ E [ X 2 ] = α 2 + α λ 2 Var [ X ] = α λ 2
  3. Normal ( μ , σ ) :
    M X ( s ) = exp σ 2 s 2 2 + μ s M X ' ( s ) = M X ( s ) ( σ 2 s + μ )
    M X ' ' ( s ) = M X ( s ) ( σ 2 s + μ ) 2 + M X ( s ) σ 2
    E [ X ] = μ E [ X 2 ] = μ 2 + σ 2 Var [ X ] = σ 2
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The pair { X , Y } is iid with common moment generating function λ 3 ( λ - s ) 3 . Determine the moment generating function for Z = 2 X - 4 Y + 3 .

M Z ( s ) = e 3 s λ λ - 2 s 3 λ λ + 4 s 3
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The pair { X , Y } is iid with common moment generating function M X ( s ) = ( 0 . 6 + 0 . 4 e s ) . Determine the moment generating function for Z = 5 X + 2 Y .

M Z ( s ) = ( 0 . 6 + 0 . 4 e 5 s ) ( 0 . 6 + 0 . 4 e 2 s )
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Use the moment generating function for the symmetric triangular distribution on ( - c , c ) as derived
in the section "Three Basic Transforms".

  1. Obtain an expression for the symmetric triangular distribution on ( a , b ) for any a < b .
  2. Use the result of part (a) to show that the sum of two independent random variables uniform on ( a , b ) has symmetric triangular distribution on ( 2 a , 2 b ) .

Let m = ( a + b ) / 2 and c = ( b - a ) / 2 . If Y symetric triangular on ( - c , c ) , then X = Y + m is symmetric triangular on ( m - c , m + c ) = ( a , b ) and

M X ( s ) = e m s M Y ( s ) = e c s + e - c s - 2 c 2 s 2 e m s = e ( m + c ) s + e ( m - c ) s - 2 e m s c 2 s 2 = e b s + e a s - 2 e a + b 2 s ( b - a 2 ) 2 s 2
M X + Y ( s ) = e s b - e s a s ( b - a ) 2 = e s 2 b + e s 2 a - 2 e s ( b + a ) s 2 ( b - a ) 2
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Random variable X has moment generating function p 2 ( 1 - q e s ) 2 .

  1. Use derivatives to determine E [ X ] and Var [ X ] .
  2. Recognize the distribution from the form and compare E [ X ] and Var [ X ] with the result of part (a).
[ p 2 ( 1 - q e s ) - 2 ] ' = 2 p 2 q e s ( 1 - q e s ) 3 so that E [ X ] = 2 q / p
[ p 2 ( 1 - q e s ) - 2 ] ' ' = 6 p 2 q 2 e s ( 1 - q e s ) 4 + 2 p 2 q e s ( 1 - q e s ) 3 so that E [ X 2 ] = 6 q 2 p 2 + 2 q p
Var [ X ] = 2 q 2 p 2 + 2 q p = 2 ( q 2 + p q ) p 2 = 2 q p 2

X negative binomial ( 2 , p ) , which has E [ X ] = 2 q / p and Var [ X ] = 2 q / p 2 .

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The pair { X , Y } is independent. X Poisson (4) and Y geometric (0.3). Determine the generating function g Z for Z = 3 X + 2 Y .

g Z ( s ) = g X ( s 3 ) g Y ( s 2 ) = e 4 ( s 3 - 1 ) 0 . 3 1 - q s 2
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Random variable X has moment generating function

M X ( s ) = 1 1 - 3 s exp ( 16 s 2 / 2 + 3 s )

By recognizing forms and using rules of combinations, determine E [ X ] and Var [ X ] .

X = X 1 + X 2 with X 1 exponential(1/3) X 2 N ( 3 , 16 )
E [ X ] = 3 + 3 = 6 Var [ X ] = 9 + 16 = 25
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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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