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270 Chapter 4 Special Distributions
4.5.9. Differentiate
the moment-generating function
1
0
pet
1−(1− p)et
r
M X (t) =
to verify the formula given in Theorem 4.5.1 for E(X ).
4.5.10. Suppose that X 1 , X 2 , . . . , X k are independent negative binomial random variables with parameters r1 and
p, r2 and p, . . ., and rk and p, respectively. Let X = X 1 +
X 2 + · · · + X k . Find M X (t), p X (t), E(X ), and Var(X ).
4.6 The Gamma Distribution
Suppose a series of independent events are occurring at the constant rate of λ per
unit time. If the random variable Y denotes the interval between consecutive occurrences, we know from Theorem 4.2.3 that f Y (y) = λe−λy , y > 0. Equivalently, Y can
be interpreted as the “waiting time” for the first occurrence. This section generalizes the Poisson/exponential relationship and focuses on the interval, or waiting
time, required for the rth event to occur (see Figure 4.6.1).
Figure 4.6.1
Y
Time
0
Theorem
4.6.1
First
success
Second
success
rth
success
Suppose that Poisson events are occurring at the constant rate of λ per unit time. Let
the random variable Y denote the waiting time for the rth event. Then Y has pdf f Y (y),
where
λr
y r −1 e−λy , y > 0
f Y (y) =
(r − 1)!
Proof We will establish the formula for f Y (y) by deriving and differentiating its cdf,
FY (y). Let Y denote the waiting time to the rth occurrence. Then
FY (y) = P(Y ≤ y) = 1 − P(Y > y)
= 1 − P(Fewer than r events occur in [0, y])
=1−
r −1
e−λy
k=0
(λy)k
k!
since the number of events that occur in the interval [0, y] is a Poisson random
variable with parameter λy.
From Theorem 3.4.1,
(
)
r −1
k
d
−λy (λy)
e
f Y (y) = FY (y) =
1−
dy
k!
k=0
=
r −1
(λy)k −λy (λy)k−1
−
λe
k!
(k − 1)!
k=1
λe−λy
(λy)k −λy (λy)k
−
λe
k!
k!
k=0
k=0
=
r −1
k=0
=
r −1
λe−λy
λr
y r −1 e−λy ,
(r − 1)!
r −2
y >0