C H A P T E R
9
Solution to Problem 9.1.
Let
X
i
denote the random homework time for the
i
th
week,
i
= 1
, . . . ,
5. We have the observation vector
X
=
x
, where
x
= (10
,
14
,
18
,
8
,
20).
In view of the independence of the
X
i
, for
θ
∈
[0
,
1], the likelihood function is
f
X
(
x
;
θ
) =
f
X
1
(
x
1
;
θ
)
· · ·
f
X
5
(
x
5
;
θ
)
=
θe
−
x
1
θ
· · ·
θe
−
x
5
θ
=
θ
5
e
−
(
x
1
+
···
+
x
5
)
θ
=
θ
5
e
−
(10+14+18+8+20)
θ
=
θ
5
e
−
71
θ
.
To derive the ML estimate, we set to 0 the derivative of
f
X
(
x
;
θ
) with respect to
θ
, obtaining
d
dθ
(
θ
5
e
−
71
θ
)
= 5
θ
4
e
−
71
θ
−
71
θ
5
e
−
71
θ
= (5
−
71
θ
)
θ
4
e
−
71
θ
= 0
.
Therefore,
ˆ
θ
=
5
71
=
5
x
1
+
· · ·
+
x
5
.
Solution to Problem 9.2.
(a) Let the random variable
N
be the number of tosses
until the
k
th head. The likelihood function is the Pascal PMF of order
k
:
p
N
(
n
;
θ
) =
n
−
1
k
−
1
θ
k
(1
−
θ
)
n
−
k
,
n
=
k, k
+ 1
, . . .
We maximize the likelihood by setting its derivative with respect to
θ
to zero:
0 =
k
n
−
1
k
−
1
(1
−
θ
)
n
−
k
θ
k
−
1
−
(
n
−
k
)
n
−
1
k
−
1
(1
−
θ
)
n
−
k
−
1
θ
k
,
which yields the ML estimator
ˆ
Θ
1
=
k
N
.
Note that
ˆ
Θ
1
is just the fraction of heads observed in
N
tosses.
(b) In this case,
n
is a fixed integer and
K
is a random variable. The PMF of
K
is
binomial:
p
K
(
k
;
θ
) =
n
k
θ
k
(1
−
θ
)
n
−
k
,
k
= 0
,
1
,
2
, . . . , n.
111