Springer Texts in Business and Economics

Moment Generating Function (MGF)

a. For the Binomial Distribution,

Mx(t) = E(eXt) = XX) eX‘0X (1 - 0)

X=0

-X

= [(1 - 0) + 0e‘]

where the last equality uses the binomial expansion (a + b)n =

P ( v I aXbn-X with a = 0e‘ and b = (1 - 0). This is the fundamental x=o X /

relationship underlying the binomial probability function and what makes it a proper probability function. b. For the Normal Distribution,

completing the square

Mx(t) = 1 f °° e-222{[x-(^+ta2)]2-(^+ta2)2+^1 dx

os/2rr J-i

_ e"2O2 [^2-^2-2^ta2-tV]

The remaining integral integrates to 1 using the fact that the Normal den­sity is proper and integrates to one. Hence Mx(t) = e^*+ 2°2*2 after some cancellations.

c. For the Poisson Distribution,

= e-X X ^ _ X = eA(e‘-!)

^ X!

X=0

X

where the fifth equality follows from the fact that X = ea and in this

X=0 '

case a = Xef This is the fundamental relationship underlying the Poisson distribution and what makes it a proper probability function.

d. For the Geometric Distribution,

1 1

Mx(t) = E(eXt) = J2 ™(1 _ ™)X-1 eXt = ^(1 _ 0)X-1e(X-1V

X=1 X=1

= 0e‘£ [(1 _ 0) ef-1

X=1

1

where the last equality uses the fact that aX-1 = ^ and in this case

x=1 a

a = (1 —0)є*. This is the fundamental relationship underlying the Geometric distribution and what makes it a proper probability function.

e. For the Exponential Distribution,

fi 1

MX(t) = E(eXt) = 1 e-X/e eXtdx

J о Є

1 C1

last integral, we get

Mx(1) = У” =(1 -

where we substituted "/(1 — "t) for the usual ". The x2 distribution is Gamma with а = 2? and " = 2. Hence, its MGF is (1 — 2t)-r/2. g. This was already done in the solutions to problems 5, 6, 7, 9 and 10.