Chapter 3, Section 9 Discrete Random Variables. MomentGenerating Functions. John J Currano, 12/15/2008. æ. ö. k. ( c ) k j E [ Y j ]. E [ ( Y – c ) k ]. k. =. å. ç. ÷. j. è. ø. =. j. 0. E [ ( Y – E ( Y ) ) k ] = E [ ( Y – ) k ]. m. =. k.
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(c)kj E [Y j]
E[(Y – c)k]
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E[(Y – E(Y))k] = E[(Y – )k]
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Definitions. Let Y be a discrete random variable with probability function, p(y), and let k = 1, 2, 3, . Then:
is the kth moment of Y (about the origin).
is the kth moment of Y about c.
is the kth central moment of Y.
Notes.=E(Y) = the first moment of Y ( =1)
2=V(Y) = the second central moment of Y ( =2 )
=E(Y2) – [E(Y)]2=2 – (1)2.
What is the first central moment of Y ?
Definition. If Y is a discrete random variable, its momentgenerating function (mgf) is the function
provided this function of t exists (converges) in some interval around 0.
Note. The mgf, when it exists, gives a better characterization of the distribution than the mean and variance – it completely determines the distribution in the sense that if two random variables have the same mgf, then they have the same distribution.
So to find the distribution of a random variable, we can find its mgf and compare it to a list of known mgfs; if the mgf is in the list, we have found the distribution.
We shall also see that the mgf gives us an easy way to find the distribution’s mean and variance.
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factoring out
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Why it is called the momentgenerating function:
Using the Maclaurin Series for ex at the right, we derive
by Theorem 3.2
using the Maclaurin series for ety
interchanging the summations
by Theorem 3.2
k
2
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Why it is called the momentgenerating function:
Thus,
et cetera.
2
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3
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6
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so
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so
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7

6
3
9
Theorem. for k = 1, 2, 3, . . . where mY(t) is the mgf of Y (provided it exists). Also, mY(0) = E(e0Y) = E(1) = 1 for any RV, Y.
Example(p. 142 #3.155). Given that
find: (a) E(Y); (b) V(Y); (c) the distribution of Y.
(a)
(b)
Thus, V(Y) = E(Y2) [E(Y)]
2
t
3
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3
1
2
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m
(
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6
6
6
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p
(
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6
6
6
y
3
1
2
,
,
,
6
6
6
Theorem. for k = 1, 2, 3, . . . where mY(t) is the mgf of Y (provided it exists). Also, mY(0) = E(e0Y) = E(1) = 1 for any RV, Y.
Example(p. 142 #3.155). Given that
find: (a) E(Y); (b) V(Y); (c) the distribution of Y.
(c) Since and the
support of Y consists of the coefficients of t in the exponents of the powers of e in the nonzero terms of the power series, Y has support
{1, 2, 3}. The probability that Y assumes each of these values is the coefficient of the exponential in the corresponding term, so
respectively.
Y = 1, 2, 3, with probabilities
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Example. Find the momentgenerating function of Y ~ bin(n, p) and use it to find E(Y) and V(Y).
Solution.
Then,
by Theorem 3.2
= (pet+ q)nby the Binomial Theorem
Theorem. If Y ~ bin(n, p), then its mgf is
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V(Y) = E(Y2) [E(Y)]
n
p
np
np
(
np
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2
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=
np
np
np
(
1
p
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npq
.
Now differentiate to find E(Y) and V(Y):
(a)
(b)
)
2
5,
bin
3
2
3
3
2
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3
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Example. If the momentgenerating function of a random variable, Y,
, find Pr(Y = 2 or 3).
Solution. From the form of the momentgenerating function,
, so that
we know that Y ~
P(Y = 2 or 3) =
P(Y = 2) + P(Y = 3) =
Homework Problem Results (pp. 142143)
3.147 If Y ~ Geom(p), then
3.158 If Y is a random variable with momentgenerating function, mY(t), and W = aY + b where a and b are constants, then
mW (t) = maY+b (t) = et bmY (at).
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t
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e
1
=
m
(
t
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e
.
Y
Other MomentGenerating Functions
1. If Y ~ NegBin(r, p), then
This is most easily proved using results in Chapter 6.
Exercise. Use mY(t) to find E(Y) and V(Y).
2. If Y ~ Poisson(), then
This is Example 3.23 on p. 140, where it is proved and then used (in Example 3.24) to find E(Y) and V(Y).