Chapter 1
Special Distributions
1S
p
ecial Distributions
Independent Bernoulli Trials
If
P
(
X
=1)=
p
=1
−
P
(
X
= 0), then
X
is said to be a
Bernoulli
(
p
) random variable. We
refer to the event [
X
]as success, and to [
X
=0
]as failure.
Let
X
1
,...,X
n
be i.i.d. Bernoulli(
p
), and let
S
n
=
X
1
+
···
+
X
n
denote the number of successes
in
n
independent Bernoulli(
p
) trials. Now
P
(
X
i
=
x
i
,i
,...,n
)=
p
∑
n
1
x
i
(1
−
p
)
n
−
∑
n
1
x
i
if all
x
i
equal 0 or 1; this formula gives the joint distribution of
X
1
n
.From this we obtain
P
(
S
n
=
k
µ
n
k
¶
p
k
(1
−
p
)
n
−
k
for
k
,...,n,
(1)
since each of the
(
n
k
)
di±erent placings of
k
1’s in an
n
−
vector containing
k
1’s and
n
−
k
0’s has
probability
p
k
(1
−
p
)
n
−
k
from the previous sentence. We say that
S
n
∼
Binomial
(
n, p
) when (1)
holds. Note that Binomial(1
,p
)is the same as Bernoulli(
p
).
Let
X
1
,X
2
,...
be i.i.d. Bernoulli(
p
). Let
Y
1
≡
W
1
≡
min
{
n
:
S
n
}
. Since [
Y
1
=
k
]=[
X
1
=
0
k
−
1
k
= 1], we have
P
(
Y
1
=
k
)=(1
−
p
)
k
−
1
p
for
k
,
2
,... .
(2)
We say that
Y
1
∼
Geometric
(
p
). Now let
W
m
≡
min
{
n
:
S
n
=
m
}
.W
e call
W
m
the waiting
time to the
m-th
success
. Let
Y
m
≡
W
m
−
W
m
−
1
for
m
≥
1, with
W
0
≡
0; we call the
Y
m
’s the
interarrival times
. Note that [
W
m
=
k
S
k
−
1
=
m
−
1
k
= 1]. Hence
P
(
W
m
=
k
µ
k
−
1
m
−
1
¶
p
m
(1
−
p
)
k
−
m
for
k
=
m, m
+1
(3)
We say that
W
m
∼
Negative Binomial
(
m, p
).
Exercise 1.1
Show that
Y
1
,Y
2
are i.i.d. Geometric(
p
).
Since the number of successes in
n
1
+
n
2
trials is the number of successes in the ²rst
n
1
trials plus
the number of successes in the next
n
2
trials, it is clear that for independent
Z
i
∼
Binomial(
n
i
),
Z
1
+
Z
2
∼
Binomial
(
n
1
+
n
2
)
.
(4)
3